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Compl.er Numbers and Functions
1. Complex Numbers. 2. Argaud's diagram. 8. Geometric representatiou of zt t 22,4. Geometric
representation of zvzoadzt/22. 6. De Moivrds theoreul 0. Roots of a complex number. 7. To
elpand sinno,cosno and taan0 ia powers of siao,cosoandtano respectively. 8. Additiou
formulae for aay num.ber of angles. 9. To e-pand sin- 0, cof 0 and siat e cof 0 in a, gerries of
sines or cosi-oes of multiples of 0. 10. Compler fuactiou Defrnition- 11. Exponential funitiou of a
complex variable; Circular fuoctions of a complex variable. 12. Ilyperbolic fiuctions. 13.
tr'ormulae ofhypetbolic fimctioos. 14. Iuverse hlperbolic functions. 15. Real aod irnaginary parts
'
of circular aad hyperbolic functioos. f6. Logarithmic fuactions of a cornpler variaHC. lZ.
.C
Summation of series
+
-
i.9 method. 18. Approrimations and Limits.
191.COMPLDC NUMBERS
DemltiOn.A″ 凛ber ofι λ
e FOm κ
+ゥ ,″ ル″″
″ガy″ e″ α
れれら
′
er.9“ ど
ご
=マ (_1)ピ s
cα ′
″αα∞
“
mplex number.
anaで 1属
観讐器
A pa″ OFCO可たけ
PropeJ康輝:`1)丁
(2)動
“
+″ mdお Writtenぉ R← +初鑢
げ
″
dyお Calbd theれ
響“7′arr
ml● ‐ ″ +ゥ
and=― ウ ●″ Satt ro"● ●」暉 2″ OFa“ ヵ。′ル几
“
'1+ケ 1=4+3y2,0赫 '1-,1=r2 ウ
ers″ 1+ゥ lα だセ +夕 2● た Saガあレ
ua″ ″力
ο
れ
"co″
"ら
“
R← 1+ウ 1)=RCr2+ウリ,ゴ .e.ェ 1==2
2・
′
α れこ
I← 1+″ 1)=【乾 +ウ D,′ yl=y2
02ど
.e・
3)S“ れ,崚
″οCο ″P′ a″
●
ぉ
● ′J館
らprOd“ ′α潤 92oι ″■r ofa2y ι
“"bers tsゴ `Jfα
π れこ
に二￨“
“
“ and箋 +ゥ 2 Le two given complex nunber8,then
If,1+3yt
o their gunt
, 3← 1+,1)+Q+らり=● 1+崎 )+'01+yリ
(■ )their direrene
3← 1+″ 1)― ←2+ly,=← 1-エリ+j● 1-力 )
j■
=← 1+3ylxr2+ウ 2)●
く
)thdむ pduct
2+:← V2+2yl) '
'12-yッ
壺範
壺姉 =寺場=揚瑞論霧=7#+j葦滞
Cの
qu●
(4)Eυ cry co"Jα れ滉b″ r″ +ゥ Cα れ●′
′
岬 8レ ρttsed i2飢 e rOra rc..SO+:3"0)
“ +″ ),二 ″
Put,
P←
・ =r co3 0
...(j)
“
atld
y=′ 8三 ″0
...(二 J)
瀾 +″ ),三 ″
・“
SqmmFmda価鳴we酎
“ ノ+y2=′ i″ =ヾぽ+yう (talging psitive square root only)
Di宙 ding(::)by(:),we tty/工
=tan O ie.0=tan lo/工
Th‐ ェ
+ゥ =「 (“ 0+I Sin O)whe“ r=ヾぽ+y2)and O=tan lo/1).
0
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)・
カ
了ず
600
墜 HER ENelNEEttNC MATHEMATlcs
rlefinitions-
The number
r- * {1r' +y2) rs calkd,
the
moduluc ofr
+
iy
and.
+ iy) or arg (r + dy).
Elde$ly
Let aU the real numbers be represented along
XOX, the positive real numbers being along OX and
as
negative ones along OX.Lc,t OA be equal to one
of measurement (Fig. 19.1).
anplitude 0 has ar infraite number of valueg. The value of 0 which lieg b€tween
^!h9
- t aad lr_ir called tlrr pn'ncdpal ualue of llu amplitude- Unlesr otherx'is€ rp""ffr"a,
to mean the principal value.
Notc. cos 0 + d sin O b bricfly writtet as cis O eronound as ,sit
6) If the conjugate of z = x + i! be Z , then
*"
,haU
'
t"L ;;;j
(iii\zZ= | z l2
lz l={F1r1;
[=
1z
1
)21+Z2=″ 1+乙
ゴ
)に 1/22)=″ 1/■ ,where■ ≠o.
(Jυ
(u)4=rr.Vz
E“ 呻 LD■
(id)
(υ
婉
酬
“
Ⅲ
疇 αだ α″ 雄
凛
“"1_cOs
α+i
週
。
Ｉ絶
Put l_cOs α=rc“ O and sin α=「
sJ″
α tor
sin O.
′=(1-COS α)2+sin2 α=2-2
r=2 sin c/2
tan O=ポ
量五=
等
Takeapoint&onOXsuchthatOZ =x (OA\.
、
Hence,
ifYOY
be a
linc perpendicular to the real atis XOX, then all imaginary numbers are
imaginary eris, the positiue ones along OY and
rcpresented by points on YOY, called the
negatiue ones olong OY.
Obr Gcometric interprctetioa of i*. From the above, it is clear that i is an operation which when
multiplied to any real nunber makes it imaginary and rotatee its direction through a right angle on the
complex plane,
(2) Geouetric reprereatation ol compler numberet
Consider two lines XO& YOY at right augles to each other.
￨
Let all the r.eal numbers be represented by points on the line )fOX (called the real orrs),
cos α=4 sin2 ω′
2
=cOt a/2
ノP(z=r+U)
..
i,r
:
W---=
i
Problem8 19・ 1
1.島りress the db向ng
&
0詩 ―
島・
日0192.
V3+ご
8.Ifα +`β
・
4 rF:丁 +「
L If″ ady
(jj)1+sin α+ic∝ α
十万 =Lム
・
4υ
2+,2)la2+ら 2)=1.
ing“ d
ttan出
“
are“ t sOlve the equation=≒
8. Find the values
of .t,
y so that - I + ir2y
7. Find what curve z, + (1 + i) z + (1 -
i), :
,eゃ日 3υ h"rnts OFェ
″
-1詩 =C
デ
“
y
(Poo■ c,1988)
md}
+
L on OX represent the real
numberr{Fig. 19.2).
Since the muttiplication of a real number by i is
on thmugh a nght
of ls“
呻
■the rOta"on
““ numbers be
Therefore,
angle.霊
let all the imaginary
represented by points on the line I|OY (called the
,maginary a,tis), the positive ones along OY arld oegative
ones along OY. I,at the point M oo OY represent the
imarmけ numberゥ
.
λoW
贔e…
“
ery"P″ 馳`″ ル
●‐―‐
●‐ ●^" `3.^
_.. "′ ″
“"19“
_"ぃ ′
"a "お
, TR0
Ca蔵
3“ n」 OrdIP3● たSα ″ 仁 ,り
:●
Yr:;:*"""'*"
Eremple lg'8. ?tre centre of a regular hexryon is at the origin and ofte uert?r is giuen
'$ + i on the Aryand diagmrn. Determine the other verticts.
by
4
may represent complex conjugate numbers.
(κ r“ lsier“ ,1998)
“
O represents?
(Goralゅ L1991)
an d 12 +
.
If (r,0) be the polar co-ordinates of P, then r is the modulus ofz and 0 is its amplitude;
Obr Since a complex oumber has magnitude aad direction, therefore, it can be represeoted like a
vetm. Hetwftet wc sltr,ll ofun refer b tlu amplet, numbr z = x + i! as
0 thc pint z whoc (,.ordinalr'Js dre (\ I or (ii) lhe ucclor z frcm O to Pk, Y.
(Ca“ c● 41988)
.
1/(α +jO),proVe that(α
positive tral numbers being along OX and negative ones
along OX. Let the point
diagraminwhicir.u,ror,.."n*r,""i3?ff"J;#t:';il;;l,,ilff.f
Erpress ttre followingin the modulus-amplitude form:
(F)―
.'
Complete the ntangle OPM_rた
ln the fom● +ib:
(:)響
Fi9.191・
Thus it naturally follows that the multipiication of a real number by r is equivalent to the
rotation ofof, through one right angle to the position Ol".
‐
言棚孵鵬
∴飾
嗣
…
Example 19・ 2.2配
げ
unit
Then tr on OXrepresents the positive real number
.r and i.& =i2x=-x is repres:nted by a point tr'on
OX distant OL from O.
From this we infer that the multiplication of the
real numberr by d twice amounts to the rotation of OI,
through two right angles to the position Otr'.
tl
Ir
(i)a(zr=i1z+),1(zt=i@-4
601
192.(1)OEOMETR:C REPRESENTAT10N OFIMACINARY NUMBERS
is written as
mdlx+iy)or lr+iyl.
The angle o is called rlre emplitude or ergunent of x+iy ond is written
(z
amp
COMPLEX NUMBERS AND FUNC■ ONS
￨、
“
5←51)“
ば
.ル ″
presenta伍 on
組
RttrrAlgaad,Ptts 18“
J′ ●
■
②
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ofcomメ ex num“
ぉmme
mm maぬ ema● cs
thr“
由the
memd“ of
603
COMPLEX NUMBEtt AND FUNCη ONS
61 fi = {s *; so that oA = 2 arrd zXoA= tan-l tzJg = 80.. (Fig. t9.B)
Being a replar hexagon,OB=OC=2
and
■
Obr. By means
-
OPz, aoy vector P2P1 may be referred to the ori8in.
Ph) when
60 amp (z -o) = q, where k and c are constants,
{Gomkhpur, 1999\
the origin (Fig. 19.6).
→
Since AD,BE,CF are bisected at O,
→
→
→
Then z― α=OP-OA=AP
.ぁ =_a=_ヾ 3-ビ
(じ
E
193.(1)OEOMETRIC REPRESENTAT10N OF zl+々
(ii) amp
F● .19■
上
凛
=rl +rz
.
l':
tbe same centre and radiqs 3.
Ilence l < l z +2i l sS represents the region outside the circle l z +% l= l and inside
(includiog circumference oO the circle I z + 2i | = 3 [Fig. f9.71.
(ii) 8(z) > 3, defines alt points k) whose real part ia greater than 3' Hence it represents the
region of the compler plaue to the rigbt of the line r = 3 [Fig. 19.81-
Il,rl=PuK=OMI
NP=r\IK+I@= LP1+MP2=y1+y2.
The co-ordinates of
P ere (rl +xz,!t+yz) and it
represents the cornplex number
z
-a.r+tZ +i (yl +r2)
= (r1 +
iyl)
+ @2+
-
oV
a circle whose centre is Afu) and radius h.
*
"f
FIg. 19.6.
^
rlr, icamp (.D 1= q, means that .r{P alwaya makes a constant angle with the
Thz;s tlu l,ocus of Pk) is o straiaht line through A(d mahing an I o. uith OX
E-omple 19.6. Detennine the regian in the z-plane rcpresenkd. by
(,r < I z+2i l a3 (d4.B(z)>3 (iii) n/6sdmp (z)<n/s.
lil | z +2i | = 1 is a circle with ceutro (-2r) and radiue I errd I z +2i | =3 is a circle with
2・
.SinceON=O.L +LN=OL+AM
(z
X-axis.
Let Pl,P2represent the∞ mplex numbers,1=,1+ウ I and 22==2+ウ (Fig.19・ 4)
Complete the paralldOgram OPIPP2・ Draw PIL,=ダ r and PN tts t0 0X Also draw Plκ
and
)IZ― 。￨=l means thatAP=ん。
i^* ri *^
品 =_品 =_2a腱勝 =_あ =マ 3-i
Giil If z = r (cos 0 + i sir 0), then amp (z) = 0.
:, rc/6 s 8mp (z) < r/3 defines the region bounded by and including the lines 0 = r/6 and
0 = n/3- [Fig. 19.91.
iy2)=zr+'.2.
Thus tl:e point P which is the extremity of the diagonal of
tlre parallelogram having OPq and OP2 as adjacent sides,
Fi9.194.
locus of
0lz-al=h;
di=2(cos 90° 十J Sin 90° )=2
昴 =2(c"150° +:dn 150つ一 t+,
→
of th'e reletion P2P r = OPt
Eremple 19.1. Find, the
∠ XOB=30° +60° =900
∠ XoC=30° +120° 3150°
→
+iJ{
represents the sum of the complex numbers P(21) and P2(a2)
such
that
I z1+22 l=OPandamp (21+t2)=ZXOp.
-" t "r" 5b, * rr-f = qL
(D Geonetric lepreretrtatlon of z1- z2
I*t Pt,P2 repreaetrt the comptex numbers rl=rl+tJl and
z2=rz+i!r. Tben tbe subtraction of e2 fmm zl may be taken as
IRe■
obe Vectorially,
addition ofz 1 to
-
§3112)l
e2.
..
Produce P2O backwards to .R such tbat O.B = Opz. Then the
co+rdinates ofE are evidently (-r2, *y$ and so it corresponds to the
complex number - t2 - iy2 = - 22.
'
￨￨(
日g.19&
197.
Eranple l}{d,.If
Ｆ
Pfv
F19。
Ｐ
Complete the parallelogram OIQP1, then the sum of z1 and
(-22) is represented by OQ i.e. z1-22=O?
=fir.
Hence tlw compLex number z1-z2is npresented bX the uector
21, z2be any
日 9.19a
tuo complet numbers, prcve that
(i)|21+z2l<lzrl+l,,lli,e,themodulusofthesunoftwocbmplesnumbersisless
than or
at thg
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most equal to the sutn of
their modulil.
ノ
(三
二)121-221≧
ル凛配
IZ■ ￨― l z2 1『・●
・ι
reα た,ぬ oれ 0,α
l*t
Z2
rnd
,ヽ多
Measure oIf
Z2
△OAPl,
lz1+z2 l=Qp.
Now from LOPP, Op<Op1+p1p, the siglr of equality
so that
nrresponding to the case wben O, p1, p are 6eltinsar.
Fi9・
191Q
lz1+22 | < | z1 | + | z2 |
and
...(■ )
s lz-z1l+lz-z2l=ta.
■
21, z2
and major aris 24
sothatSp=
We know that SP + S P
=Z40P2+40Pl=02+01
た
sをお虚
=Ny
mOdulu3 0fC“
lO?-Oj1 =Za.
lz1-z2l=2oe
or lz2-z1l -?ac
whencee
diagran, show that
・
V′
and
′
Idft"*mciningvertcea.
0
F19.1911・
lz+3l<{
2
Z2==2+″
g + i, 4 +
l&,
19.2
.・
2=r2(C0302+i Sin 02)・
(r1/rD IC● 3(01-Q♪
tlv
complex number
C)I Z1/Z8卜
IZュ
1/1■
+J Sin(01-0り 1.
z1/z2is represented by the point P' such that
￨
(jj)amp(21/22)=amp(81)=響
p(22)・
Note.rFpl(21),P2● 2)● P3d P3(23)“ Cay tin"p●htt rル a
thmぬ
(Jj)IC2))2
04ad■ L1999)
￨￨
■/2.
・
fた
“
“
F19.1913.
.P re,rosents the number
Hence
Undhru, 1990)
I
a rlZl● 221=121-221,pve tbat the direrence Ofthe amputtdes of21 and 22お
Sin 01)
and ZXOP=ど」ちOPl=∠ AOPl― ∠AOP2=01 02・
_ g + Ei and _ g _,U form a
(jυ )12+21+lz_21く 4
1912
綴=畿に"=洗 =亀
=lz1_z2ltfut-
tadthetocru giveaby I e - 1 | + | z + | =3.
' ?. Whet
domaia ofthe:-plane is represeated by
0職 )″ 3く口mp● )く ″
0+j8LOb―
F:9‐
Measure or OA=1,construct triangle OAP On OA directly
Sim,lar tO the tnangle OP2Pl(Fig.1913),so that
sihEt€d et the origin. If one of the verticee of the rect agle is t + l{s,
find the coaplex aumbcrs
represeotiag the other three vertices ofthe rectaagle. Fiad-alro
the area ofthe rectaajle.
a
trirDgle coastruc"t d ia the courpler'prrne has its one verter at the poiat I + d{g,
Firrd
[| TI!e1!
uro
oomprer ,.u'bera reprcreating the oth6r two vcrtices,
o the origin being it' ci;ue-cenEe.
5' the-ceotre ola rpgutar hengoa ic et the otigia aud one verter is given by I + d on the
Argand diegran.
(02<
2)・
ty.
21=,1+3yl=rl(co8 01+二
Z'lf lzrl=lzzl andamp(21)+amp(sr)=Q thea show that z1atd,z2square.
are coajugate comptex
numbers(
lggr)
t' 1t*r";8le ir coastructed ia the complex phne and its sides parallel to tlre axes Gomkhpur,
aad itc oeatne is
-..6.
-:
rs zl,22 1S
“"bοlZ2ト
●1′ 力ar(三 )181231=IZll
(2)Goomo""representation of z1/22・
Let Pl,P2represent the∞ mplex numbers
t=-Za\
Problens
'
by′ 10POf.12山
ly
ぶ讐9計 i:瞥盟∬響盤鑑 1￡路鵠 :T器乱
_zl1 ana*SZ
= lz_2211
1z
shich is tbe desired equation ofthe ellipse.
Also we know that SS, = lae , e being the eccentricity.
Lu an ArgaDd
OP=OP1 0P2=rir2
j.e.
∠AOP=zAOP2+∠ P2οP
lz-z1l + lz -a2l=2a
1.
along OX GiS. 19'12). Construct A OP2P on OP2 directlY similar to
(Jl)amp(211り )=amp(21)tamp(■
わ
(i\ be aa-y poiat ou the given ellipse (Fig. 19.11) having
.Le:t P
r
r
09
C器
Nso find. its eccentricity.
(22)
(cos 01 + t sin
″れ
Heict,た ●´´
。
ごあ′
・ f`あ。cο 凛メ●
Obc. I z1+e2+zs I S I zr | + | z2 | + lze
l.
Iageueral, lz1+22+...+2, lS lzr l+ I z2 l+...+lznl
E:anple 19'7. slou that thc equation of the elripse ta oing
foci .t
rl
∴P represents the nunber
‐
′lr2 1COS(01+02J+I SIn(01+0231・
IBy C】
I
OA=l
OP/OPl=OP2/OA
‐
...(f)
lrr l=l (21-22)+z2lslz1-z2l+lzzl
lzr-rzl > lzr l-lze
zra
zz=rz+ iY2=r2(ca 0t l i sin 0,
and
lql=OPylzzl=OPz=PrP,
atS(21)andS
REPPESENTAIION OF
Pt, P2 reprosent the compler numbers
z1 =11 + iY1 =
Complete the parallelogram Op 1pp2, zo t}Lat
flence
Agab
Thus
(l) CEOMffilC
19.4.
″ιs OF`ルど,ル ″ ″C● of′ ゅ co澪
れ:枷れ、ぉ
ヴ
“
“
れο
3′
′
ル α,ル ″ “
ιor'ルニ
′れ配 Jjl.
"ル
“
""わ numbers 21,22
“
[,-et PbP2 represent the compler
Oig. 1g.10).
bci
605
COMPLEX NUMBERS AND FUNCT10NS
604
亀
お
hQぬ ¨
ぶ識憲ヽ
1914,we have
+J
PlPr=zt'zz and
コ .2_9_0'S)
Pfl=zs-zz
eぃ
0
Fi9.1914.
の
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襲
H,GHE,[NC:NEEP!NC MATHEMA■
a=1障
=ampl景
〕
coua:xNuus€nsANDrurcroNs
6s
Problems 19.8
l.
If z = I + i, finil zz , z3 , l/z and plot these on the Argand,s diagram.
2. lf zy z2bn tvo conplor aumben, slrow that
￨
=岬磁 )一 a璽こ )
=β
―α=′ IP2P3・
nttmple 19.&■
drλ e rOc“ s orttepOin`み
lz1+2212+lz1-22 12=2{lz1 l2+ 1zrl2l.
lElat' Lot zr = r 1 cia e1, z2 = 12 cis 02 so that z 1 + 2 212 - r 12 a. rr2 + 2r y2 ce
lz1- z2l2 * rr2 a rrl - 2r, r, oc (01 - Q).1
″ぬ.2
“
l
2Sra庶
hJ奏電
鰭智奪暮島露鮨ilililttti鑑ぎ
ra C°
(j)Since l′ ―●
￨=APand
3. l4lhtarethelocigiveoby lz-3
4. Fiod the locus ofz given by:
(D lg l=
5. FiDd the locus ofz :
i熙織
1嵐
.
(i) when
8. lt zy
i澪
雷織出
お凛
場s卜 1,″ ヵ′
・
tttlt:鷺
・″
、
“
“たJac“
t慧
_3凛漁翼罵I翼糖
喧
甜￡‖ 樹富せ鯖ご
晦 mp障
)=ZAPB=∝
″
擾甚墓墓万●4PBorri● ル″ヵたんPas“ s
X、
、
ア
/イ
′
ffi
22, zsbe
島
::I」 ii」 iFilli:二 Jhilt鷺
Lmpl1 199.ffzl,22,23 1e tte ve咸
。
1-^一 …
f12_ィ
+″ 32_′
2● 2′
掟es
aFa″
8+Zれ
2Jaた ra′ rrl● agre,
"″
.*f,
(ri)
lsr-l
01 )
;
(ii) when
l:-3l.
l=
z_
zE
ir purely imaginary.
212,232+222=221Zl+23)・
(Gor● 力
λク r.199′ )
“
195.DE MOiVRE'S THEOnEM・
′
｀′
ヾ
Statementrrf″ れ ril α
tt fれ ′ ●
響 ■PasJ′ lυ 2
0
C″
F19.1915.
●(oo3 0 +iSin Oメ :=C。 ●■0+id●
0,■ agarじ υ
α″uctあれ
υ
οο
r“ ega“ ve.ο れ
,psビ α
θoFrAF υ
α es
■0:
Or(co3 0 +i3■ は0)′ tis co3 nO+i gin■
0.
`“
P瞑渕臨餞二 WЪ
α
′
antta几
"P3お Psiri“
By actual multiplication
cis 01 ci3 02=(∞ 8010902 Sin 01 sin 02b+J(Sin ol cos 02+∞
8 01 Sin 021
―
=COS(01+021+ISh(01+02),J・ e.CIS(01+Q分・
Similarly ci8 01 d8 02 CIS 03=Ci3(01+02)CiS 03=CtS(01+02+03)・
Proceeding in this way,
Aral
T品蠍柑:織 1響識阻腸讐
紺鳳淵瑞ュ
Liζ
-
the vertices ofan isosccles triangle, right angled at 22, prove that
cf″
i:κllli通 1::
i"
(caticut,tgEill
(82
l=Llz+ I lfort=land2
lz-zl.
IZ― bl=BP_
赫
m
』
蹴
柵
憾
鐵蝉
Iミ
n￡
607
cis 01 cis 02 CiS 03…
……… 。C鮨 04=cis(01+02+03+・ ……+Oa)
Now putting 01=o2=03=・ ¨¨
・=0.=0,We obtain(CiS O)a=CiS● 0.
P夕 ′
)
Cas●
Fig 19 16.
I二
た。2p2お α
"agariυ
●JP3reger.
.
Let"=_,where,"is
a+ve integer.
“
proue that
￨
‐
ヽ
∴
0"メ
(P● ●■α,1990)
thMI盤
譜露
よ
鑑鷲∫
脳無 L:蹴罐王普l Ptted ttЮ uハ 6● 面面d“ Ⅲ
諷 oコ =赫 =轟
“
8"0-3sin a0
(∞
ughan∠ oil」
1:1∫
rltiply 1lI=1111量
it by cos 0 + i aiifT51numttrttЮ
0.
(by case I)
∞
8"0+jSin 20)(COS● 0-`8in aO)
IMultipbing the num.and denoln.by(cog"0-I Sin"0)1
∞s“0-l sin“ ■
=∞ 8“ 0-′ Sin"0
=032屁
0+sin2“ 。
=COS(-70+三 SIn← 靱0=CiS C-40)=CiS 20
Casc tt Wle2 2お αfraCrあ ■,pOs2″
ο ■2garJυ e.
“「
_
1・
.・
="=″ 】
Let Pa=P/g,where c is a+ve integer and P is any integer+ve or― vet
FIg.1917.
' One of tho remarkable theorems in mathematics
Moiure (1667-7?54), a French Mathematician.
www.Engg-Know.com
;
called efter the name of its discoverer Abraham De
‐
―
608
H|GHEA ENGINEERING MATHEMATCS
Now (cis O/qf = sg (q . 0/q) cis O
=
Taking qth root of both sides cie (0,/q) is one of
the g varuee of lcis 0;1/e,
"'
i.e. one ofthe values oflcis O)l/,,
cos 2●
=cieglq
frlf
*
L9.10. Simplifu
(cos 30 +
(cos
have.
(cos 0 +
i sdn g)-,
cos n0
Erample 19'13'f sln a+srn p+sdn1=cos o+cot fl+cos7=g'
siz 3c+sin gF + slz 3T=3sitr (c + p+fl
prove that
'and
(cos 40
i sin 40)D
sin 40)r (cos 50 + r sin dOf '
i
-
Tben o.rI
30+d sin B01a= cos 120 +i sin 120 (cos 0 +i sin 0)lz
=
(cos 40 - i sin 40;5 cos 208 _ i sin
200
=
= lcos g + i sin 0;_20
(cos 40 + j sin 40y3 cos 120 +
i sin 120 = (cos 0 + i sin 0)r2
=
i sin
(cos 0 +
Erample Lg.lI. Proue that
0メ
sin
i'
0)11 (cos g +
sin g;12 lcos 0
+c■ 甲 o_J覇 ■0・ =ノ
i sin 0f21 ,
-=
QfD ''
"in
♂ 0/2)i巌冨 πo/2)l rJa"″ 鴫 J998)
.L.H.S.=lr(cos α+ビ d■ α)la+lr(cos α―
lsin a)1・
=′ I(cos α+Fsin α)・ +(∞ sα ―
α 〕=′ (cos nc + i sin
)■
=′ 2∞ sれ 0
32・
'ユ cOs″
`sin
(J)2 cOs rO=/十】′
ざ、
ぬ eme
==
j
(J)Ta■ ing the● ve sign, =′
=ザ
[Substituting the vaiues ofr and at
0土
=
tcos se
-
-.-
6 and C = cis 0, show
0競 =減 an∵
O「
?0 +
(cos 0 +
-
i.sr! ?,0f
i sin
i cos ('lc + 5p)'
ω
= 1.
"
ttii)
l'coe0+isin0l=cosBe*isin80.
lsin
0+
i
cos 0
J
=¨
躙
2β ,c=cIS 2Y and d=cls 25,prove that
掏
…
4. Ll r, = cis ltt/ 2'), show that Lt
5. Find the general value
or
11
‐
編
中奪
島 Os
・
lrI・
i i.il"tt
¨
lKanpur,
:2 13 -.' :n = -1.
7996't
the equation
".ti"nes
i sin g) (cor 20 + i sia 20) ... (cos nO + i sin n0) = 1'
6,Provethat(i)(o+rf,)''l' +@-ibl"'/"=21a2+b21"'/b.."[io"-t!']'
dI
n
.Jihopat'1998t
f
(ti) (1 +
=(cOS O+l sin O)″ =Coe rO+ビ sin ro
0)o
that
慄
(coe 0 +
lcos
= sio (4d + 5p)
ffiffi
Ifp = cir
5et2
- i sin {8)e
(cos 40
(cosa+isincr)4
i ein
-i-t cos
-----,/
u,!/4.
- i)"" = 2'
7. Simplify lcosa-cosp+i(sinc-sinp)|tr+
idn a
=(∞ 80+F sin O) r=cos rO― l sin「0
Addingコ´+1,I=2 oos「 o.Similarly with the―
ve sign,the ttame result follows.
0丸
cr)3 +
3.lf`=cL 2α ,b=cls
io * cos n0 - j sin no)
0メ署チポ警鵠
・ lMad疇,2000S)
2
md
2.
proυ a′ ぬ
α′
=c∝
c=cisc,6=cisFaodc=cis1.
r+i (sinq+sin p+ sin 1 =g
+b3+c9=fubc
l. prove,r",,.,
,,..
tiil
(0/2)cOs(■ 0/2)_
E=ample 19‐ 12.fr2 cOs o=x+二
rα ,1998)
【 rttisた ●′
“
Probleuc l9'4
Put 1+cOs O=「 cOs α,sin e● ・′sln α
"SO+:J■
r2=(1+cOs O)2+sin2。 =2+2 cos o=4 cos2 0/21.e.「
=2 cos O/2
2 sin O/2・ cOs o/2=tan O/2ia
tan α=Iだ
二
α=o/2_
::I卜百
.・
{」
a3=(cos c+cos p+cos
*,
+Ic●
9'
cos (cr + p +
(as d)3 + (cis rs = 3 cis a' cis p' cis y
cic 3c + cis 39 + cis 3Y=3 cis (a + I + $.
Hence by equating the real and imaginary parts, we get the deaired results'
.
nd
o3
(cis
i0){ - crs 200 _ I sin 200 = (coe 0 + i sin 0)-20
The given erpression = {c.os 0 *
α土
.'.
31=3
cos 3d+clos 30 +cos
L.et
(cos
cos 50 +
.'.
{0 +
i sin 30)-{
-
蔦訂万瓦菖薫〒冨高雇百薇再::s.-lo
- --!9!=n!-t?-99c-Eqll!-g!-tq0 (2 cos nE + 2r sin n0) cos t' - I 0
cos ill
+0n)
Erample
We
=万
n-
2. (cos 0 - i sin 0)n = cos n0 i sin n0
S. (cjs rn0)r = cis mne (cds nO)m.
=
0+だ tin 2″ 0+1
乙こ(22-1)0+じ Sin(21-1)6+COS O+I sin 0
(1 + cos 2n 0) + i sin 2n0
0 + sin 0)
lcoe 5-IT e + cos 0) + i (sin
Raise both sides to porper p, then one of the values
of (cis e)p/c = Gts 6/qy
cis (p/g) A Le.
one ofthe values of(cis 0)n cis zO.
=
(by case I and II)
lr*s the theorem is completely established for au mtionar ualues of
l. cic 01 .cis 02 ...... erc 0, =61s (0g + 02 +....,.
Cor.
609
COMPLEX NUMBERS AND FUNCllCl郎
i)'
8. Provethat (i) (1+Bin0+icos0)"
メ
-{
,,,r
(llarothlcada,
+ (1
f|l{iltiffs,f
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=
.*
[T
lcos
tr-cos
p-d(sinn-sinp)1"
+(1+sin8-icos0)'=2"+
-
.").
,
"t"
(T - ,")
I co6"
1S'Guiarat'
(l -;)*t [:f f ]
l*)ol
19901
翔１１
9.If2 10s e=ェ +1/.and 2 cos O=y●
1/y,shOwthi● ne ofthe val● es Or
プ+赤お儘叫
①チザヽ
2c"ぃ o_り
0ノ
2c¨
10 Find the nodulus and pnnctpal Val■
11.IFα ,o betヽ r∞ ts
Ofェ
Example 19・ 14.nπ d`ル ε be r● ●rs OF″ 西ry
・
“
ね′
ル Argaれd dι 曜
.
"れ
Ifχ be a cube r∞ t oFunity,then
o+″
2_2r+4‐
(κ α
ap●
4r998,C● ￨"ba′ ●昴ら 199′
(M●d7as 200θ s iJ●
e Ofthe argument oF(1+J V3)“ /(V3-:)::
0,prOve that
α″+r=γ 'lc“ aェ /3.
,f9“
"な “
611
COMPLEX NUMBEFお AND FUNC■ ONS
)
)
'=(1)1/3=(cos O+I SIn O)1ん
∴ The three values ofェ are cis O=1,
(″ .mtrp“ ら」9“ s)
cis 2nl3 = cos 120" + i sin 120" = -
・
f"6S)
iil::iiii::[,li
五留
鮮灘
鮮露
灘
ェ
1+y ｀
+2 1=。
fl;:T舅
19.6. NOOTS OF
iず i離
Jf[i
y
i -, f,
1・
(Iに IP″ ら
r9")
F19.1918.
鷺常滋寓 i篤比渕社ガis嗣 .,
There are q and only q ditinct values
of (cos 0 + i sdn 1tt/q, q b"ing on integer.
Since cos 0 = cos (2nr + 0) and sin 0
(2nx + 0), where z is any integer.
sin
=
cis 0 = cis (2nx + 6)Ry fle Moivre's theorem onc of the
values of
.'.
Givir:g n the values o,
(cis 61lzo'
o'11r/e =
(2ni + 1l/q
"is
l, ,, a, ......, (c - r) successirety,
∴ (1/2+ヾ 3ゴ /2)3/4=I(∞st/3+J sin■
...(
∴T“ 士
∞
produd=ds僣
fottυ
_“
2_.+1)=0
← +1)♂ ―ノ +″
15=- 1 =cis
ι.a
"o#,itiiHalues
Moreover,
or
q + 2 etc. are the mere repetition of the values
s
the g values Bu"T bv
clearly distinct from each other, for no two of the
i.?
o. aiff". Ur. _"iaiii"
zr.
"q"ut
"f
Hence (cis 0)l/q has g and only g distinct values
given by (2).
ob' lcig 6yz0 ere olo is a mtional fraction in its tilDest
t2rms, has alcot1 and onlv <1 d.islincl ualues;
which are obtaiaed ';
by puiting n = o, t,2,..., q _ r .*;;i"&
in cisp l}ntr + Dl/q.
Note that (cis 0;6/15 hss sn1, 5 distinct values
and not 15; because 6/15 in its lowest tcrm zs.
=
;. In order to find the distinc, ucrtu.s of
lcis gl',q alllay, se that ptq
iaits
angles involved therejn are
*:
+I=ο
(P● Pldlc力
.
0r
n =cis (2n +
cη、1998)
ェ5+1=0
l)n
3r srn
:r cos?+r
,t srn;,
3n sln 3r .I, cosE--.
,
---tr -l .-,-!
sm
cos
cos;+r
E.
5
E-,
5' Rejectingthe value - I which corresponds to the factorr + 1, the required roots are:
3n
cos n/5 + i siln */5, cos 3r!/5 t i sin 3t./5Erample lg.l7,Findthe 7th roots of unity ond proue that the sumof their nth pwets always
udnishes unless n be a multiple number of?, n being an inbger, and lhcn the sum is 7'
7=ds午 =← S等
Oν 7=Os猜 +関n ttν
We have
l・
Putting r
is
lowest terms.
ulefgllr be ernproyed for edracting any assigned root of a given
qurntity. We have only to erpress 91
it in the fonm rfcos'g i';r, el and proceed as above.
Notc' The above discussion
=L
5 = [cis (2n + 1)tlr/5 = cis (2n + 1) n/5, where n = 0, 1, 2, 3, 4
Hence the values are cis t /5, cis 3nl5, cis n, cis 7x/5, cis 9r/5
get a value or(cis o)r/c to be cis (2't +
o)'zs, which is the same as
for n = q, q + 1'
+¥+警 +等
ret rλ θοЛP"′ O SO′ υe rλ● θ9“ α′EOn
Multiplying by← +1),We have
,)t/q
ν4=(cis lt)1/4
)=dS4■
ノ_,3+ノ
of (cis
/3)3〕
=lcis(2ぉ +1)π l1/4=cis(2● +1)■ /4 where a=0,1,2,3.
/4,cls 5■/4 and cis 7■ /4
Hence the requred values are cls 71/4,cis 3■
l)
we get the following g values of
ntinu“
:jil'j?:'JrYi;"11,1f
(Bangalore, 1993; Rewa, 19901
Put 1/2=r cos e andヾ 3/2=r sln o sO that r=l and e=■ /3
Ex3mple 19 10.ase De■
rn"
=(CiS O)1/3=(cL 2■ ■)1ん ≒dS 22■ /3 where n=9,1,2.
A COMPLET NUMIER
lcis 0;l/4 =,lais 12nr' +
a
′
だoィビ
these three cube roots are represented by the points i
A, B, C on the Argand diagram such that OA = OB = OC and i
,.
./AOB = 120 , zAOC =2413.
.'. Ttrese points lie on a circle with centre O and unit radius
・・
・‐
‐
・‐
such tlrat ZAOB = 4BOC = ICOA=L2O'i.e. AB =BC =CA.
・
・
・
・
・
・
・
む・
Hence A, B, C form an equilateral triangle.
FilenProvel[「
.
'愛
aπ d tt。 ″′
λαιハeyf● r"α ■●
9“ JJα ″ra′
P=l
g2tt/7.
.'.
=
0, l, 2, 3, 4,5, 6, we find that 7th roots of unity are
Sumsofthenthpowersoftbeseroots=I
②
www.Engg-Know.com
+P'
I'
p, F2, P3, gn, ps, p6 where
+9b*.'....*p6'
"'(i)
摯〓
612
COMPLEX NUMBERS AND FuNCr10NS
=
'€
being a G.p.
'1- {;,
P"
with common ratio p
Ylhen n is not a muhiple of
Z, p?o = 1p7yr = (cis 2tya
= 11-p7" = 0 rnil I pr *0, as n is not
a multipleof
i.e.
Ttrus
19.7.
nwltiple ofl
=
(i), s! I +(p7f +(pr\u * ...+(p?)@
E'lhple lg'l&Fizd t'rre qaation whoJc = 1+ l + l + I + 1+ l + I = z.
rrrlts are 2 as x/2, z cos str/Z,2 cos stcl7.
g
or
= s6s
Then
* ; sil 0, where
e=
il.l, 3x/7, ..,, LSx/1 or
-y7=(cos-0ri sin B)?=-gs6 ?0+i sin ?0 =-l
g + 1) (yo _f +y{ _y3 +y, _, _
ii
O
or
y3, (ys +
yz +
I =0
_,t, * t rt\- * i
Hence the roots of
l.
Dividing (2) by t l),
3 0 sin3 0+■
aCl∞ sa_1 0 sin O―・
C3COS・
亀 c∝ " 5。 sin5 0_ .
tan n8 =
cos" 0 - nC2 cosn - 2 0 sin2 0 + "C4 cosa -{ 0 sin{ 0 - ......
,grz/7,11ft/7,lgn/1.
i)il
,3-12-lr*l=o
where x=y+1,/y=2cos0.
Now since cos lBnZ? cos 1t/2,
ens lhn/?
=
(ii) are 2 cos
f, ,*" ?,
2
and dividing numerator and denominator by eosn e, we get
nC1
tan 0- "Cg tan3 e+ ^C5 tans 0 -......
+a-a | nCztan2 o + "C4 tatr{ e -......
・ (J`)
.。
= cos Sr.z?, cos gn,/Z = cos
"os
-
S,:/?
I9.8. ADDIIION FORMUIAE FOR ANY NUMBER OF ANGUS
We have,'cos (01 + 02 +...... +0o) +l sin (01 + gr+...... + 0r,)
f.
= lcos 01 -r i sin 01)(cos 02 +
Pnblema lg.S
2, lf u is a compler
cube root
3. Find all the value6 of(
" Iillrl'"-"ts
5. Use
ofunity, prove that I
(u} (-l+i)z,i
,iu) 6 +,v3):,r
+ ro + ur2 =
= cos 0r cos 02 ...... cos 0n (1 +
= cos 0t cos 62 ...... cos 0n [1 +
tBongalore.lgg0\
+
(I _ ,.;g)r/;r
0,
tCoinrbarore, t9g0 S\
- l)l/6.
on the Argand diagram, all rhe valucs
of
(l + ivi)r/5
and verify rhat they
(i)rs+l=0.
(ii)r?+r{+r'+l=O
,.:re i r
u,ttt'+x-_u-_!=o
(Madras,20Mpn (iu)(.t_l)i+t5=O
fom
a
(Mamthuada, IggB\
.
8' hove thet the nth roots of unitv form a goometric
progression. ltso show ,t*
mots
ih:
thc values 0, r, 2. 3' .-.,
tf |
a positive integer are
Z
as
|
(
tMadros, lgttDl
I
+
nethi,
Zt / 7, 2 ms 4n
/7,
Z
cG 6r./ 7.
i
cot
1994
01 + 02
nrltr ), where
:
i tan 61)(1 + i tan &r......(l + i tan 0n)
i (tan 0l + tan 82 + ...... + tan 0a)
;
Erample f9.f0. If 0r, Q, 0s be three ualues of O which satisfy the equtitton
tan2i=l.tan (0+c) ond such that no two of them differ by a multiple of r,, shou that
;oJ'rf'JH::'l
-
r-L
- Find the equation whose roota
lO.
are
-
i sin 0o)
ltan Or hn e2 + tan 02 tan 03 + ......) + i3 (tan 0r tan 02 tan 03 + ......) + ...... + ......1
= ccis 0f coe 02 ...... cos 0n (1 + is1 * s2 - is3 + s4 + ....,.)
where sg - tan 01 + tan 02 + .-.... + tan 0,,, s2 = f, tan 01 tan 02, s3 = f, 36n 61 tap 02 tan 03 et .
i
Equating real and imaginary parts, we have
cos (01 +02+...... + 0r)= cos 0t cos 02...... cos otr (l -s2+e4 -...)
sin (01 +02+.,.... + 0n)= co80l cos 02...... cr130, (c1-s3+s5-.,.)
& Fird the roots common to the
equations r{ + I 0 and-16_ i
=
=0.
7'solvetheequation.rr2-l=0andfindwhirhofitsrootssatisfytheequation.t{+rz+1=g.
8' show tlrat the mots ofthe equation (r
- l),, = :,, a being
0,, +
+ i2
De Moiwe,s theorem to solve the
following equations :
is zero and their product is _
t
i sin 0z).,....(cos
Nowcos0l+isin0t="os0r(1+itan01),cos02+isin0z=cos0Z(l+itan02)andsoon.
.'. cos (Ol +02 +..-... +0,) +i sin (01 +02 +...... +0r)
Fitd all t}re values of
(i)(1+i)r/{
(Itladras,19965't
(ii!) (- I + i.,/3)3/'
...(2)
expansions ofcos n0 and sin n0,
_t o
=
l(y+L,ry1s-Bg+1,v1f- Kr*'ttr1r"_zl]tt*t/yt_L=o
*'
...(1)
Replacingeveryain2obyl-cou20in(l)aoderery"os20byl-ein20in(2),wegettlredesired
...(r)
=
Ltjtitn
(cos 0 +
Equating real and imaginary parta ftom both sides, we get
- "C2 cos r -2 e.in2 0 + "C4 cos"-4 e ain{ 0-.....
nCr
I
sin n0 =
coqn - g sin 0 -'Cg cos' - 3'6 sin3 0 + nC5 coen - 5 0 sins 0 - ......
=,r,
gety6 -y5 + !4 - !3 +5r _y +
I = 0.
$e
rls roots are/ =gis 0wheng il7,Stt/I,5fr./7,
(d) by
sin n0
cos n0 = cos^ 0
= o.
Lcaving the factory + twhich corresponds
to
Dividins
d
=(cosn0- lCgcosn'2 0sin2e+.....-)+,i(r,CrcoC-r0sin0- trC3costr-30sin30+......)
From
Iety
qnd lon 0
(by Binomial theorem)
(aayl
Zp
-
e
(by De Moivre's theorem)
i sin 0)a
rC2cod!-30(isin0)2+aC3@s,r-30(lsin013+......
=cos'0+ ^C1 cosn-l01ising)+
We have cos n0 +
?.
S= 0.
Vlhen n is a
613
expond sin n0, cor n0 ond lon ng ln power ol rin 0, cos
respeclively (n being o positive integefl.
To
+
03 +
o is o multiple of n
Given equation can be written as
Reraa, 19941
・
Ｑ
or
L13 + 11,
www.Engg-Know.com
-
2) tan o . t2 + (2
,ft
=
-tr) r -
I iffi
l tan a = 0
where , = tan 0
一
一
一
¨
”
¨
¨
¨
¨
﹁
一
´
一
．
¨
．
COMAEX NUMBERS AND FUNCTiOi\.IS
.・
.tan 01,tan 02,tan 03,being its mOts,we have
61=Σ tan 01=―
λ
2tan
32=Σ tan 01tan 02=2-λ
E-oaple lg.2l.
α
33=tan α
輛Qサリ=号 =旦鶏鶏芦
_o
=―
Thus 01+o2+03=れ ■ α ,whence
=
1.Prove that‐
Follows the result.
.
(Mcd“ 嶋 200o Pつ
O“ a pO“ 口
Omial h cos O?
砺
ェ
=2“ 0
1999′
′●■8カ θ ,1998)
`‐
(Mだ
"t19“
1= 211 6diana except in
8. If tan r:+tan-ly+tan-lz=t,ehowthatr+!+z=xlz.
g. Iftan-r: + tan- t y +lan-l z=t/2, show
o,
s)
If
a =coe
srnn 6
cort
0 ln
oae
particular
l/z
= 2 cos 0, z
- L4=
2i sin 0 ; zp +
(Kanpur, lgggl
19'20. Erpand coss
o terle3 or {ne3 or corlnea of multpres ot
*"
".ff*
"ransion
l/*
=
0
5.32 sin40.。
in a *ries of asines of murtiprcs of
*-i)
L/1z2)g
ofsiaa
-
14(2r siu 20)
Ir'o lff)_4 sin g0 +3 ein 60+g sin40,r4sin20r.
0 cosn 0 is a eerie€ of sinee or cosioes of
multiples of
0 according as
n
is odd
0àd詢鳴 199`S)
cdいは、19θ の
(Mc““ ″´″″,199θ
“o.
)
`λ
y1998S)
=豊 .h70-3 sln 50+sh 30+5sh o.
7.ExP飩 ■
d COS5 0 sln7 0 L a Se五 ●
s Of8i¨ s ofmdtlples ofO?
3.If∞ ●50=A cos O+B cOs 30+C cOs 50,flnd sin5 0 in tem“
″c陸 ″
lP● 屁
αごra,2∞ 0)
ofA,B,C
“αdraら
●И
2剛 S)
9.If.h40.。 s30=Al cos O+A3CO● 30+A・ co8 50+A7COS 70,prove that
(Hamirpur, 1996 s)
A1+9.{3+2545+49A7=Q.
Madms,
1998 S ; Bangalore, 1990)
19 IO. COMPLEX FUNCTION
■ ８２
′
１コｚ
＋
*i
-",
-
s20=cos 60-2∞ ●40- cos 20+2.
6.sln5 0∞ s2。
l/*
セ
=
-;.))t"
3.c∝ 70=豊 (m870+7● os 50+21● os 30+35●
cu p0 - i sin p0
■∫
O
l/z)(z + l/z)13 = (z - L/z)4 V2
1.32 cOs6 0=c∝ 60+6 cos 40■ 15 cos 20 +10.
Defirition. f for euh ualue of the compler uariable z (= x + iyl in a given regian R, use haue
otle or nlore ualues of w (= u + iu), then w is said to be o conpler funcliot of z and we write
w = u (*, y) + iu (x, y) = f (z) where z, u are real functions ofr andy.
Ifto each value ofz, there correspoads one and only one value ofu, then ul is said to be a
single-ualued function of z otherwise a multi-ua)ued fawtion. For example w = !/z ig a
single-valued fuactiou and ra = {z is a multi-valued function of e. The former is defrned at dl
points ofthe z-plane ercept at z = 0 aod the latter assumes two values for each value ofz ercept
= (2 cos 80) + 8(2 cos 60) + 28(2 coa 40) + E6 (2 cos 20) + ?0.
Ilence cGE
(poona, l9g0\
4.sin8 0=2 7(c。 880-3● os 60+23∞340-56 cos 20+35)
0.
L€t z=cos0+isio0, sothatr+ l/z=DolsOandzr+ l/t =2rursp0.
;.
(2 cos 0)8 = 1z + L/z\E
8cp7 .lt ,"nt
,r*t .
,cern .)*t"*,
.** ,"*,
=28 +
)t
)*
8cr1r6* Ltz\.+ iCr1za +l/z\+ ECsftz
=1zs +l/z8l+
+L/zzl+ ,Ct
Q
g.
2.26sh7。 =35 sln O-21 sin 30+71■ ●50-70.
= Zcos p0, zp - l/zp =2r sin p0
These results are used to erpand the powers ofsin 0 or cos 0 or their products in
a series of
sines or cosines ofmultiples of0.
Eremple
of muttiplcs of
Prove that(1-6):
case.
0+ j sin 0 then L/z=wsO_ j sin 0.
By De Moivre's theorem, z{ = cos p0 + i sin p0 and
z+
si,us
Problems 19 7
that:7 + yt +zt=1.
cor'6 ol
o xrizs of
(Bcagar● L19"s)
Ofthe equattOn t3+Px2.9r+ρ =0,prOve that
tan- I p + tan I
t9 9. ro €xpond Jna 0.
in
sin 100 - 4(2t sin 80) + B(2r sin 60) + 8(2i sin 4e)
o,7 be the r00“
tan- I o +
:.
= 2i
CV電 ′tlら 1997)
■ 2(1+● ●
S80)=lr4_ピ .2)2 where x=2 cos e
1013+´
a tan 50=・
whe"′ =tan a
7.r■
J(レ
…
O
i)-. [,' -i)., [.' -i) .' (.' -i)- "? - ))
Since
i7 = -i,
.,. sin? 0coas 0= -rr!
Pr"e that(3■ ):
3.sln 70/sh O=7-56 in2。 ,112 sin4 0_“ 8L6。 _
,whe壽
- L/z)a t(z -
=tl'-
60=32∞ ♂ 0-48∞ 340,18 cos2 0_1.
4+竜誤
;=lx3_=2_2+1メ
1z
=(, -*, *,
tan α=tan(■ ■
Problens 19 6
a bress血 60/8in
Eryand sinT O .",ss
Le;t
z=cm0+isinO
so that z + l/z = 2coe 6, z l/z
= Zisi[ 0 andeP - L,rzp = % si\ p0.
:.
(% sia e)7 e cos 0;3 = 12 t/z)7 1z + ttz)e
-
IBy§ 141
and
615
[cos 80 + 8 cos 60 + 28 cos 40 + E6 cog 20 + SSl.
atz=0.
〓
０
www.Engg-Know.com
616
617
HCHER ENelNEERING
2. Separate into real and imaginary parts
(ii) exp (5 +
(j)eXp(5+j● /2)
1911.(1)EXPONENrlAL FUNCT:ON OF A COMPLEX VAR:ABL
When F is real,we tre already familiar with the exponentlal inction
∞
♂
+≠ +… ¨
=1+子 +手 …
・
oc∝ 2=2c∝ ≒ -l m畿
+…
Similarly,″ e dcr"`働 e ettac几
′ルれCr20P2
0Fne conρ
た .。 0だ α6た
2=工 +″ ,aS
o=1+争
イ+… +手 +… …
i・
4. Prove that(1)iSin(α -0)■ ●
=[イ
e' = d -
Thrrs
A.tso
ei)'
障威‖
…
・
4J石 [ギギー
卜
…山
a Glven:=去 +CA+1,Where L R R are red,exp"“
y
(1)DdL frκ
。 C。凛P′ 鶴
be r2。 ′
「
.
Thus slnhェ =♂
ai"_"-,r
--E-,
COSy =
t&+e-1,
and
SO
:お
-
z,
sec
z
and.
lt
Z,trnz=-
ulz + e-
sln z
col z as thair tespectiue reciprocals,
+e-''
cosz+isinz= ett
---t'
Cor.sinh O‐ 0,CCISh O=l and tanh O=0.
(2)R● 2α 琵0■ S
.et'
-e-t'
' -E-
=
where
""
r
-:
団 e缶』
“
+ iY'
.{Iso we have showtr that ei' = coe .v + i sin y, wherey is real.
Thus eio = cor I + i da 0, uricru 0 ic nal or anpler. Thic it callzd thc Eglerb theorem..
Cor.Z. De Moiore's thrr,rem for compht numbrs
Whether 0 is real or complex, wc have
(cls 0 + r'sin 0f = (/0f =C'ro = cos n0 + i sin n0
and
find the
r*l
and. ima,ginary part, of erp
zl
(r * u,2 - 12 -,v'] +
=
exp lz2l = e"
p. 296.
"'
=
-
""
and
" ;t':" =
-12,-"
.
"'2
い
e dn
2qr, +
.'.
Cor.
i ein 2rytl
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m“ ∝ 0=写
ザ
Ⅲ =ギ
“
.z
C
t
=9-
--
ir
tan ir
sinh
ir
corh it
tenb ir
=
cosh x
=i
tanh x
= i .i! t
= coa r
= i tan
x
1・
メ0=が 'X=2 rl
i. "' --t-'t = i "inh ,
.o"o=#=coth'r
sin ir = i sinh r
cos
hzl.
2ry
一
〇
fmt note
2'"'
-
Thus
S・
cjrc.`α rfancrt。 ●
VduesOmⅢ n e=写
-
Problems lg.8
z = x + iy,
たαPta
″晦parbο ′
“
be′ い
∴PttmgO三れ
Thus De l\toiwe's theorem is true for all 0 (real or complex).
IHint_Since
C i and cOsh==♂ +ご
haね
」
…“
∝
購T鳥 =写
ふ詳器 =等メ
山 =蕊 =ル ;… 臓=轟 =ふ
on.
Cor. l. EulerJe theotem. By definition
See
"」宙ng the vdues o「
crv = cosJ + i siny and e-iv = cosy - i siny.
ciranlar functions of na! angles can be written as
is. therefore, natural to define the circular functions of the complex uariable z by the
lf
m〃
Ж
andお W“
dゆ ed“ 貯
Otギお
"en as dnh x.
"rbouc dne d
C"hメ・
た
●
蔵
置いhypeFbOliC cosine d X“ dお ″
)キダ ″
。
“
エ
ei'- e- iz
-,
srnz=---Ef-icosz=
l.
p h theお
1912.HYPER30uC FUNCT:ONS
=er (cosy + d siny)
equattons :
cosec
“
A andO.
.r +
Stny =
urth
J“ ∞80■ Jn釧 ■
L還 1富 I府よf五撫 l屯 :>hoJn山
ltan a
Since
It
°
ュrzギ l轟"ぬ乱
舞―
i! =r(co6 0+isin 0) =re,o
-'. Erponential form of z (= t + D) = re'o.
(2) Circulsr fulctionc ofa conpler variable
.'. The
′
・
3m01″ =Sin''α ご
)
Puttingェ =O in(J),we get
=1+暑 +γ +ギ +響 t…
`●
=d2
(jjI)sh&=3 sin 2-4 sin3 2.
轟
`jα
e2。 reXp
3i)2.
fSIn 2 and cos2,prOVe that
3.From the deflnitloい。
…(f)
…(ff)
..(IJf)
…{`υ
..(υ
,
)
..(υ :,
下
I9.I3. TORMUTAC OT }TYPEREOUC FUNCNONS
b\ Fundamentalfurmulac
Or
(1)cosh2r-ginh2r=1 (2)aech2:+tanh2*=1 (3)coth2.t-cosech2r-1.
(4)einh(r+y)=gir\rcoshytcoahrsinhy (S)cosh(.rty)=cosh.tcchytsinhrsinhy
t tanhy
(6) tanh (, t ?)
= I,h\t
t td[hr tanhy
lc\ Functions of
19‐
r
∞nsider only thdr prindPal values.
t
r * I = 1 + 2 sinh2.r
= 2 coeh2
(
(2)To s■ oω
(Jハ
1
r
ld,\ Functions of 3x
(10) sinh & 3 sinh:+ 4
!
(11) cosh
&
= 4 cosh3r
- S coshr
sinh.t + sinh y = 2 dnh
(15) cosh
r
+ cos6y = 2 co"b
f
costr
f
r
rnr sinh.t
-
sinh y = z cosrr
16) cosh.r
-
- coihy = 2 sinh
=
=
+z
'
(l/i) sii i (, + y) = - i [sin & cos iy + cos ix sin iyt
= - i I sinh r . coshy + cosh r.i sinhyl = sinh.r coshy
Otherwise : sinh .t cosh y + cosh r sinh y
1.r
+y;
f
sinrr
lf
ri"f,
￠`=2士
!2
f.
- e* - e-k
+ 2t =
「
_22a“ -1=0.
ヾ(4z2+4)=2±
x sinh y.
or
2u
Example
Z __=t__=sinir(x+y)
19'22.
If
=log[1::)*f,"rr""
" [1 -,,J
1''u
=22tt
(ri)0=- ilogtanF.;)
(,c′
99イ )
た
',「
弔囃
鋼
e●
/2 _1+tan O/2
― /2
。“
1_tan O/2
卜﹁
は
0∞8C― ∞
8D‐ -2血
∵ dnV
="u
follows the result.
弓卜午
Similarly we can“ tablish the fOmulae(7)to{11).
Putttng c=と ,and p=り ,∞ s=― cosゥ =_2 dni工
● =tan(1+:)
“
By componendo and dividendo, we get
tanh&=守
+Jez + t)]'
+=
(Madros, 2A005\
=辮
We have
[z
u = log tan Qt/ 4 + 0 / 2\, prooe t hat
(i\tanhu/2=tan0/2
Putting A=r:, tanro=3tanr,r-t4n3rt or itanhBr-3(itanh:)-(itanht)g
l-3tan'ir
l-S(itanhr)z
マ●2+1).
eu -e-n
z=V;7.
Applying componendo and dividendo, we get
+ cosh
Similarly we can establisfr the formulae (5) and (6).
軸鎖
2=;′ Og争号
.Tattng the positlVe sign only,we have
i.e.
r.
d _e-F .er +e-J , d +e-, et_e-.r €J+J_e-lr+.fr
_
=_T_
⑫
′
剛
r)
=:ぽ ―
=z +'J@2+ l) or u =log
Similarly we can establish (ii)
(tii) Irt tanh- ! a = z, then z = tanh u
=
Z
(JJJ)'aa力
en
1
(eI
-t'*C)' | -l:i-l
=l="-'f i r" + e-k
Similarly we can establhh the iormuhJ t2t'and t3l.
(4) sinh
+セ 2+I】
22=● “-1/● “ Or e2●
.・
f
(
&oofs. (l) Sioce, for all valuej of 0, we have cos2 0 + sin2 0 l.
=
.'. Putting 0=Lt, we get cosl r* +sin2rr
= 1or cosh2r sinhlr
otheruise: cush2r-sinh2
,",-
=′ .gレ
nsaremmytJu中 .w¨
“
This being a quadratic in c“ ,We have
cosfr
f
r′
`C)sれ
gレ +ヾ ●2_11,
●
cosl ″ z」 ′
。r
1+3tsnh2r
(e) ( 13)
た
r■ α
C)Let dnh 12=“ ,then z=sinh“
sinhSr
=
3 tanh : + taP\3 r
12) tanh ,, -
ザ dnhザ
14.:NVERSE HYPERBOuC FUNCTIONS
2
(9) tanh
2)=2 slnh
COSh y=-21isinhll争 Z)(J SInh壼 ≒テ
=盤
“
if棚譜証腋漁L“
lI亀 l」器品
2z
Zr= df
tqn[z
COSh,―
Similarly we can establish the fomulac(13)to(15).
(b)Ail.ition formul*
(7) sinh 2, = 2 sinh r cosh
(8) cosh 2r = cosh2.r + sinht
619
COMPLEX NUM3ERS AND FUNC■ ONS
bg珊
+ysh jザ
0
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ノ
/
621
rUNcTloNs _
o+ip=5sc1aair; then o-iF=sec(r-dr)'
coMPrEx NUMB€RS3\I-D
(olLet
Problems 19.9
l.
Eliminate z from p
:
cosech
lr
2. If-v = log tan r, show that sinh ny =
|
{tanl r - cot,,rt
cosh (€( + p)
(ii) sinh (c
* c6h (a -
Prm thar (i) (cch 0 r
sinh 0)' = msh z0 + sinh n0.
8. Express cosh'0 in terms of hyperbolic cosines
7. If sin 0 = tanh:, pmve tfiat tan 0 = sinh :.
8, Il tan r/2 = tanh u/2, prove that
(i)
tanr=sinhuandca:coaha=l;
9. If cosh r= sec 0, prove that
(iltanrh2r/2=tatlet}
lO. Show that ran- I
ll.
,=
;
(ri)
fl{+***i
of mulup,I;;:t'i
= coeh 6e +
sidr
∴
ee.
鋼
卜
/4+0/21.
sinn-r:=cosh-li{l+r2)
=tanh-l\(l_rz)
' = =lcos."h*1
Z
Gaipur, l99gl
α+Fβ =tanh
"二 O+ty)=(1/1)tan(Iχ ―y),
o
に
0)=log tan(■ /4■ 0/2)
(id)
tanh- I
=5
(`)sj●
(ェ
。腱。′α鷹ごJmagι れα[ソ ρarrs Or
+″ ):(′ 二
)c¨ ←+ケ )(J力 )ね ″←+ty);(Jυ )oο ′
(r+ty);{υ )sac(x+″
,r,88s)
Example 19 23.rFCOSね
);(υ
j)Lt
Subtracting,
α+ビ β=tan(ェ +″ )then α_jβ =tan● _な
2α =tan← +ヶ )+tan(χ ―″
)
=論
車
椰=
).
Since
ゴ
)cο sac(工 +ly)
∞t● +ウ )=電
=COSh(“ +jυ )=COS(￨“
F+″
― υ)
〓
cos t'+I Sinh``Sin
=COS j“ COS υ +Sin I“ sin υ COSh“
υ
・ Equating real and imaginary partS,We get x=COSh“ cos l;y=sinh:`゛ n●
=∞ Sυ
蕊
=鵡
and論
=“ nυ
Squa五 ng and adding,We get the rlrst result.
Agdn
=鵡
.。
量子もテ讐:増号:
=∞ Sh“
嵩
and計了=gnh“
Squaring and subtracting,we get the s∝
Ex3mple 19 24.ffra●
β=鵡
Similarly,
(“
998,
′
o+l■・)=sin r cOsゥ +cOs,sin″ =sinェ cosh J+J cOS I Sinh y
Similarly,cos(ェ +け )=cOs x∞ sh y_J sin,sinh J.
(:ゼ
cぬ ar
+lυ )=x+″ ,prο υ
こ
磁れ"ω Fメム
リ
・
+∴ ヨmd法 _∴ 封
話百
rみ
Pr●●FS(f)Sin
Adding,
sinh
=晶
β=品
(Pο ο
aα
15.REAL AND:MAC:NARY PARTS OF CiRCuLAR AND HYPER30uC FUNCT:ONS
(1)To sep● ′
are
2r'
.
-= co" zi, + co"2r':
iv - cosh 2r + cos 2v
=;t;;;A;;i;;G;,
蝸=―
(cos 0) = cosh- I (cosec 8)
(jJj)sech 1(sm O)=log cot O/2_
19
$/i\
Subtracting 2jβ =(1/j)itan(と ―y)― tan("+y)〕
---li(l_r2)'
13.Find tanhェ if 5 s砧ェー●
●
shェ
- iy) = ll/iJ ten (d, +y).
Itan (ir - y) + tan (dr + y)l
(1/i) sin
sin (uu - y + i* + v)
iB = tanh (x
c
12. Show that
:(ta■
-
2a =
2n(l+r2t
(f)shh‐
:量 1凛 :II∬ li:li
SmlhHy,c∝ ho+ι
(jjj)If
then
Adding,
ts.Gujarct, lggol
I
"ヵ
nh∫
ir*ff
(di) tanh-r r = sinh-r
報
:rぉ
(li)u=lqg611r/4+.r/21. (lfadms,Igggs)
(iiy:=lo6tanrn
=継ザ
轟
調
COSec← り
軌
2 cos.x cosh v
cos 2.r 't cosh ?-v
″
②
ハけぃ
Ъ37[1窃「 11:露f聴子係
“
PrO“ ‐
)Jnhけ初
Sh,“ nJ
―∞
い
二
二
認::K翼晰ヨゎぃ ∞
1劣溜∫
Prove that
(i)
椰=鑑器器齢=調
た
+ p) ccsh ro
-
=調
=SeC O+ty)― Sec← ―ly)
Subtracting, 2 jβ
p) = Z sinh o sinh p
- pt = | tsinh 2c + sinh 2p).
4. If tan-v=tanatanhpandtanz=cotatanhp,provethatten6,+z)=sinh2pccec2n.
E.
α=
た
3. Prove that
(ii
--
+ 9 sech a + r = 0, p, cosech e + q. sech z + r, O.
=
(0+jO)=♂
ond result.
°,Siο ″ぬar
O=い 1/aπ/29π ご
o=:I喀ぬn“ /イ lα /a
Since tan(0+jO)=COS
α
⑫
一
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+:Sin
.
α
∴
lMam`力
tan(0-Jo)=COS
r99o)
αd∝
“
"94;PoOlta・
α ― I Sin α
丁
tan 20二 tan I(o+JO)+(o―
=
Jo)】
As i is to be beh″
COS,=+ヾ
l.
―idnα or :芸
:を 41ポ ≧ 裏
野
「
=
(j)sm2。 =士 sh o
(jj)ノ
[」 llililα ndiliilillllin(11lil,t。
Then α_jβ =tan 1● ―ゥ
).
。
。
s)
).
y)
← +む )+← 一二
2α =tan 1← +ty)+tan 1← ―ウ =tan-11_←
+け )← -3y)
Adding,
1 性
α=:tan-1_ノ
nh_1百
l●
=鶴
+ι ッ
)=sin
)●
―む
)
or
l_sin2.=sinh2 J7,I.● .∞ s2″
=sinh2y.
mム
∬ 露冨
tan‐
l,士
A tt B
"な
と,1999:Aaご λra,1990)
0)
(li)cos 20,cosh 2o=2_
+y2_2y∞ th 23● 1,0..
23=y sin 2A.
(置 )=sillh
(J●
(Moご
"s,1999)
α,prove that
(M● dra.s.20θ り,′ セ
●
′
″,1998)
"dtc力
(Madra∫
5.lftm● +“ )=tan。
‐
a rtm仕 +“ =shに
+わヽprOVe that轟
j“ cヽ prove血
■
…
htt th‖ arヤ
lO, Separate cos* I lcos 0 +
ち
ザ
+,―
訛
●∞
"=士
≒
,1996S′ νara`λ ●dc,199`)
“
r99θ
(C● Jalbarott
・(Eと
t″ 2 and 20=← +:)・ +α
磁 d"・
=量詩
(α
+■ )● SeCに
脆山
"00S′
“
一
)=デ
゛
静
:1'総協r脇』
12 1Fcos 1← +")=α
:10g tan←
_封
i sin 0) into real and imaginary parts, wherc 0 is
ノ _ェ (1+“ 2+υ 2)+.2=0.
)
ni r"ハ
i19989
(Nogpur. 1997)
a positive acute angle-
sin2 a and cch2 p are the rmts of the equation
+Jβ ,ShOW that
c),2.∝ 2α _y2∞ seC2 α=1, (:i)ェ 2s∝ h20,y2coseCh2 β=1.
1'Prove that(I)sh 1")=‰ ェ+j log{ヾ 1+ェ 2+ハ
.′ 99o)
∼
力′
(■ ι
,r99`s)
(Bo■ gar●
(j:)sm 1(cosec 8)=■ /2+ilog cot O/2.
= cos .r sinh
7
…(:I)
)
1916.LOCARlTHMiC FUNCT:ON OF A COMPLEX VAR:ABLE
lll Dcfinition. If z (= x + i1) ond u (= u + iv\ be so related that e"' = z, then
logarithm of z to the base e and is written as w = logc z.
,
have Sず 0=嗜 4らに螂 2,=dn Oい cm“ o bang,pdt市 e“u"
tall lB=tan■
4
犠
l cOshJ7+ゼ C“ I sinh y
sin 0
=sln2″ +sinh2y(sin2,+cos2ェ
菫
Sin
ll. If sin- t (u + iu) = a + ip, prove that
1=sin2 r cosh2y+cOs2.血 h2y=sin2,(1+Sinh2 y),cOs2 x sinh2 y
譜
01.
9.Reduce tan 1(cos O+i sin O)to the rorm.+fb
・ぬ
n-1ル =:餞 nh 1ヨ
に
百争7=Jね
争7
∴
=:tanh-1=争
β
‐OS 〒
SCPa磁商
O病
・
亜
:淵勝鯛
″
棚鳳
」観
=m-1ビ
and
o)
載y=bg N(dn O+■ 1+dn
&Ifa+jb=tanh(υ +:r/4),proVe that c2+02=1
)=tan lm
+lX+ウ
.'. cog0=sinrcmhy ...(i)
Squaring and adding, we have
論
dnhβ ,pm"that s∝
_,2
Subtracting, 2:β =tan 1← +″ )― tan 1● ―
ウ
Let sin 1(cOS O+lsin o=ェ +ゥ
Then
cOs O+J sin O=sin←
議
■Ifx=2暉 α
cosh O,y=2sh α
)・
∴
(gin 0 30ぬ
4 1ftan lA+IB)==+ヶ ,ProVe that
2+y2+2ェ ●
●
t2A=1.
(j)ェ
l
M
+論
Ifcos(o+ゎ )=Cos α ttI
=(珊
α+iβ =tan l lr+ウ
r工 =COS∼ 1マ (Sin
r cosl.+蝸 )=′ (COS O+iSh Ol,prOVe that e20=Jn(α
r Ex3mple 1925。
Let
(Sin O)。
i8) = r + ry, prove that
論
黎書方墨帯無携
°
(,tt +
=二十
1
芸==器 =:発
e4●
If sin
ff≡讐
By componendO and di、 ″denoo,we get
。r
623
Problems 19 10
ねn21=ほ nЮ +ゎ ■ 0-￨lbl=
21=T=烹
___
The“ lation〈だ
ぃhen,JVeS dnhy=イ
0=(2+1/2)π /2
20これ■+■/2 o「
れanh
FUNCT10NS
oon― ■/2 and π/2,therefore,we have
2coso
=
0
Also
COMPLEX NuM圧然 AND
AIso
u is
saだ
roル α
...(:)
1 ... .2ル
:マ
=11
"ro+2in*,-"to.e?jnr=z
_..{j`)
logz = w +Zinn
i.e. the logarithrn of a comple, number has an infinite number of uolues and is, therefore, 4
mu lti - v al
ued,
funct ia n.
The general value of the logarithm ofz is written as Lag z (beginning with capital L) so as
to distinguish it from its principal value which is written as log z. This principal value is obtained
by taking n =O in lag z.
Ttrrrs from {i) and (il), Iag (x + it\ = Zit-r + log (r + iy).
www.Engg-Know.com
丁
︲
625
COMPLEX NUM3[RS AND FUNCT10NS
Obs. l. Ify = 0, then Io6'r = 2inn + log r.
This shows thaa lhe logarithm of a rual quantity is also multi-uolued. Its priacilnl ualue
.is ral while
all other ualues are imaginary2'We know thol tlw logarithm of a negatiw guantity has no rcal wlw. But we can norc eoaluag
this.
ag.
lo&(-2)=lo路 2(-1)=lo路
2+l● gL(-1)=lo&2+J■
l・
=←
いnd
これπ+log← ο )
'°
ond imaginary parts o/1o
1I
-
e(x
= e(r
=
=
i ("2 * y2) + i
I
l2nr.+ tan-
+
i.t)l2itrr + log to * rptl
=/
(I〕
エ
:;五lli「 lli」
fi‐ e (イ
sin B.
Hl-77/猜
′
1・ ′ ′
2.
"os2
,
Parts
and
0=
鳳
け封
i cos, siohJr)
= log
r
(cos 0 +
d
sin 8),
=
{
1;,*"n
zr
-
cos
2rr]
tan- I (cot, tanh y).
Thus tog sin (r + dy) = log (reio; = t6g 71;6
`力 `π
η ―C∝
21+Ⅲ n-1い t=ねnh"
λο●
Юra。 ′′
9“ 。
``0■
“
Z=● ■+(-1メ
∴
(″
「
log sin (r + iyl.
sin iJ)
sinh2 y)
(￨)sin z=cosh 4=cos 4j=sin(π
(A,ど力rcI.r99ィ :Mα αrrltrodα .′ 990)
'R′
t
coshy and r Ein 0= cos: sinhy,
*
Lample 19 31.Eàd α
[Put0=rcosO, -l=rsin0
f sothatr= land 0= -il2.
=21PIπ +log〔 ″(∞ sO+'Sin O)〕
α′
iosh2 y
(:)sfa z=cOsh`
I
=2PIπ +10g′ +lo=2,Pa π+log
= sinr
:10g[:(COSh
I
Log(― f)=2i"π +10g10+J(-1)〕
a rぬ
that r = J 1sin2 r
0
こ
=5r logr +x {2nn+0).
/
rcos
imaginar!
cos iy + cos
=
+rttl2ial+ logrc"l
- rtx. i.vrllogr.i r2ar +erl
"(r rB
s:4' - 4 (cos + i sin g).
.'. The required real part =
cos
and the imaginary part
Erample 19.27. Find the general value of log (_ i\.
r
= log (sin:r coshy +
where
O/r)l
Put c= rcos 0, I = r sin I so that
r = r 1cr2 + Bly and 0 = tan- r p/cr.
o' ilt
t:'t lag
19.80. Seporcrc into rcal and
log sin (.r + iy) = log (sin
+ rp;t *,."
+
whereA =: logr -y(2nr!+0) andB
Example 19 23.PrOυ
/→
n*ample
so
`ケ
= Zinr + log r +i0 = log
(0 + ip)' *
O=tan・
:::】 =C03[(230)l=cos 20
言
蹴州寡=解
=2'81
「
]願「
載H;4dn。
。
+ウ ).
=2j″ π+log lr(cos O+j8in O)1
(3) BeoJ
-1=● ●●■●
'6hえ
=0.6931+i(31416).
(2)Rear a"dj"agtた α″ Parrs Ofぃ
LOg(″ +ly)=2ゴれ■+log tr+ly)
・
Thus cOs1810g〔
(1:)sinh′ =j
/2-4)
(■
Ifsin
/2-4`)
0 = sin
a
then0=nr+(-l)"4
)ゴ =dnh z=等
・2L[:I::]inlill[r」 iifl・ i[:;:=2..+j./2=j(2"+:)π
Problems 19 11
+log“
:ι
"1=2:(2inκ
=c_211R+::og!eゃ
`:=●
+,(iκ/2)=2_(2■
・
=e-211罠
1:Ж
′
2)! 1 ・・
′
1′ 2)●
f i cis■ /2=exp(I■ /2)1
1.Find the general value of
(1:)10g l-3)
(f)log(4+3だ ) (ranirp“ ら19“ S)
2.Shtt that(:)log(1+itan α)=log(3eC α)'fα ,where αis an acute angle
Taking loga五 thns,we get(II).
EttmpL D".Pr● ●cЙ 峻
′
(鍔 )=2ね
「
Putting α=r c08 0,b=r sin e so that o=tan 16/α
聴
畔
嘔
)封器
出
‖
ルた。 α ″
“ 午 …
a「
レ
(鮮
(R●
距
鰭♂ ‐■
=log e2ò=2JO=21 tan lb/α
.
.a prove that
】
a,199o)
“
,we have
1麻鴛li鷹 J二
")tan 1≒
+tm-1≒
+… +tan
i≒
=tan―
:景
ment of
4.Find the modulus and a■ ロュ
(j) (l + i)r
(ii) iix tt ' tr'
φ=α +Iβ ,proVe that α2+β 2=e “ “'1'R0
5。 If F・
-'
ar/ =A+認 ,PЮ
www.Engg-Know.com
“thtt tmサ =景 and.2+328e ■
lCr・ h詭 rore,1"ハ
(Bangalore, I*)31
|Poona,1990l
026
7
H:GHE,[NGINEC,INC MAIHEMATICs
. lf (a
+
ibf
' 'r, prove that mc
= mt
2lrio,-r b/a
tan
+ dy) =
c + i6 whero o, +6!
+ ry)
-
log (A +
Jv2
.?- --"-=
sul- u 6g'u
ir),
(3■ ●
Pa′ .1998)
* I, show that
cosz=2
(3anfarott r990)
show
that
lwarangal,
(κ
p.rts log cos (r + iJ).
l9g9l
″ ヵ輌 erに ,r"8)
“
(ii) tanhz+2=0.
o*orro/t
o*or+aP*..;...tanter,,s=ait*,
i.e.
+¨
)十
ol
IC+お
ε.`手 If
is a series of sines (or cosines).
=Icos
α +じ Sin al+ェ :COS(α +β )+J Sin(α +β 】
then
C=α 。∞sα +α l COS(α
+β )+α 2∞ S(α +23)..…
C+β =● 。1∞ sα +1
=● 0●
(″
:α
+α lど
3in司 +α l lcos(α +β )+J Sin(α
に+')+α itと ,2β 〕
r― 轟
+妥 …
∞
…
=“
…
・
・
Erample
sine e
+......
,,
l十
′
T′
+甲
.'.
「
メ
+_ ."=(′ ●ェ
=
「
"=(1-F)‐
C + jS =eio sin
“
0 rttu -'t rin
logr
r=:frI「
争
|
r
+
i
sin J0 esnr O -
sin 20 . sin2 e +
|
I
- "2io
sjnz o
*
t"ti'"
6.
si"40 sin{ 0 + ......
30
sin
sin3
t
-......
(cos
c
+
i sin c;
= log re'n =
Iogr
Q
tHantirpur,
- ...'.' o
0-..'...6
:
-
s●
れ
,ο
Let
and
S=0-:sh O+髪
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binomi」
ピ
列εttjry orぬ ″ r3● S
“
oo「
(―
...tr
t
ifrOm(jl
te3
)
ω
・20-:i}:い 30+… ・くつ
C=1-:∞ 80+減温20-:幾 cOS 30+… "
;COS O+赫
r sin cl
+ icr.
=tan 1(鵞
Serle● depending o口
“
Sl
't sin F)'
loirrithoic rcrier
series
0+
Ex3mple 19・ 34.Find,ル
I―
1996
pt
i sin 0) sin 0l
= log (1 + eis sin 0) = log
+ cos 0 sin 0 = r cos u ; sin! 0 =
1
+
0l
I
sin2
0
sin
0
log
+
cos
[Put
[l
=
=
-
-]
-4o
g
[1 + (cos +
=・
∴ ….‐ ‐(1+xメ
rin El
C=cos0-sin 0-|cos ze sin20+ico63g'sin3
″
1+“
・
“
卜け半#リー
甲
- f,
sinzg sin2
Equating imaginary partst we have S=α
+f,f+… "=れ力
ェ
(5)Binomi」 ●曖ね8
ノ+甲
ＦＴ
一
=J・
ia l+薦 +甲
―
…∞
=ω SX
¨
エ
…∞
=む方
手+暑十
２
工
，
〓
ｒ︲
ヽ
…∞ g(I+ェヽ
・―
手+手 …
たェ
ー
が七
手ギー……た
i・
and
tlu
10.88. Sr4rn
Let S = sin 0 . sin e -
1-岳 +♯
+爵 +i+… …
ェ
“・
- rr o
u
"""
+
[cos (a
erie●
,
=・ J・
"iaaxto
F' ito
+r sin 0) + i sin (a +, sin F)l
G" 0
sin (c
parts
from both sidea, we have S = cr
Equating imaginary
+Dl+.… …
l+=+手 +妥 +… ¨
"=♂
(2)Sine,cOsine,sinh or cosh●
qp
-
t dn pt
Serier dependiag oa
J)Sum up this last series using any Ofthe fO110wing standard series:
(1)Exponendal"de3 1α
ia
′
ria . SO
-
p +
(s+ 2p) + i sin (a +2p1l
.*-#-
t."ira*z6r*......-=r,o[r
=cia+j(o*Fr+
.
(Jj)Multiply the se」 es Ofsines by j and add tO the series ofcosines,so that
fr t-,
*
=d
S =oO sin s +at sin (o + p) +a2 sin (a + Zl]) +....,-
(Re″ α.f998)
n(α
cos (o + 0) + d2 cos (o + 2p) + ......
ProgcdTe' (r) Put the given serics s (or c) according aa it
then
- write C (or S) = a similar series of=cosines ior sinest.
t l.I.
.
-
oo cos cr +
I
∞
れ
+雨 (o+β 」
α
“ :轟 0+2詢 …
+お +D+手 Bh(α +2D十 …
Let S=dn α
…∞
and C=cos u+,cos c+D+手 COs c+2o+… …
"
,C
19.t7. SUMMAIION OF SERTES
+ iS,MEIHOD
This is the most general rrethod and is applied to find the sum ofa
series ofthe form
a0 Ein o + al sin (c + p) + o2 sin (o + 2p1 + ....-.
or
-=*
{iu) Fimlly express the sum thus obtained in the form A + i8 so that by cquating the real
and imaginary part6, qre get C =A and S = 8.
Scriee depcnding on exPoBc[tiel rerier
た●wries
Ex3mple 19・ 32.S“
"′
s,″
l, wherei{2+82 =e}'.
Separate into real and imaginary
12. Find all the rootr ofthe cquation
(i)
(襲1■l r991)
losll+y2y= 4
t_a? _*'
10. If sin- I (r
ll.
+ t6g 1c2 + &2)
,
COMPTEX N|]MEENS AND FUNCTIOI'Is
(6) Gcomctric scriec
of-ylr is
,- /g-&I - zoc
t;+,.6J.|=p:*.
8. Pro,e thst tan f,
9. Iftan log t:
of the values
0く
dn 20-:器 dn 30+… F
628
HCHER ENCINEERING MATHEMA■ CS
C,お =1-ン 0+釜
`20-:器
0・
0…
・
…
COMPLEX NUMBEtt AND FUNCT10NS
E=ampL 10艶
一昨翠 ♂
%。
■ 脚 ′
距
ギ
尋
i璃
di基
取
菱
豪
…
::乱
)-1′
2 cos。
∴
3.κ JnO―
:′
il″
_ギ c“ 3β +÷
n3●
:∞
(H● ″
:′ rp● r・
α+cos(α
+β )●
∞s(α ,2β l●
10. 1■ r cos O+ェ 2 cos 20+ .¨ .+■
1995S)
(ル :υ a,r99o,
′
:terms
=ユ
.t●
(漁
`r:`■
●
Os 3α + ……
“
″ l cos(2-1)0.
2rr● sin'α
、
r".1998)
力″
(Ret●・
●.19θ り)
1918.APPRox:MAT10NS AND uMiTs
ExampL 193二〃
S“
=器
。″ 唸
ルどα
・
eギ =1-轟
=評 whぬお
We know thac
∴
:凛 αセ
"“
ε″ o● ″ごねに
ryaご ″ ,I"ヵ
w small.
"山 L∴ omぃ t“ ぃ
輌 ←鮮だ仁
¨
“
nearけ
¥=1-ギ
+讐
Omitting 04 and higher pOweぉ ,we have
Ψ=1ギー
轟
=ュ
02.1:5 Henc。。
=
°=50マ
7 radian=嘉
x5729 degrees nearly=397.
h¨ that Oお r譴 Ъ
・ G"n¥=器 ´
“
"
2r∞ S,=器 Ⅲ
祠0叩 ma"け
「“
3.So市 e approximatdy the equa“ oncosI:+0)=049
.
3"+景 c∝
9.sin αcos。 +sln2 αcos
02 and hghcr"woぉ o10
Problem3 10 13
‐
―
霧 c"“ ‐…一
“
■"Jn α
+Ψ J.2α +型≒翌 Jn 3α ‐…ニ
3.cOs
al―
h:… …:蒟 ←:[― ゞ
t卜誓ー
t]
Ⅲ日 ‐
ト手
Hence the given equation be∞ mes,
(Rcψ ●4r998)
s5β
dn 5α
/6,so O must bc vory sma‖
:+50=051 or O=轟
"
∞ …"
5.csh α
+ゞ轟 ●
」 +… Ⅲ
4.eC∞ sβ
629
/4
、_
sin 20+計 3 sin 30-…
___
o)=ひ 5・
J“
=:+50,Om競 ing
°r
2“ nα _T+Ψ
__
Since o.51 is nearly equal to 1/2,which is the valuO OFsil■
Sun the following se五 es:
1.c‐ 0+sh ecOs 20十二曇 ∞s"+…
app―Jmaめ
__
卜
Problem8 19 12
ヽ
S● Jυ e
‐
鳥ra・ ans.
⑦
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tN●3R
AteEBM IDETE● 悧 :岬 N｀ 1剪 RC5
33
which rs a bloct of n2 .r"z4n& arrelgsd in the forE ofa squarc alo[g n rows
atil n columns.
Tho dragonal throuah the left hand top corae. which cootsi; the et"ir""t" i1J2,.",
......1, i"
elled rhe leoding or prittcipal diaonal.
_2
(2)
Linear Algehra : Deterrninanrts, Matrices
Miaorq Cofectors
fLe mhor ofca
eJam ent i4 a d.teminant is thL d2terminezt obtained b, deletitu€
the rou
n uhich intersect iD that ebri*,k .
“
inゅ andlllg by 21 11 e lst ro■
of
"″
=
- lor
lar "'l
c6l
sOFa″ rO″
餞″ 読姉
rarで 07“ ″ n,
v、
‐CA… t cICI訥拷日も1性鰤訛
矧
ら3C2) Oila2r3 Cr2)+。 lla´ 3 α 3b'
Simllarly expandlllgby c,(1`2nd column)
2 2 DETERM:NANIS
△
=lrl+ら 222+甲 3 ら
獄
唸
品:朧職鼎還凝塾後声
纂
S姉職し
a″ れ
硼
を十
たの
“
`″
ルれだ
OMer.鳳
△お滋
of
the
h
¨Ｌ
『
P輝
:髯
。:::￨“
卜
2に
3に
:ll
¨ イル ″ ″ 油 ″ c21・ rOp ror c.7“ 凛″ けル
“
α
メ 32+C3C2 α
2+￨●
oder,6 de\otdby
艦総盟翻譜・
彎
認離器競脇%1蹴軍題
:裏
eg,hヽ
ln geaetal, a detzrminaat
312げ
織
、ぼ 9…
“
Vhlch are arranged in 3 rO″ s and 3c′
懃 ■s
1に
=― らllar3 α 3CD+ら 2● iC3 α 301) 631alc2-Qol)
and aq"Щ dlng by 23に ● 3rd mw),△
=o。 49+ら .B3+COCa
l
,Ⅲ b鷲撒
sign. Th2 sipn
み
え
ら
″
π
d arPad“ 昴
θ
"読
織雛渉嚇鵬げ″慮鰍げ“
=● 1(ら ′3
檬
■
〓
岬
剰
劉caa&`"鰯
4凛た
﹃“
:
●gl● ●.●
コ
(3)LPLcebep●
CSμあ‐
miiot uilh tlc pmper
The cofactor ofan element ia u-sually deuoted
miror ofD" i."- a"
C2-臓
SInllarly,
'臨
x
ｎ
°
織
蠍
鑑雌鷺」
2
its
血
Thur tie cofactor of6i = 1- 113,
21 :NTRODucIIoN
h
らゆ
￨￨
(- l\i ti.
〓
Tha @l,.ctaE ofarry elarnzut in o &tetmina8t
aa eleme* in thc iti row orrdjth column fu
by the correrDoadiDg cspital tett€r.
﹃％
ｎ
ぼ
α
=￨を
ｄ
For instance,In△
３
﹃”Ｃ
colu
Ｌ場祐
and. the
1n genenl,■
81驚
“
c●
=漁
・。
::卜
F::1-C31亀
::￨
Cslalわ
3 ●3。 1)=0
when,≠ ブ
'Naned ater a great Frenぬ mathenatclan ac‐
32
_SPa滅
α
″凛″
″
″ ル rOrs″
″″
“ “ “″ ●
=― α
」blc3 60rl)+belars-00Cl)―
+●
+qら
=△ When iづ
今 ■
‐O
CObndge mdt h硼 .ゎ r“
c●
να
勲
C“
1749 18271 He made
'aon
,,
"“ "ユ
:環盤∵議驚踏出
欄鵠:僣轟:協薔
ぶ辮増撮
F° "pdmtul.tt mdastmnomy Whい
―J
〇
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“一
2・
５
３一
…Ⅲ 麟
[;‖
HIGHER ENGINEERING MATI]EMATICS
UNEAR AIGEBRA : DEIERMINANTS. MAIRICES
l.ExP祠
=α 02● 3 。 3C2) a2(ら lC3 ら 3Cl)+● 3● IC2
l・
ExPanding by 21, △
md臆
[:よ _;ll11_11群
_11)=abc+2なん―CF2_bg2_Jし 2.
出
珈 J“ ″
北富 :筑″
h ttlJⅢ
Interchangin]宅 fillll湾 li背
・°
-2× 2_3xり ●
3c2xO_3x釧
171lL高」湯
二
I螺響
ふも
l■
2l
一
in the aetermina* :
l_
rs-i oG^
o o, I
,1r.1 "
|3
|
t
:
,
ara"読れ
鍼
,…
equy範
ゞ２
４２
５
メメ２
４
２
・メ２
３
１
o=
=
Thcn
'
lo, bl "rl
lo, bz cl
l"q b3.bscT)
";l
a1(b2cg
-
tExpand by
+al● 2b3 a31リ
Xpandhgけ 2」
-
b{a2cg
-
lor o.z os
a'= lb, b2 6;l
co
agcz) + c1(a2b9
-
a362)
I
icr
"el,
fExpandingby.Rll
、
娠Ⅲ刷脇鶴粧椒酬潔二九
4:J… 04.
J)・
"商““
.
._r“ p・ r.ar“ ha,ル
“
た
"""arivと "ビ
,こ
●
`
.
器棚翼盤脂詳
胤:翼l酬譜躍器常地獄憮 i=譜 ep面°
a rik"″ ″
ヴ″″ル銅2ル軌ル″
わ
た&¨ ″お
れ
″
勝轟1錫臨
us
prぶ
た撥隻Jヲ臓をJ
"α
For on‐ pandmg by C2,
L.IIS.3-メ
♭lo2C3
● 3C2)+pら 2● lC3 α 3Cl) ′ 03● IC2 ● 2Cl)
blJl+0232 。 333)=R.H.S.
鍋軌臓キ引錢
￨
11■
撥
ん
あ
昴
ぉ
た
処″め
″
・ “ ゎ
め競魔肌悦颯憲
=pl―
^R;l
`α
…′T分
2C3 α
.げ
Let
dun“
れ′″ れ,構
Йε議物
ム
LA″
れα″ υ
α ぉλ‐ lr加 ′
e′ ′
′ s are認しπ′
た配
・剛ゼ′
“ in which two parallel
“lines are identical.
Consider "渤
a determinant△
2.3. PROPERTIES OF DEIERMINANTS
P" following properties. are proved for determinants of the third order, but these hold good
for determinants of any order. These properties enable ue to simplify
a given determinant and
evaluate it without expanding the givin determinant.
l' A dcterminant remains unaltered by chonging its touts into columns
and. colutnns into rows.
-
終
田
撥
ど=←
´
＾
´
、
＾
¨
¨
一
一
．一
．
Ｅ
縫
０ごごｄ
為岬
品
…°
ド
T… 転
IF●
mち麟艤
"α
:‖
Lg●●
enl,rα Fy,hな ″
αれ
=
１
",ro"
orr
c
fi _!
' lo
一一
3. Find for what
D,
Gr・
﹁
l. Evaluatcthecofacrorofa-*"**rrrrranr,
1∼
創
Problems
面
PtheommcOno_Ing the row3 ra…
れふふ島協曰
2. Calculate the co-factors ofa,
′1)
1ら
=α 10208 ら 302) 。 102C3 ● 3CD+Cl● 263 ● 3bり =ム
.ル
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J“
“"“
e出
"“
"●
in山
“
`"″
mぉ
゛
ι
けЙ 読
c●
LINEAR ALC[BRA:DErERM!NAMS MATRICES
HiCH[R ENCINEE'INC MATHEMAT CS
″
:.数 L指距………
;馘鰈
W If n each elcments of o linc be addzd equi.multiples of the corresponding elements of one
onnorz pomllol lircs, the determinants rcmains unaltercd.
L励
6。 nsider
the deteminant
Δ
=￨::
￨:
t,,t
:￨:i[:::￨
Then
each ofwhose th力 d cOlumn elementB∞ nsists ofthree tems.
﹄
﹄
引
引
﹃
一
一
＋
＋一
〓
br +
0r
cr +
yrl lo,
D1
.rl 1", Fr .rl lo,
O'- f l = O in whbh a,
As
-each
of two determinants.
D1
rrl lo, Fr
abc
of two tsrms, we express
it
:￨￨+￨￨￨:
:￨
::￨+￨:i::
followingaotatlon:
::
:￨￨
.
oddp times the elements of the fust row anil g times
,int 6lumn and - a time the
t
i:￨
△=lli 「:: ￨ ::￨
lExpand by Cュ
= I.
1■
as a Bum
Exampb 2ふ
-I
21=2R21
:: :il=0
,ル
OperatingR3 C71+R2),We get
lЪ =i:
墓
from
0 as a, b, c are all different.
ｂ
ｌ
♂′ ′
α ス︶Ｃ
*
1
″
あ
れ
釘
″“
“
::多 ::Jを :il=a
睦
￨:i
1+:1
1
of the
Openttng21-22 ■ 4,22 3R3,23 274,the glven de"minant
yrl
1::
[Passing C3 ovet C2and C1 in the second determinant]
￨:1い甘贈¨
n tines t.bc elerenta
鳳藤
″
崚
な
。
助
鷺考
;
o, 6, c from 8t, 82,.Ba respectively of the first determinant and
ofthe second determinantt.
二
:￨
=―
_;l
=・ bCII
::￨=￨::
mad olunn-
i: fl.i* ft rl
l" o2 ot-rl l" o2 orl l" o2 -rl
lr o orl lo o2 ri
ia b2 rr-rl=la bz ari*io bD -il="*li ;;rl-l; i, ii
l. ,2 "'_rl l" "2 ;rl l; iz
l; ;,1 l; ? il
[Taking co',mon
C3
￨:
Suppose to the elenonts of the rccond row, we
the elemelt of tbe third row ; then we say:
p tzrms rzspectiuely, the
b, c are different, shout that
1" "' . if-iJ
term of c3 ia the given determinant c.onsists
-
ar=￨::::￨::I:
elementa of tha
lo o2 o,-rl
16 52
"rl
Sirnilarly Operate 'C"+ mC1- nC2'
meals th.at to the elctnents of the third crlumn add
l:: t;tr: ::rH:l:: i:::1.l:, l; ::l-l::
Example z's. t1
Ds
７６２１
α α α
lo,
I
Operate X2 +p&r + g&s.
Further, if the elements of three parallel lines cottsist of m, n and,
determinan-ts can be exptessed as the sum of m x n x p dtteiminants.
For iostance,
l"r
we ahall employ the
・
物動︱
しい
・
1〈
cr
1", b; ;l
.
△=● 1+dl― el)● 263 α 362) ● 2+ご 2 ● り●1ら 3 α 301)+● 3+d3 eDta102 。
α2b3 α 3b2) CXalb3 ● 3bl)+C3Calら 2 α ψl)]
"1)
=に
lot Dr
rby μ co几〕
=Δ +0+0=Δ
Ob*This pruprty *:pety ueful for simplifiing dekrmra4D8. To add equi-multiples of parallcl iines,
Expandlng△ by C3,We have
.
o=
(Operate E2 -.81 aud 81 + J?s)
灘i:￨=0
警
fr tl121=。
∴ ∵‐
哺 :_JJI_
r ←
。
ヽ
Now exPand by Cl.∴ ―← +1)← +a(3ヒ +8-5)=O
Thus ヽ
κ=-1,-1,-2.
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+1)l・
+2)￨￨
or-3cI+1浄
°
針 181〓
+2)← +1)=0・
弯
．
ｒ一
ｄ
＋
＋
Ｉ一Ｃ
＋
ヽ
Ｉ一ら
＋
こ
Ｉ一●
ら
α
〓
Exnmple狙
2+c2+′ +ゎ
」
ca2+。
Hence x― α is
Put'eir,ga
=b,Et=RzaodhenceA=0.
０
０
０
０
１
０
１
０
The leadiog term iu A = o362c.
2+。 2+.2+d2+λ
Hence
.'.
a-b isafactorof
Itre corresponding term
dll=t・
a= (c -b)(o -c)(a-d)(b -c)(6 -dXc -d)
Erample 2.9. Factorize 6 =
λ
▲ 。
青ご
二
λ
Multiply Cl,C2,C3,C4by α,o,c,a respectivdy and di宙 de by abcd_Then
a.
on R.H.S. = hazbzc.
lt
I tl
'
laz 62 *l
bt "'l'
l"'
Putting a =b,Cr=Cz and hence A = 0.
.'. a
... (a
cttλ
- 6 is a factor
- 6Xb - cXc -
of A.
Similarly 6 - c and c - d ate also factors of A.
third degree factor ofA which itself is of the fifth degree
o) is a
as is judged
from the Ieading term 02c3.
." The remaining factor musi bc of the second degree. As A is symmetrical in o, b, c, the
remaining factor must, therefore, be of the form k (az + bz + c21 * l1b" + ca + ab).
ttλ
l
Takingらぁらご∞mmOn from 21,22,23 and R4 reSpectively,we get
A=(a-b)(b-cxc-@) lk(az +b2 *r21+11b" +co+db1l.
If ,t * 0, we shall get terms tike oa6, b4c etc. which do not occur in
'-
biλ
clλ
li:λ
7oF△
力 =1.
)
わ配
1:S a Fac`ο
)・
Similarly, a -c aird o - d are also factors of A.
Again ptttingb =c,R2 =Es and hence A =0. D -c is s factor of A.
"'
Sinilarly 6 -d and c -d are also facttirs of A.
Also A is ofthe sixth degree in a, b, c, d and therefore, there cannot be any other algebraic
factor of A.'. Suppose A =i(a - b)(a -c)(a -dl(b -c)(6 -dXc -d), where i is a numerical constant'
(1+:+:=:+:)
l:λ
“
.
ｃａ
α
IOperate C2 Cl,C3 Cl,C4 Cll
Eil][:liliillIII!l1 1:λ
Δ=α
r“ れ ●鷹 △ α″ ルれcだ。2S OF■ α dr″ ο′ αra′ た ′
iccJ″ 力ェ=c・ ″曖几← ―●
″△bec● コにおたみ′
“
".=“
::111
撃
e
lα
3oFadc″
ａら
“
dおぬ
庶t権ヒ露器器島瀧霊盤i糧選
棚歯1織戯:胤勝競胤澄
″
=島
″
２′２♂
Ｃ
・
l
2+biic2+二 +“
.
daた
oF●
Example 2・ 8 Facわ磁 CΔ =li
‐
ｌＦＪ″
１一
，●
一︱
＋
一α
＋
１■ 一Ｃ
ｌ
１
F。
a Factor of△
OL rFiPa凩 J助″ 蔵 s
み
::11
= abcd
θeレ ″
滋れ,た ″″陸 =α ,rルれ,_oお a racror ar△
"ェ
“
△=′ 0)
“ Let
∴ ′(a)=0・
Since
△=O whenェ =a,
J.● .● ―
α)iS a Factor ofFO).
１１１＋
ｄ
-1+b-1+C-1+d-1)ljl
.Factor Theorem.Ifぬ
rtes becoЙ
α ドざ１
．
・
ｄ
わ
０
こ
△
♂♂占♂
α
＋
♂・
・♂♂
＋
・
・〆♂♂
Ⅷ
Operate Rl+いら+R3+R4)and take Out the common徹ゎ r frOm Rll
【
=α oCd(1+α
3(α
￨l i : :l=λ
Let J be the given determinant. Taking a, b, c, d common fmm E5 82,.Rg, i?4 respectively,
we get
3,
UNEAマ ALCEttA I DETE'M:NANTS,MATRICES
、
Ъ
れ励
臓「lЪ ヨ
HlGHER ENGINEERINC MATHEMA■ CS
¨
38
.'.
ご11
*
6X b
-
c)(c
-
t
must be zero.
o)(bc + ca + eb\. The leading term in A = 62c3.
The corresponding term on R.H.S. = I b2c3.
Ilence A = (a - DXb - c)(c al(bc + ca + ab).
︲
置
珂
λ
Ψ‖づ
…
…
嚢劇
，
Operate cl+(c2+C3+Cl)and take Outthe common fadOr from Cll
〔
a = I(c
A. Hence
-
:. I
=l
・
o2
*
o2
bi'")2 (c+O2' b2 l=Wtr(c+6+c)3
c2 (a + b)21
| ,'
︱
l1a
I
Erample 2lro.Prouethc*l'-
︲
卜
の
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u N.r
IJ
,
tsss\
,'
40
同GHER EN●
,N〔
ERlNC MAIIEMパ lcs
uNEAn ATGEBRA
:
Let the glven de"minantbe A rwe putα =o,
21封
イメ
..
"
l.
Agan rwe puta+ら +c=o
2手
l
“
erefore,
kovc tf,c
み=2
folo'iq
l"--.
| %
■厚,1偽:.1)1睡 “
……
topcIate
cr +1
rcsolrr : (8 to
| 2b
;
lil)
l'Y'ol
lr 'u I
,* 11
l, c {I
I a4
cal
-# ln"oo,.'*
,-hu
-2o2 t-"'-El
d
(N. Bcngal, ]gag)
*ecal
rf *"a"1
C +da6l
d."ul
-,*-.
ど■た ,'99θ
C●れ
ｃｌ
あ
1, Ira+o+。 ⅢQ“ふ綸
:ヱ・
:=
lα
“
"藻
",4-●
.I",6b電ル■
SOIVe the em..nliキ
●199,)
“
②
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)
1 各￨:
。
c:ェ￨‐
● ｅ”
1・
7021●
fG)
♂ ‘１裁
♂
メ０１﹄
that ech erthe roll_ng det_nanta van8hea ll● 2):
ｌｃＪｃ
.
ｉｂ′
￨】
１・♂ ♂ ●１”
４
・ｎ
￢型
■.M●_t…
● =26c e+0キ cゞ
Pmblm● 22
1 11=2
I{ 5 6 !l
tDtPald by Asl
=226●
li,
|;;:l "liii
Frctcir. ..ch oftl! fotloiing d.nlrEbEt, : (lf b
+ら ,c)21t・ +c)● +a)―
llz
atlii
*l: f l1l i..n,e,.o-"
―
町
ca, c2 +
"'-rl
i i il
lrrerl ",^'"r*',
,,.]t'.;);u'
mmおmら md倒
:圭 :.1)l pperate亀
rl
%I
b-c-a 26 l-1a+6+c1t
2b c-n-bl
ro I z.
_・
l, .'
l" e
tRd*h', tssol
"'!'s a[ dif€t€ttt snd 15 61 6'-ll=o'th€Dshoralt'6c(5'+6+@5]'o+]+c'
"I ifl=liiil
Henee△ =2abc la+￨+c)3
0the■続 3● :operatlng Cl― C8and C2 。
3,We have
+ら +。
1"" b'
,.*.,-",rfi
:￨=27ん
α 211i三
ョ
6, c
la
D cI
llic "i" ".rl = -1o-6xb-cx.-dxo+6+c).
?l
l1 2 s .l
=c_1,then
ic
srrow rrret
s' Ifd'
Or n“ tぬ
鍬l瑞1霊ずhら acぬ e… hhg亀 “
。r54=27た
ll
il="-."-r*-,
tl
“¨
Л
コ¶﹃州︲ ¨
To dete‐ Ineた ,put c=ら
i:
られ９
Thus△ =極 lc c+0+cゞ
1l
∫′♂♂ “枷打
い。
鑑冨臨
lr I n
sh"*,hdli
a!
メ♂Ｆ♂ (剰
MATRICES
lo o r
∴ c is a Factor of△ SImlarlyみ and c are its Factorg
_ぽ
DEITRMIMNTS
‐
隣転111
‐
tdた rtl'999iル
υC.I"Ol
１
・
１
︰
！
︲
︲
Ｊ
ｎら
００１嘲
22.12To 21o 2cio
型２．¨ｃ
ぃ
一
］
ド鋤
〒
こ
一
［
︶
範２
・
︼
一
２
鴫
一
一
・
一
´
一
´
Щ 政
・
一
一１
３
３
鉾
銅
﹁
﹃
紳
﹁
2a1liエ
Hence the
rule for the productr of two detemlnants is
biclこ
4abc
クし
輌ら 991)
"出・
x!-nnpre 2 rr . Eudtuatel:i:
By the rule or
where
eご
dp=
(a2 +
LzX-c) +
dgt = 0,
1ab
dB2 = 0,
d,$ =
メ
△
〓
Δ
△ ０
００
Ｏ
r\fr. "'l
〓
u"
△ ００
i)
３８
凸Ｃ
２Ｃ
３
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lir^r*"
I
＋＋＋
の
＋
２
‐´
。
as tiesqilanof odrrrr*inolr,
I
βメメ
＋＋＋
喘嗅噛
凸疵働
.
＋
“
亀Ｑ﹃
ＱＱＱ
＋＋＋
■島島
＋＋＋
α α α
占九疵
凸凸凸
〓
２角
へＡ
△
′ 3凛 2)]=△ lx∼
︲
Ｃ．０Ｃ．
1(′ 2れ 3
ｍ
′ 322)+●
＋＋＋
"1+′
where A" B etc. are the co-factors
(Rα ″Cたら1989)
l2bc-az A
bz
cz Zca_bz o2
Erample2.ls.Erpressl
3凛 122 ′ lπ 822 ′ 2れ 123 ′
3● 2211
れ 322)― れ1(′ 2● 3
)3.
Hence A'=A2.
l・
=△ l yl(″ 2● 3
2)
lc2+λ
Ob* A'is called the reciprocal ar e{fugete detetaiaent o/d"
271V3)=― ′
2● 120・ △1;(υ 372Wl)=― ′
3″ 2ar Δ
△=△ 1『 1″ 2● 3+′
2れ
１
８
０乾Ｃ
Thus
:‖
β ノ β
ｂｂｂ
＋＋＋
1・
〓
Ｍ
″ 3al.△ 1;(銑 vIW2)=′ 3″ 122・ △
μ メニ
ａ α α
3島
￨ヨ
α α α
l vanlsh.The rmaiingsa detemnants are:
荘解"2撥
。
2+。
。
ら
げ
袴
軍ぷlLi"北 li
=0
〓
△
￨ィ
+1,2\,dzs=g.
+ b2 + c2 + lu2r.
△=￨な
Ｌ場﹄
1"2れ 3優
when i = l,J =,t 2, we
=
Ｌｎ
Ъ
吻
=厳
筋装 i‖
σ
l%b=犀
メ・
i醗璃μ
￨ョ
I
I
+b2 +c2 +}u2,
2)ん
_ =が .2+ら 2+c2+′
Similarly(●
-:l
g
=
^laz
Tc2+λ
A18001L均 =1撃
l_l i
-6f)(-6)=l{c2
+ cL)L+ (ca _ 6l.p
+ b2 +c2
°2+ら
HenCe
Similarly(吻 73Wl)=′
;:ll,
dre = o'
dzt=cj,du=llaz
1n thls wり suCh 21 dete取
mants Out of27 vぼ
tr
*# "*ruol*ff *., ;:*1,J, :*#r,,
dlrr=(a2 +t'2)\+(ab+cl,)c+(cc
■ の﹃
２﹃
﹃れ
歯ぷ拙鑑識瑠出retem富降常篤瞥席F躍鞭臨
rr,r,,
m
"*, ;r:"^l
ld, dp drsl
t=ldn du drrl
lo, dez ,rrl
2 4 MULllPuCAT:ON OFDE■ 閣 :NANIS
Qち
:
cll l\ ry nl
+b1n1 +crrtr, diz+brm2+crn2, a1l3+61a23+c1n3l
fc1
fc1l1
b2
c2lxll2 *2
la2
"rl=l"ji*ogrr, *"r"r, aztz+bzmz+czn , "ilr-b;;;";;;,;il
f"a 6a csf lts -s "31 l"li*orî*"i"i',
"sq2*o;*;*r;";:,, ;il;-;;;;_;:;:l
D1
44
MATHEMA■ cs
側綴督綴縄驚[I二 i盤
raL0 0ut(-1)common l
for, . or, ...ou ..."r.l
lo4 azz ..azj ..."u1
""
Thus.{.=I ""' qiz
...aij ..."*
| ",,
t..-....-l
Lo-, amy anj ..."*l
I
:=』
I
2
: il=△
ie a matrix of otdzr mn. It has tn rows atd n eolurnns. Each of the nzn numbers is called an
element of the matrir,
To locatc a4t particulnr element of a matrdg the elements are dznoted, by a tetter
folloued by
ttoo sufiies whic"h respectively speci$ the rowg and the columns. Thus og is the element in the
_03+ら 3+cs_3abc)
Hence the given deteminant=A2=(α 3+。 3+c8_3●
i-th row andj-ft cilumn ofA Ir rhis notation, the matrirA is denoted Ui t".;1.
A matrix should be tnand as a si.agle enti$ luith a numbr of ompotunlq rather^than
lc)2.
a collectioo of
numbera' I'ot' erqhPlc, the co-ordiratss of a point ia aolid geometry, arc given by a set of tlrree numbers
which can be-represented by the matrix L;r,y,il. Uutite . dit"r-io*t,
irrui* irroot *auce to a siDgle
auuber aad the quec6* o163,liag the value ofa oatrir uever arises. The" difiereace betweeo a dgtarminalt
and a matri: is brought out by the fect thet an intarchange of rowg and columns does not altcr the
determinant but gives an eotirely difiereut matrir.
PrOblem8 2・ 3
1.勁nd the value ofthe dete… taゅ
Where A=￨￨
I
,=l-,
; , -,1
0
Ir,
一一一
(1
)
Dellnition - A
system gf
装装
剰……
le detenninant
考1
SI
A=[
]口州
幸i]
if
-. Thediagonal of this matrir containing the elements 1, 3, 5 is called the leading or principal
d,i.agonal. The sum of the diagonal elements of a square matrjr, A is called the tr,.ie oi a
A square matrlr is said fo 6e singular if its determinant is zero othenoise non-singular.
Diagonal matrir. A square matrix all of whose elements except those in the lead.ing di.agonol,
ore zero is called a diagonal matris.
A diagonal matrix whose all the leading diagonal elements are equal is called a scalar matrix.
For example,
[s oolol aua lsool
lo -2
lo a ol
Lo 061 Lo03i
br*i;;; ll ;':;;;
are the diagonal and scalar matrices respectively.
Unit matrir. A d.i4onal marit of ord.er n uhich hos
" "i"Sl"
an m byn matrix ; which is written as
Lpital letter.
unity for all i* d,iagonol elements, is
called a unit matrix or an identitlt matri* of ordzr n and is denoted by rn. For example, unit
ぁ
︱
UV
lz1
I
square matrir. A nwtri* ttovi.rtg n mws and n rrilurntrs b calld, d quan matrb of order n.
The determinsnt having the so-e elements as the square matrixA is called thedete'rrtrinant
of tha matri* and is denoted by the symbol I A For example,
l.
￨12=4●
mn nutnbers orranged. in a rectangurar
formation arong m rows
m x n mdtrii.. A matrix is also denoted
matria e.g.,
Row and column matrices are sometimes ccll ed row oectors tnd, mlumn uectors-
262c2
25.MATRICES
and, n columns and boundcd. by-th"
rout
Lsl
′
臨 …い、
静
琳挙￢
…
〇０に
I:f
A matrit. haoing a single column is called. a column matri.x, ?.c.,
ががが
枕 FactonZel[:II
35n.
４
antrレ
●1+
Row and eolumn metrices. A matri* hauing a single rcut is called a
３
l・
1l withoutevaluatiogAaudBindepeodeotly,
=‖
1+Cl
戴慟
e…mtl替
4… the d―
11
rl
÷割
2♂
li l
(2) Special matrioes
２
3.PrOve th
￨￨,
′
みぬ “
2.ShOv7 that
_:
rMこ 3A2+"+32
matrix oforder 3 ig
０
０
０
１
０
０
０
１
www.Engg-Know.com
１
￨
46I
HiGH€R ENGINEERING MAIIEMAIrcS
-
denoted by
lI
.€.,
;
thc eleirzr.ts
d
a
is
a ourr
-,t
i,.
called.
a null
or zero maarix a,,d is
DreDple
tu thot all thz tedding diagonol
ehments
34=[1:
an .etu, th.d the natri,
corrforEAblc.
i.c.
(A+B)_C-A+(B_L\=B+(A_A.
i3) Multlpucetiotr of rEstrix by
natri,
uhose each elenent is h
rhus
The d,istributive
a
r ,calar. Thz prcd,uct of d matrit A b! a scatot h is a
fimzsihe conesponding etentnti of A.
.rl-fr"r tbr rcr']
frr ar
oz cz)
hb2 hc2l
laz
law holds for such prcducts , i.e. h(A
+ B.t =
hA + hB
=lczt
l.--...t
lt
lc^r c^2 ...
is defined as the m
xp matrix
"rpl
czp
I
cdpJ
\rYherecr'=o bU+oip2j+ai363j+..+oiab^j,i.,.,theelementintheith'owandthojthcolumn
ofthe matrix r-B is obtained by weeving the ith row ofA withJth column of8. The expression
for cii is koowD as the inur prod.ucl oflhc ith rcw \pjth tbeJth column.
Post-tuultiplication and Pre-nultipli.dlion. In the product A8, the mdtrix A is said to be
p*t-multiptied by the nat,rix B. Whereas i.o .B4, the matrk A i. sstd to be pn'tuultiplied by
B. Io oae case thi product Inay €ist aad in the other case it may not. A.lso the ploducts iD both
cases eay exist yet aay or oay not be equal.
In partilular, ifA b€ a squat€ truEir ofthe same ord€. !s I
rheD
and
ob.. L If AB -
Lkaz
...
f"r, crz
czz ...
呵判引
Addition snd subtractior ofilAtlices is arao.rariur.
′´”“
d.
ν ″ ¨″
lubttact
十
qtutl;tioe,
十
cot
＋
3_
is
A+E-B+A
＋
i.e
肩 ″ 滉 π
as] lcr+c, ar*arl
Ouly mairices ofthe slme ord€r cai he ad&d or
１ ” ¨“
１
1.
2. Addition of matrics3
〓
AB
d1l-]c1-c1 b1 -dl
["r 6rl_f"r
bz)
Ld2
lcz a2|1- la.r-c2 t2_a2)
Ｂ
arl
Similarly A - a ie definzd es a ntatrir whace ehmcnts are obtoined,
by subtrd.ting the
ofB from the corresponding elements ol A_
Ob.
xp coafonnable rDatrices, then theiI product
6, +7,1
elements
rLus
In general,lFA=[i
￨
減
,
を
＋＋十＋
[￨￨:lil:麟
％
％％¨
[cr
ル 2・
・・
be two m x a aad n
bzl+lc2 a"l)""*", t),
laz ar.l
1ar
“
嘘
勉勉¨
Thus
B.
f,r 6ri [cr d,l [a,.6,
`り Ⅲ
if
agd lii) edch elernent ofA is equal to thz correspondin€ elen2nt ol B.
. (2)Additiotr o.d subtracuon ofEtatrice* IfA, B b. tuo motrbes of rhe e(.me order, then
thcir sum A + B is d.efinad a.s the ma,ril: each elemeat of uhich
b lh' sum of the coffesPo^ding
etementt ofA .,nd
ＬＦ
ｒゴＪ ′
α α α ‘
お
Ot MArnrcEs
Tuo ,natri.er A and. B drz said to equat if and onlf
6\ the! drc ofthe sanp ord.er
be mulripli.ed onl when thc numbcr of
the second- Str,.h matrices dre said to be
￨
are upp€r and lowe! trian8ular matrice3 respectively.
EOUAUIY
it
the fiftt is equsl to lhe numb$ of mus
ls n c1 and 11 '.ri'-dl
s....ol
fl
12_5
1l
[o_. 9_,c]
Ll
(r)
一
il
llo.-O
2.6.
1司
MultipucsaioD ol E,s,ttlr.o& Tuo mottices can
columhs
square matrir att of whooe elemelaLs belols the leading diagonal
are
-^-.TTfi:.lT_1"-!O..A
zem
rs ceaed an uppet triongvhr f^4trrr. A square matrix
all of whose elemeats alrove the
leading diagona.l are zero, ia @,lled a tabe. ,riangulzr matrk
Tltus
:
〓
(4)
&生 -4B=[1:二
３可
ｄ引︲
。
lq t, el
Io o -gl
Fl and l_+ ^
la 6
fl
,)
Lz f
lc { oj
i
二
a Bheu-ilrnmztic ,narrE. E:aDples of syBmetdc and skew-s,,lBmetric
matrices are
respectively.
．
８．
．
c@lled,
o* j
1 1:だ
We have
一
is
o1, t
j,
０
If at, = - o, p"
for all i a^d
ム =『
said to be
48, u)here
０
a6 = ay
-
〓
ulun
2.14. Euolrdte 3A
降Ｆ
SyEEotric eEd ukew{yliDetric eabices..,l squ.lr' l,l,L,:ir A=Iai} is
symmetric
Obs. All the ln*s of ordiaary a\ebro Lold for the additioa or .ubua*ioq of mstncee and their
multiplicatioa by s*larr,
口Ｐ
9lr
i
are zeftr,
MNpCtS
一２３
一
２１
・
[o o o
looo"l
natri,
tlNEAR AtCESPA:DETERMlNANlヽ
日粋
fo
N.,I daft&
IA=N=A
OA=AO=O,whercO
O,
it
^=[l ]]*.r=[i
www.Engg-Know.com
det
not
êet
-l]*,e"
dil!
i5 a
null merirL
imPl! that A or B is a Lull t@tri,. For inslance.
tl," n,.a""tea is
snu
matlix, artho,r8h ncitherA
il
no'8 is a nurl
UNEAR ALGEmA :
DdRMNAtflS.
rvtATmcEs
=隅l'翻郡 ]fiキ
al ttr=Ac3o,″
μttre21aれ “
r■
Letム
=〔
ル
4島 C"鷹麟∞
ゎれル畿
り bo ofOder"xれ ,3=lbJ beOfOmera xP and c="が
‐en43=LJ Ibl.l=Σ
0
be Oftter of′ x9,
ttb″
1・ 1
r=lilttb21・ ICrJ=lril二
∴
γ
鼈場
lγ l=￨二
鉤dhし , BC=ll」・り =Σ
′・〕‖ α
彙
00=し
qII=IΔ
￨`:lら
we have
１
m=[i:lil:￨:ま
l=￨'ユ
IJ:lattb‖
ε
ヶ
‖
i勢
工
…
豚
erirt Firrdre'rdy.
4_lLぬ
山
?
h man叩3T[」
11=F珂
紹
工y]
古
にこニユ,19"拗
col--.. ltrtrirB huy rovr and ll-y
columns,
IrA+B=F f]-a r-a=[ 1]*","*"product,{.8.
BodrAgand&{
(LM.I.E, fign
ｌ
e
WhiCh 8蒟
二
・ [21"
]=[I 瑚
'L llatrir.{ harr rorr aadr+5
降
５
可
１
〓〓
２
９
ａ ■
¶
７
３
ｌ
４
ｎ
鳳
覇
７
=￨:￨￨￨:￨: :I11111: ￨:illi::￨=￨:i
η
”
ユ
ｘ =障轟 =[:
』
瑚
C・
o。
L b whatValu_ご亀缶
3[:
ニ
Ｍ
Ｌ︲刊創
刊
５
+III=Q″ ル″ A=[ :
劉
﹁
３０２月
刊劇
２
０
i : 4andB=[ ￨ ]働
等『
3‐
２
４
３０２
２■
間に︲
︲け
３
ゴ
１
０
＝
ｌ
ｌ
１
＋
一
０
４
３
２
５
３０２
１
４
一
１
１
６
２
０
５
７
３
８
■﹂出ビ
＋﹄“Ｌ︲
一
一鉱〓
“
２
３・
ＡＡ
Ａ
:7.P70“
め″ A3_412_麟
Ｊ到
Ex3mple 2・
刊
可可
EvidentlyJお ≠m.
ttr=郎
i驚二
tJttJ・
一
l _il:::::::￨￨ .i:￨:￨:i::::￨=￨:]
Hence
￨,16″
￨￨
Problem8 2・ 4
Agdn considttg the rows OfB and cOlumns Of`ヽ
:
司
「 ::i:::::::]=￨三
９２２
AB=￨:￨::::il:: ￨:liiL_:
︱
Conside五 ng rOws OfA and cOlumns OfB,we have
蔵らル
′
blJ%
′=1
.
riα
lhaluateoLザ
]and
σ
ゃ
[11野
Ｊ
Ｌ
¨
・
コ
．
t31
www.Engg-Know.com
t51
]Verit that“ F=■ 90
0‖
0)日 X“ 5刻 x賤lxB
dA3or m,w轟
a
1_司
.
mdAO● 0=“ +κ
XttX膨
手
/21shwthatr+A=ば
α
―
詢 :驚』
し
2 鴇
lL IA=ド〕
lB力 ●
F● み,991)
[∬
],3bW that A2_m+7=o,where rお
dfLぬ e
1《
unlt m・慮
JOrderぃ d“
卜
1&FA=[i llB=[1:1恥
Ⅲ 饉
S」 t“
"A2,M+鉱
+m+湖
.
+A'tisasymmetricroatrix,
matrix.
of a symmctric and s}cw-aymmciric
andashew-rymmetritmatrir
呵ｑｑ
一
ｌ０６
５
０
０
ｔ
３
０
一
５１
０．一
＋
︲
一
；
二
ｑ釧﹁
１一
一
４
凛
た
た
λ
α昭
α
はんむれ
ぶlttJ器島勝〕
掘賠誕
」
γ競朋ml協縦
５
27.RELATED MATRiCES
rua
d symmetric
５３３
４
３
１
一
“
＋
＋
Ａ
〓
Ａ
tHanflar mtr量
tlre
〓
A+A′ =l_:
fllli鴛
u
Ａ
Whe“ ιぉ b“ rt五 m四職
一
hb・ ebrmι tt
ｄ
m.五 XA=[1」
Ａ
“
Then
ton“ 鮨
r=11
h"ri
n ttS a"J優
14W“
=f,a
０１２
盟 1:澱
We haVe
]
”
︲
ホ
n
Ｆトレ
´bWぬ訛 A・
1ぜ ‰
一二
五 =[:
押
l.二
司
r.e.
Eranple 2.lg,Erprcsstlla motrirAasthe sumof
胡剛劇﹁４﹁
︲
︲
・
ヮ
刊・
﹁Ｕ
．
リゴ
ｄ
硼引
﹁
・許
・
:澱
“
et =a
A'=(P+QY=Y+q=P-Q
ard e=;6-A')
whore
thtt fA=[11{司ぬ mA″ =[1蹴飯
+
-rl
P=i6+$
３“ ８６“
HfA=障
Tlren
firug
２３０．
By Ma血口MId lnduc● ふ ,pmt
ly' wl = - tt = - c ie. c = - ttir a Ekew-symmetric
la
!t'
" = [;" - -,j = -
r$=f;u'..t t'y=!l'
whidr showr thet theg ir osa md oaly ouc way of erpraocing:{
metrix.
+メ
…Ii‖ …[11,
1ユ
As.in
8'=[}<a
that.A=P+Q.
“
+め 2=″
II
B=f(A+A')aod"-iU-^',
HeoceA can be e:gresaed as ttre sum of a symmekic and a s}ew'slmmetric matrix.
To prove the uoiqrDnessr essurne that P is a symrnetric matrir and Q is a rkcw-symmetric matrix such
a unit matt Jsemnd oder
Miユ ,I"θ 〕
12 rA=降
5l
kt
,.
(RCaCあらf990)
la rA=降
MAIRICES
UNEAR AI-GEBRA: DEtERilhlAM$
ｄｊ
ｍ０
ｍ
¨．
〇
ＯＬ
団
一
ｎ
ｉ
Ｓ
“ Ｏ
ｑｊ
句血
嗜耐
9. Prove that the prcduct oftwo matrices
HGHEN ENOINEERING MA]HEMATICS
騰
50
―
￨￨
2)」鮨増oint Of a square Hna価ヒ.The determinant ofthe square matr破
〈
"が
ThuS the tranSpOSed matriX OfA=￨;
:liSA'=[1
1
A=li:
乱 ξ器
′
Ψ
農躍粘
P"ぬ α
ヴaC a“
Xれ
m壺破.」 ∞
the transposoごぬ e
ぉぉЙ
●
ttα げ
ねrr"■ ,″ “
た
●
カル
ρ
"d′
`ね
LB〕
恥
.屁
1>2 Eι
甲
″
斡
“
"ぬ
Caみ
"●
■● ″ exPres・
“
△
=￨::
￨:
:￨￨
9ed os_“
forraed by the cofactorc of the elements in A ia
[At 81 crl
le,
42
Lo, Bs
Lcr
cz
le, 82
″濯蹴 rtt― _ル
“FoT,the elelnent in the fth row andJuh COl.Ofク ′
忠
nけ pttLttCd驚 :蹴￡
席勝ll=鵠喘脱嚇撚R総_議 ′
商
::l iS
"げ
ht■ L the」 v曖 3寧 2К tta鍮 ,the■ ■‐
,・ 1,:・ ―A与
:●
.
αり
""α
ガCα ″d α slew‐ w凛凛肩 c
C2l. Then the transpose of this matrix, i-e.
"rl
is caIed tbedjoint of the matri*Aaad is writt€u asAd'i.A.
Thua
tte adioint ofA
k
tl:a
troinspo*l matrit of cofactors
l81
Bz
島島Ｑ
lbe matrix
,aI戦譜繊 lttT鵬
￨:
:]
of A-
好
礁善鮮網撒撃幾輌 :輔鏡獲
"02‐
SfP3g“ raな ―
www.Engg-Know.com
54
LiNEAR ALCEBRA:DETEPMINAMS MATRC[S
HiCH[R ENGINEER,NG MATHEMATICS
=￨::
亀乾﹃
nant OfCOeFlcientS be△
Ｌ場﹄
rthe d…
hmple2・ 21.Sor“ ′ル
lL+y+2z=32r-3y― z=-3x+2y+2=`
眈 MIE.,I'9θ 〕
-3+a「
]=Щ
Ц11?=8
Ex“
ng b)剣
1●
d by Cll
倒鋼
↑
一一４
＋
１一△
〓
一
一
一
０２
３一
2)+
―
―
ll=-1
:
〇︲
２引＝＝
１一△ １〓
一
一〓
″
・
Hence
11 。
1-1叩
鋼… 尋[{
==1,y=2,z=-1.
No麟腱The use oFCraner8 rule mЧ 己ves a lot oflabour when the nunber ofettatlons exceeds fOur_In
such and other ca305,,9 nummd methOdS glvenln o 22・ 4b22 6 are preFerable.
＞
一
〓
一
〓
Ｃ
Ｑ
〓
Ｃ
〓
Ｂ
● Ｂ
△=alAl+α 242+・ 3A3
・
Al=-1,■ 2=3,■ 8=5:31=
ゆ戸
ｌ
３
Ｃ ● Ｃ
〓
２１１
・
‰
﹄
，
機動
斜
＾
縞
昭﹄ ■ ２
〓
Δ
“
Ｌ
１
sr
こ″S● ル ″あ′MC″た●
=あ
３
﹃の。
１
ｗ
・
Ｏ
≠ １
﹂場鈍﹄場﹄・
詢
倒 ω ｊ ω 珂一一〇
・
﹂ ¨
・
け．
Ａ
♯
ｌつ﹃山
ｃ
Ｌ場鈍祠
髪
２” ﹂場蝿
鈍Ｃ
ｗ α
ｌ
３
■場﹄ｏ
％
ｌ
ｏ
２
２
﹃．
ｑα
﹃α
・
艇ｈ﹂ばｒ
一
一一
一
′
・
・
¨
・
″
ν︶
Ｚヽ
・的 ″
詢
ｍ
ｂ
一
ｉ
Ｓ
Ｍ
¨
ｄ
ｈ
専︶十１計一
カ動力Ｌらヽ４あぬ﹂場鈍︶
=lili l :il10peratecl+yC2+ZCJ
呼り呼４亀亀３﹃Ｑ﹃ヽ
・％・
“
=1-:―
３
降］
︲
Here△
then
zarあぉ
滋“
s.
￠″凛α′
"7‐
たmj2● 2rs
(J)SOr“ lio2 by」
・
:2● 2歯
by rap dセ
3.
X[]
[￨=:￨:1
窪
s
-l-r
=11-s
tLi r
-;
Hencer=\,y=2,2=-I.
ヨ
sl I sl .l-s-g+zol ttlzl
zlrl-sl=*l
-e-3+281=l
Efzr*rs-++l
-rr.] L;J
L.rJ
Problems 26
Solve ttre followiag equations with the help of dcterminants (1 to 4)
.r+y +z = 3 iz + 2y + 3z = 4 ;* + 4y + 9z=6.
:
(ル ra′ 。,199J)
●,199θ
●亀らご力′
t.
2.
r
+3y
+62=2; & -Y +&
=
9; x-4Y +2.2=7.
+3a=152.
- 3. :+jr +z-6.6ix-!+z=22;t+2y
1. ?zs ty = e8', fztx = ea ; *tza - L.
6. Zvu - wu + uv = Suvw ; Xow + fu u + *to = lguow ; Gttut + lwu' uo =
Solve the following system ofequations by matrir method (6 ta 8) :
2y+z = 2 ; x + Y + z = O6. 2:-+ 5y+ 3u = I ;
7. x+y +z=3;r+2y +k=4,2r+3y + tb=7.
-r+
.
'
E. 3r-y+z=6 ;* - y + 2z =l ;fu3 -Y +z = 4.
9. r - 3y - 8a + 10 = 0 ; 3r + y - 4 = 0 ; 2r + 5v + 6z = 13'
―
D
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)
17 uuw'
I■ ● 2000)
(ェ
tA M_■ ユ,1997)
lAnd■ rc,1998)
rad“3,I"θ
(■
)
爛︲
，
57
lo.Jyz=Sz'=ifz=2.
rr.
By
6adrsr-1,
6crnt4 tggls)
sorve the riaear equatioa
AX= B,
*h*"
e
=[l
i
j],
"= [i]
*.,
― 」
li :
=
(aM',E'
:997 w\
r獄
靱
dttl胃
淵
脱
躍
おmd
byttS2“
輩
Υ
∴ pの
:
(ガ
(Bhopal,
=2 Hene therank● fthe givm matrix IB 2_
)Given ma鐵
l99ll
‐
―
"idifr;;;;oith"
l:]」
matrirA, dereting all the other rowa and
i".*t" i. Jr"a' tn" i"r, e of order
"r difereat
sa-" order, got by deteting
Ю penting
than
i
aiaor
of order r-,
its
rr* is;,.
the
marit
allthe2udorderminors,""lrll
orderr* I ar.
its ooii" < r.
ratrir.d indl be d€,Boted by""-,
p(rt).
H"-*t8ry trelsf,ormatioa of a mrtrrr. rhe foltowing operatious,
-^.-12.)
three of which
€ter to rows and three to columna are Laowa
u
t*rcformations :
"l"iitory
I. The intenhange of any tum rouc (columu).
II. The multiplication of ony rcw (column) by o *n_n*
IrI' The addition of a o,nstant murtipie if thc
rutmberof any
ekmcn*
iciinl.
Consider the
o lJ
matrir A
lo I
=
a unit
ol
lo o
rl''
Lo,
il
b1r
lot 61 "rl
I
az bz
czl
６
６
ｂ
１
１・
１﹁電
ｂ
〓
ら
陸Ｆ
わ
Ｘ
１
１・
劇﹁可
6s "rl
ＬＩ″
にＦ
f"r
ｄ利劇判
００
ｌ
Ｆ” Ｐ
Ｘ
〓
Ａ
ｎ
)lill
”
Ｒ
tiiolrriiirr*-T'no
rurrh of the foltowing matricec:
Opat r"1)(・
ll=-r*o.
ofA is zero. But, of
:. p(A\=2-
(6) Theorern. Etementary mw (column) transformations of a matrit A can be
obtained
pre-mukiplying (pst-muttiptying) A by tlu arrusponding elemcntaqr matrices.
■
ＩＰ ′
Ｆ
:40カ
AIso wery 3rd order miaor
ltool
lrool
l-rool
lrool
r olarea2r=lo
0 1l=cB;tR2=lo
-- I ol;R1+pa2=lo
-' i 0l
13=10
l_o
to
?"tnr-g"*
Eraraple 222. Dacrminz the
10pmting C8+302,C4+C21
i.s zero.
Ilence the rao&. of the given makix ie 2.
(4) Elemeotary uatrleee. An elementary ntdtix ia that, which
Ls obtaincd finm
mdtrit, by subjecti.ng it to dny of thz elcnzntany trunsfurmations.
trl-e'loles of elementar5r matricea obtained from
mut (corumn)
the
corresponding elements of aryr otlwirow
Notrtioa.Thcc&rncalcrymutmn
ftrmatroaswillbcdGootedbythefollowingsymbols:
(d) rRg.for the intercbenge
of thc ith and..;ith mwa.
(n) lP; fc mulhplicatioa
ofthc dth row by t.
(tu) P; +pE; for rdditim
to thc ith row,p ri6e3 tlhclh rw.
lihe corrcrpoadiug column hansfornaEon will
be deooted by writing C iD place ofg.
do rur char4c citrur ttu ordt, or rr,,tlc
..of a natrit whilc the varue of the
urors may 8Et
'hrng€d by the traarformati* r ,"a rr, ttt"lriro c non-zcro clraracter ruaaiDs unrfrectod(3) Equivdcatnahir?to-oma trbesAandB are saidtobcequiearcnt
ifonecan&.d,t*ir*d
tlu-otl*r
-Y order and thea *quence of
equivalent rnatrices have the
rme
sane ranL Ite synbof _ is-u#Tir eqr.i"alenoe.
*^*i,l
l: i l ]=A・
Obviously, the 4th order minor ofA
of a
y
1,24 RJ
ay)
lill
of
-
10perattngR3 ■
IOperating23 m2,P4 R21
uanUhe"_
9d"ny' ilu mnh of a matrix b ttu tauest ordcr of an' twn-oatishing mi.rwr of
If a matrir har anon-z:m
If all aiaorg of a mafir
l1111
C3 Cl,C4 Cll
―
Dsl. A mafriz b soid, to be of ml& r wlun
(i) ir ias et Least one rwn-zel
minor of order r,
atd (ii) euery mbur of ord,er highcr
llyry
8り
MAIPIX
If we select any r ro,s an!1 colrrr-s from any
columns, then the determinaal formed
urtn-"." , ,'.
r' clearlv there rin be a ounber
rorg and columns &om the sane makir.
lhc ralL
80 that the givenコ腱雌
il=A〈
組
Usiag the loop current m€thod o a cirarit, the folloriog
equrtiotu are obtaiaed
7i1- 4i2= L/,- 4ir + Uri2 &r = 0, &1+ l&1 0.
=
By uakir method, colve fc i1, i2 aad i3.
2.9. (I) RANKOFA
)Operate 22 21 and島 _2に 1
[_i]
12 Ia a givea electricel rctu;or\ the equations for the curreltr
ir, d2, r3 arc
3rl + 4 +is = 8 ; 2dl - 3r2- zdr _ B ?i1 +
2r2_ Ei3 = e.
=
;
Calculsh jl .-d is by Crame/a nde
l&
(′
so a pre-multiplicatioa by 8zr har ioterchanged the 2nd and 3rd rowg of A. similarly
pre-multiplication by AE2 will multiply the 2Dd row of A by t and pre.multiplication by
81 +p82 rill result in the addition ofp times tbe Zud lw ofA to its lst row.
Thus the pre-multiplicatiou of A by itementary matiices results in the corresponding
elemeatarSr row ft'an*frnpaf,i6s ofA. It can easily be seen that poet-multiplication will perfbrm
は Mニユ ,1997騎
the elemeatary column transformations.
0
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THEMATiCs
型聖堅旦￡里壼聖⊆堕ヽ
肌雷ぷ霧獄 L糖 :雷「lα tt°肌∬窃1脇脚
競解鰤制胸舞 α
け席熾
糧継躍
R`島 _1_¨
2J島
: ∴ ‐
0r
_電
T総
du“
“
Aゎ rresult iOm FOm雌メ面。n
1=鳳 -1
(D
_1… ¨鸞メ lI=4 1
t '.'
Hence the result.
AA-r
=I
{t , the two matrices A and, I are written side by sidz dnd, the sanrc
row trunsformations are peformed on both.
A"-;; csr4, rs reduced. to I, the other
matrit
represents A-!,
Eranple
2'z,8, usins the GaussJordan
methd,, find the
'
inuerse of the matr,x
Writing the same matrix side by side with the
unit matrix of order
Ir 1 3: 1 o
[i -: i; 3 ; !j
in eumpl.e
,s. pater,r996 s)
S, we have
o-]
Euery non-zeto matrit A of mnk r, can be red.uced by a
tmnsformatians, ta the form
ft 3l*,*orlr.cnormalfornof,rl
C,ot. L. The rur* of a notrir A is r if ad only if it an be rcdued, to the normal
form (i\.
Sbwe eachelcnuntary trawfurmatiancan beaffecud,byprc-multipliutionorposr-mu|liplication
with a. suitable ctemenbry matri.r and eeh ehmentaty n"t it * non-siigular, thcidore, we-haw thc
fullowiag rcult:
Corntpndiag to eu! motn, A of mnh r, thzre erist non-tingular matrices P tnd Q such that PAQ
equab (il
IfAbeanrxnmatrir,thenPandQareequarema?ricesofordersmandzrespectively.
F-n-ple 2.24, fuduce t[c followinA matri*_into its rcrmal fonn and, hence find its mnk.
曜
ｒ
Ⅸ
Ⅸ
ド
０
ｐ
一２
ｌ
Ｆ
１
ｍド
４９
ｐ
ｏ
ｌ
ｏ
ｌ
ｏ
ｌ
０５４０
ｌ
Ｆ
１一２
１一２
０１一２
１一２
一２
２
れ
０
は
Ⅸ
ｒ
・ほ
ｒ
１４００
Ⅸ
１０
Ｘ
Ｉ
Ｘ
︲
ｖ
０
Ｉ
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２３９２
・
は
ｒ
０１一２
１
” ” ヽヽ一ヽ
０
３一２
一
[Operate Rl+3R,,22 :R3and(― :)221
２１３０
曜
０
１一２
(Operate■ 1_22 and R3+22)
一一〇５
＝
”
１
A
４１２７
４７０７
︲
・¨・¨ ・・・鯛樹可
劇司叫９引司
劇司呵
０
fe
(Operate:22 and:R3)
．０３９２〇６９００６鍛０
．．
．
ｏ３９０一
・
-rl
12 3 --1
r,
i
-rrl
^=l', 3
o-zl
(operateE2-81andRs+2R1)
...(it
Cot.2.
０
０１００
︰
．６１
一２
１
■
-1
１１２１
・
３ ■２３
１２■ １
１ほ Ⅳ ｏＪ
１００
１
ｌ・
1
Nomd fom of e nstrir.
sequence of elementary
For practical evaruation of
2'2o-
Hence tbe irverse of tbe given makix ig
Rrr=′
_1¨ … 222JA
R凸
ぬ
L4.sscrn 1999
;
Mysore, t997i
lby R12
町
R2 2bR3 3Rl,■ 4 6RF
iby C2+Cl,C3+2Cl,C4+401
1by R4 22 R3
rby R2 R3
iby R3 a2
Problen3 2・
け
卜
２
Ｆ
卿
７
６６ ■６
・
４
３８８
η
”
磁山町 1990S
一
Ｑ
認
町
harrc
6. Ure GaugJordan me&od to fiod the inverse
t *l
００
胎
lz 1-rl
(Di6
"L; i il <caûbqrgsn
-s
4l
lo -1
,J
t}e saruc anl
of the
follwilg natricer:
[e431
(n)12 I 1l
lt
-
3.
r!-rmpre2.25.
ind, aon-cittgtthr llno4ri@ p ard e anh
tlut
０
１
０
可
ｑｑ可
1990
e rudl that
PAQ = f, where
:
tl
9 -l ll
;
; oi
-i
t-8 -1 -3 lJ
l_z 2 E
9. Itnd noo-aiqular matrices P aad cuch thai PAQ il iu thc nomral folzr for the matrices
fr-i-il
fr 2s -2.l
liilA=12'2 I 3l
6o,atd, Islgsl
trlA=lr r ll
-"'Li
i ij
t, o 'r 1l
G.jî.r.Li..2tr0r (doA=l
0.|
s)
Q
ｌ
・輔辮＝
１
:
可
2 ro. (t) vEctons
Any quantiry haring n.compoueots iE called a uector of ord.er n. Therefore, the coefticients
in a linear equation or the elements ia a rm or column matrir will form. a vector. Thus any n
oumbers 21, 12 , ..,..,,4written in a particular order, constitute a vector x'
一。
司
．．
．
＝
(2)
The vccbrs t1y I3.,...., Ia are said fo 6e linearly dependent'
1q.....,l" ,rot dl zcm, strcit thot
Lirear depeadence.
there rlri.st
。
r
nurzbeire
\,
1,111 + 1212
．．
．
︲
⋮
⋮
＝
１
。
If
no
If
11
a4h
+ -..
+1'4
if
= O.
numbert, other than zeru, rz;ist,
tto uators
ore
sid. to 6e linearly independenl
* 0, h'anapoeiag l1r1 to the other side and dividing by - 11' we write (i) in the form
司
。
一
．．
＝
１
０１０
ｌ
一
一
一
一
一
一一
一
珊
ｌ
xr
= pGa2 + pSr3 + ...... +
IrIr
then the vector 11 is said to be a linear combinatiou of the vectora 12, 13, "',
rr
〓
一一
解
罰
︱
Ｑ
．一．
ｌ
０〓
-rnple}2&tvttllru*ctol:r,l=(I,3,4,2\r2=(3,-5,2,Z)an'd,tg=(2,-1,3,z)tinmt$
w .f;d,rcss otu of tlurg os a linar combination of the others. (KemLa, 1995\
l1r1+12r2+\r3=O.
lterelation
i.e.,
l(1, 3, 4, 2) + )"r(3, - 5, 2,21+\a12, - 1, 3, 2) =O
is equivalent to i.1 + 312 + 2fu = 0, 3lr - 5ta - Ie = 0,
41,1 + 2)q + $fo =O,Z?!i+Zla+*t=0
dzpen&nt ? If
ｌ
０
１
，
。
０１１
〓
Ｐｎ
“
Ｈ
Ｆトレ
t*[?
ｌ
瑚
ｌ
ihich i8 ofthe uonnat
。
。
．．
．．一
．．一
．
ｌ
+I2,
P snil
aad vorify thatA-l = eP'
followitg matricee to ncnrel form andhenca find their ranlg
ieo 31 I2 6t
2l
@aipir, l*)B; Bangalon,
ｌ
ＦｐＰ
Ａ
ｌ
ｏ劇
０１０
１
・
“
一
Ｑ
鈍Ｌ％
Ｑ島Ｑ
Operate R3
trtlrir
eaeh of the
(Dl
row (colurur) transforharion of the product by subjecting
9":vlg.el3meatav
(pGt
factor) ofA to tle sFne.
the uoit
8. Rcduce
pAe b in clu normal.ftnn
ｌ
Ｆ ” Ｐ
〓
劉刊
１２
A.
ＬＦほド
¨嚇 ¨
*r:P
,.- prelactor
ne
ra:rrl. of
軋
〓
Ａ
撼
前
恥
frene find, tlu
Iis
rorttunat*e,=l1t I 31,
L0 -1 -lJ
r"Ltarwator2'lse7t
21J
7.lI A=lz -3 { I fnd A-r. AIm fiod trqts g6a..ilgulrr EAtrice.
Hcnce p(A)
9
…
]
1
& hove thEt t.he r@ €quivahnt Drtricls
I-s
7
４２４
ｌ
Ｆ
卿
聰
叫
ｄａ割刊
11
風レ理Ｆ
＆
町
亀
００
Ｑ
００
１０
Detemine the― ■oFthe f馘 ¨ 角 g natne(1巧 ):
L‖
３
１一
３
００
００
町 C3+6C2,C4+302
００２０
００１０
０１
ｌ
００１００
００２０
１０
ｌ
ヽヽ々ヽ
００３０
０
︱
61
」NEAR ALCEORA:DLl● КMINANIS,MATRICES
´
・
・
l
__泌
■
J
〇
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As these are satisfied by the values lr=1,&=1,&=-Z which are not zero, the given
vectors are linearly dependent. Also we have the relation,
uNEAR ALC[8RA:DETERMiNAMS MATRICES
'
r1+t2-2r3=0
by meaus ofwhich any ofthe given vectors cao be erpressed as a linear combination ofthe ot-hers.
obe Applying eleraedary m* operatious to the v€ctors !r, q, ro, we ."" that the matrices
['rl
[
rul.
Ilrc rank ofB
being 2. the raak
qprcslcd er a lincar onbi'auoo
operations aod for
aly uauir In gaeral,
of 11
*a
ofA
12
is
rko 2. Moreover rr,
12 aro
'.'
*
a,r,".
[ n = ] trr +r]
linearly independent rnd
,r.urrs will hord for corumn
t}e folloring raeults :
if a giwn mahix l:ot r llnear$ indzren&nt v*ton (rowa or alumnil and, the remaining tnttora are
linzar cpmbinatio* of tlese r ua;tan., thzn rrorr.k of tlu mor',riu. is r. conwnxly, if a mati is of mnh t, it
contdins y li;uarly indepndzd vdqc and, ,emaini^g wcbrc (if any) cii k erpresscd
as a linqr
mmbilutian of tlrae vrr:tolr,
we have
2‐
0れ四 L】
'
Olr+bda2+."-"+b2Pn=lr,
0,.t1 + 0-r2 + ...... + b,;cn =
0.*1 + 0..r2 +..'... + Oln = I.r
"r)
Consider the system of"linear equations
∬ま111矧
f^)
ill[￨=￨:]
。３１０
・
２． ■ １
www.Engg-Know.com
争レ
=[:￨
２
bperate=3 Rl+Rル
一ぃＰ
row operations, be reduced to
0"rate:Rl,5R3士 R%
博Ｆ
I.n11hofa=ranhof K=r(r(thesmallerofthenumbersmandn).Theequations(i)can,
Ｌ０
︲ｌ
'rarnh
Proof, IYe consider the following two possible cases :
Operate 22 Rl,
２．
・
amn
Opera撼認 1,田 2,
９
hzl
1:￨￨,￨=￨!￨
０２
dtn rrl
av.
2:
臨Ｆ
L"^, an1
F
９
atz
702=5'
We have
(d) are oonsistent (d.e,
A is the co'efftcbrt matri*. and .( is called the ougra ented, matris ofthe equations (i).
(2) Rouche'a thsoreu. The qilem of equatbna
6) is anaistent if and onb if the acfftcient
matri* A atd, tlu augmentd rwrit K arc of the same
otlururi* the systim'is ineon'iistent.
,
by suitable
+ 0-rr, = Q'
If r * r',
5r +3y +72=4,3r+261, +22--g,7x+2y +
"ul and x=la..r. azz
l"'...11'.....1
Lo-, ana
"^J
"-"
(iil If r = f
$ii) If r = /
５５７
[o1
1,
are of the form 0.r1 + 0,r2 +
- (r + 1) equationa
Example 2'27. Test for consistency and. solve
containing the n uDktrowns ;r, r!, ..., ro. To deternine whether t}le equations
possess a golution) or uoi we consider the ranks ofthe rnatrices
orl
l,
tl:e equations are incowistent, i.e. there is no solution'
solution'
= n, the equations are consistent ond there is a uni4w
< n,the equations are consistent andttnrc an infinite number-o;rsolutions' Giving
arbitrary values to z - r of the unknowns, we may expresE the other r unknorvns in terms of
these.N,
(i\
2‖ .(1)CONSISTENCY OF uNEAR SYS「 EM OF EQUAT10NS
atz
i)
lko6edure to tect the conalstoncy ola cyrtem of equationc ln n un}nowns:
Ftnd tha ranks of the coefftciznt ,nttrix A and lhc atgmenbd. mutrix K, by reducing A to the
triangular formby eL*eoni row operotbns.Irt the rank ofA be r and that of Kber'-'
3.xl=(1,2,4),82‐ (2,-1,3),■ =(Ql,2),■ =(-3,7,2).
an
;:
Hence the equations (i) are iuconsistent.
1.(3,2,7),(2,4.1),(1,-2,6).
2.(1,1,1,3),(1,2,3,4),(2.3,4,9).
f"r,
;
Clearly the (r+ 1)tL equation cannot b6 Batisfied by any set ofvalues for the unknowns-
8
Are the fouOung V‐ ● Lnearly dopendont.IfB● ,and the relatiOn between them:
・
.l =l au
t
- r equations being all of tlre form 0-t1 + 0r2 + .-'... + 0-r- = 0'
(ir)
will have a solution, thougb n-r of tho unknowns may be choeen
equations
arbitrarily. the solution will be unique only when r = n. Hlnce the equations (i) are consistent.
II. funh of A fi.e- r) < '",nh of K. ln particular, Iet tbe rank of K be r + 1. In this case, tlle
equatioas (i) will reduce, by suitable row operatione, to
byq + bgr2 +'..-.' + btdz = lY
and tbe remaining rn
Problem●
:';
::.:.:.:.:
Ihe
I
have the same
ilft ;; ;
...(ii)
and the remaiuing m
I
',
a-l,zl and B=l ,2
L'il
l,,*+-rnl
13 caD be
b1,q + b16r2+ ...- + 61,,rn = l1l
Os1+ b2g2+ ...... +b2,rxn=l2l
{]‖
=1鋼
11慟
_lll;lLI:￨
lAndhm'
1998;
Nagpur' 1997\
&
IIIoIIER ENGINETRING MAT}€MAIICS
rsDls ofcoefrcieot Datri: e.od au$ncDt d Eetdr for
tbo Les! set ofequatio$, arc both
consirt4lt. Abo tbe gver, sJ,rteo ir equirj;;; '-.-'
--I'he the equatioas
^ Hence
2.
are
*+3!
+72 = 1,
∴ ,こ書+壼 and“ 1-器 z
lLr-z=9,
where z is a frara.Eotaa.
卜書y=imd″ ‐
0,L a par_lml口 8du● m
301ut101t,the coescent man“ ■
e_"田
Ю
ttf臨」
L騰譜“
He疇
―
υ
a″..Fλ ●●d“
ttt
r307“
３ ″″
p"r颯 22:R“ 島 :ム
Ю per“ ingR3+:R2
4-2‐ 2 CIven、績 en io equlValent to
+み +2+鋤 =0,2■ ″ =0
r Wehavez=― ″ and,=-2-″
WhlCh 81Ve an赫 ● ■ulber oFnont● 宙a1301utlon3,工 and″ belng the parameters
ple2 m a■ ″ υ ges crλ For″ たた1`ル ec“ a10■ s
Nulllber ofindependent solutlon8‐
を
bornogeaeoua linear
―
-1レ ■`ル
“ +2'`=0
(3λ +rン
(1-1)工 ■(4λ -2υ +(λ ■3ン =0
(λ
〈
■,)
2■ 13ヽ十コン +3(λ ―コ)″ =0
br rdirciDg it to the Ei.Dgu16, fort! by eleilentarj,
C C02SIste■ らcndfttι たg"aas
“
uhprl
\
fhe
is
(a
_r) eean8, if erbitriry
velues are
ifl
*q.i::,11;I1;_":.v€riabre€,thevaruesor*,"i-ii,riiiiji"""J-#,iliqoayr",,a
. ne.
ure cquullolr,8 t4r) wr have aa tBfinit€
nuftber ofsolution6.
or
OFr:ッ :z″ ゐ λ Aasル
“
has the grpdtet of thesa udlues"
SiveD equatioDs
- rr linao.rlr indzpetdenr /e,lutioas.
iDdepeodent solutioDs
, =! = z =0.
︲
︲
５１
＾
﹁
綱
﹁
・
〓
ｎ
２／
・∽・
・・″。
００
--_,1",:rr.r-b. "f-Fearly
hau. (a
:
お 2 whlchく the nulnber ofvanable(le rく ■)
qyo6o* (iii) haoc onb o mtial zera soluribn
:r =12 =...... =rn = 0
If r < n, ,he .q4oti4ns (iii)
００２
and
２
6
+ awJ,]n=
eatrit,l
１
l=
Ｉ
lhus if
Ｆ
"the
喘
co€fEcieot
﹄
3112
r
9'
可
F =
E.hir
﹁
unles!
︲
３
2
Ｏｌｏ■ ゴ判
¨一
一
一一
一
一
,._ #".U.*J:k
Ifr =a
rart
bave aD ta6.dta auDbcr ofsolutions.
α α
＋＋
d6rtt+ a^*2+......
l.
aotof
ol |iaerr hoaogeneou. oqu.tior.. cooeia.t the
]-tl-l*t
^-.
equatrona
１６ｂｒ ●
﹄︲
︲
If),=5aidF=9,thesr,6t D will
︲
９劇呵
３３
!sl!e EnlL But rt i3 ofrai_k
syste6-;;;;f*'- " -^.* 2 End'trir
叫Valuc rλ _■ ぬ
o“ ¨湘
lhⅣ eno
５４５
３
〓
レＰＰ
ｄ
錮
κ
Ｊ可
〓
五
レＰＰ
■ce3
321
10peratlngR3 7Rl-22
Drlrober of variables (r'.c. r = z)
.. the equatioae bave only a trivial solutioD
０
≠
一
わ
節
７２
〓
５２λ
２
乳
騒出蹴棚糧盟
OperatlagR3
〔
ir 3 which - tie
(iD RâBk of the coefrcieDt
are Dot of thc
g. the
rr,
[
ｌ
atnx iS OFrttk 3嘔 s
reql[lli[11111lilutlontandOnlyittheCOendelltコ
sdu砕棚
● 19")
一．
一
．
減
""。
“
(bゼ
+u=0.
１
"“
２２０２２０
一
．．一
tttπ
\:i) 4x +2! +. + 3u = 0,6. + q +*+7.o = 0,21+ !
(i) Ra.DL ofthe coeficient rrrahir
.0● s
ｌ
“
け
２４０
ぁ
ぃ血読思
蔦島臆漁
“
ractぬ ″
up"rt● anptt mt転
MAnCF3
E- Vlhen m <t (i-e. the nuraber of equations is lese than the Dumber of ver.ieblos), tI€
soluliatl is dluals oth2r than x1= x"= ...... = ao= 0. The numtrer ofsolutioDs is infinite.
II7. Wen m = i (Le. tle Dulnb€! of equatioar = the truEber of va ablss), the
dnd
êcessataof the
suffi.ieLt cohdition fu sohrtiorts other th&n xt = E2 = ,...,. = ,n = q i that the d2teminont
coefficie$t motri. i.s zem. ID this csle the equatioEs are stid to bc coasis,€nt and such a solution
is cell€d aoa-tduiaZ sohtioa. Thz determintnt is calLd lhz eli,î,,,.nlt of thz equ.otions.
E-r'ple 229. .So&,re ,hz equz;iont
Gl z + 2x + 3z = 0, k + 4t + 42 = 0, 7, + 10y + 122 = 0
(082● 2iα ,7999)
引司
ple 220.L略崚
uNEハR At● EBRA:0日餃 MIIIANTS
if,
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lr-1
lX-1
I z ..
lr,- 1
lo
Iz
wiu b€ consistent,
3t*1
4I-2
ql" r
3t+ 1
t-3
2l
I
3(r- r)l
I
3.rl=o
a).+1 3(x-1)l
2“
_WIc′
降 p“ns
lod,lil4t,
1994 t
if
t+3 l=0
2x
sacrle6′ 。
r"ese“
[Operrt i2
-]tr
Ioperate C" * 4:,
UNEAR AL6[BRA:DETERMIVい饉l MATRICES
lL lI bI = am - n, cm = bn - l, ut -cI - ralprove that I * o' * 6r * = 0
"s
+2r2=ls3
lll Show ttrrt the 5yttom of equstioDi 2l1-b2+xg=)ar,ztl-3r2t2r3=k2,-:1
posseasaaon-trivialaolutioooolyifl=1,I=-S,Obtairthcguoralsolutio"t"*c.}Iff;,
ca1
rrrrr,
13. Determine the values of tr fm which the followilg set of {uatios may possess ao,n-trivial solution:
Arr *re - f"r = O, &r -2rz - 3q = 0, 2ltl + {rr+ lrr = O.
For each permissible velue of t, d€t€tmiDe the geaeral eolution.
$<etald, 199 ,\
l{ Solve cmpletely the sptea of cquations
r+!-U +&D=O;t-2y
+z
4x + y
7y +
- fu + &o = O ; fu -
-
w =O
b - w = 0.'
2.T2 PARTMON MEII{OD OF FINOING IHE
INVENSE
'if
According to this method of fiading the inver:e, the iDverse of a dutrix A, of order n is
loowu, tben the iaverge of the matrir Atr + I car ea8ily be obtaiued by adding (n + I)th row and
(n + l)th colu-. to.,L.
: Az1
[n,
[x, :
... :
and A-r=l ,.. :
Let
A=l
L^"
Xd and
o,
r
xz1
... I
Lrr"
,
J
yg6fs15 andAg', X3'are row vectoro (beingtra[spos€s ofcolumn vectors
s6llnn
are ordi[ary numbers. We
whereA2, X2 are
eg,
, "1
... I
aleo aszume
thatAf I ie known.
['' : o'l[*' : "'l-['' ol
; ;ll*, i ;l-Lo 'l
ld
A1X1+A,,2XL'=Ii
l.hen.{.A-r =In+ri.e-
gives
-..(i)
A{2+A1r=g
As'X1+dts'=O
Ag'X2+ot=l
From (ii), X2 =-Ar lAgr and
Ilence
r
"'(ii)
"'tiiit
"'(iu)
rAd- r
using this, (iu) givesr = (u*Ag'Ai
and then X2 are given.
Also from (i), X1 =.{,i I
(I,
and usingthia, (irO giveeXs'=
ltenX1 ir
-
AzXs')
*As'Ai
1c
-As'411A il-t
=
*Ai
A1r
x
determined and hence.4- r ia co-puted.
Obr ltis is ElEo koowu rs the 'Esc
iaverse rnaEiceoAi I aad (c
&tor
-As'lir 4;t.
'&.nple
r
nctlpd'- For evaluation ofA- r we only need to determine two
Z-sl.Ilsing thc poriitian mcthd.,
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fnd
the
iru"r"".fll i
ls5 3lil
Let
so that
一
Ⅶ
一刊い一
Let
︱︱︲
刊＝︱︱︱︱ぃ
ｍ
HlGHER ENCINEERINC MATHEMATICS
′
UNEAR Aば訓要RA:DErERMINANE MATRICES
｀ n〈 J)is
A more general transfomation■・
il二 :∫
which in matrix口 otatio●
..{￨ピ
1:り }
)
Is[1=[: :][]
Such transformations as (i) and (ii), are called linear transformofions in two dimensions
Shjlarly the reLtbns ofthe切
Ａ
Ａ
２
２．
¨
﹂
︲・
linzar tmnsfortnatinn from (r, y, z) tn (1,
y' , z)
¨_(Jjj)
in thrce dimensional problems.
cr trl [".l
frrl f", br
e'l,x=1"'l .'.,i,'
u:.':
Ingeneral,therelationr'=AXwhere"=11:1,^=l:
:::
cn a"-l L"l
l'".1 lo^ bo variables
-..,yn i.e- lhe
give linear transfurmotion from z variables ,1,rt,...,r^ to the
!t!2,
transformatioa ofthe vector X to the vector Y.
This transformation is called linear because the linear relations A(x1 + x) = AX1+ AX2 and
A(&& =eAX, hold for this transfor:nation.
If the transfornation matrirA is singular, the transformation also is said to be singular
othenrise uon-sing1.lar. For e non-singular transformation Y = 4X, we can also write the inverse
transformation X = A-l Y. A non-singplqr tranrfornation is also called a regular transformation.
Cor. If a traasformation fmm (tr,
:2]
of
12, 13)
to (r1, J2, J3) is grvea by Y =AX and another trensfonration
12, rJ) to ("t, 22. zi) is grven
(yr,rt, J3) ta \zr,22, zt\ is given by Z = By. theD tlre transformation from (rl,
by
０
一一
”
〓
︲
ｅ
︲
ｃ
︲
ｎ
ｅ
∴﹄ ‰ 嘲Ｈ
give a
" ::II:li:iJI
Z = BY =
=IBA\ X.
B{"AX!
観
(2) Orthogonel trensforaetiot. The lineor transformation (iu), i.e. Y =AX, is soid to
ortlwsor@l if it trcnsfonns
!t2 +!22 * ,.. +!o2 into ,tz +x22 + ... +ro2.
The matri,r of an orttwgonal transfortnation is called an orthogonal
PrOblems 2 10
be
matrir'
["t.l
we have
and
‐
＋
Ｏ喘
ｎ
ｙ
鈍 ¨
′
十ＯＶ
Ｏい
“
一
〓
一
凛
嗜
i趨髄
A押出棚蹴1咄漱榔
= [pg2.--..-tnl,
sirnilarly YY = ltz
.'.
213.(1)uNEAR TRANSFORMAT10NS
.fx
If
" =
t.-l
l
I
rr' *, 22 * ... * *oz
Lol
+
yzL + . ,.
+
! n2 .
Y=.AX is an orthogond transforuration, then
XX =Yy = (AX)' QaXl =114617 which is possible onlv ifAA
=
f'
But A-l A = I, tberefore, .{ = A-r for an orthogonal transformation'
Ileuce a sgucre matriz Ais soid, to be orlhogonal if AX = A'A=I'
I,rA is orlhogonal, A' aad
Since A is orthogoual , A' = A-.t --
ObE
1.
A-L
an als oahogonal
.
...- (A,),=(A-t)'=6'fr i.e. B'=B:r where 8=A
I
Hencb 8 (i.e, AJ is orthogdaal' As A' =A'r, A is also orthogonal'
Ob* 2. If A is orthogornl, tlen I A | = x L.
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(Assam, 1999)
︲
町﹁
譜e l鼻 :rプ
Example 2・ 32.S施砂読ar
+ r.2 +
2rg
t
7.
the transforrration is regular.
X=A-r Y
= 2.h
Ir
2yz -
ta ; t2 = -
is the inverse traneformatioa.
4y 1
-1
-ll
tA-^rt=l
-r]
+ 1yz + B!
ample 2.33. Proue that
and B are orthoSonal
lorr
e
; rs = y r.-
the folhwing matri.x, is
y2
- !s
-
rAエムE,f"7W7
ol
-f
li!
ortrros"."r v,:ll's.t=2/7,m=3/7,n=6/7'
rratricls, prove tbatAB ie also orthogonal'
L
atz
au
-
LI, where
f
is the nth ord.er
I
iil=o
3 "u...':::
or^-ll
dn2
| "",
is called t}le charuteristic
equation of
A- On expanding the determinant, the characteristic
equatiou takes tbe form
orthqonnl :
- t + hz)!-z + ..- + Io = 0,
the
of the elements c1;. The roots ofthis equation are called
term8
where [,s are expressible iD
characteristic roots ot latent roots or eigen'values of tbe matrixA'
(-l;"Xn
勁
ニ
ユ
Ｍ二
に
お
″ν ″
＋＋＋
””切
二＋
９９９
／／／
２２１
川
一
川
利
﹁・
一
一
〓
ｈ
触
酔
恥
１つ２９９９
ｆ降隣
-
'z6
︲
︲
３
３
／
／
２
一
．
¨
一
¨
¨
¨
¨
¨
３
３
／
／
・
２
Itus.t1
o-=1.1
IfA
I l.^ ,
"*l : ? -j
L;nl-ol
0露
2.14. (r) CHAnACTERISTIC EOUAIION
IfA is auy square ngtrir of order n, we cq. form the matrixA
unit matrix, The detBrrninant of this matrir equated to zeto, i.e'
transforuation ia givea by
where
3]o"*t"etrensrorrnationrrom't''2'xttozv22'21'
ｉ
β んん１劇ｌｄ﹁
・
２つ１
■ ３１
* r""-.,rljr,*0"";:1"*
４
﹄ピ﹄Ｆ
︲
ll=_,
％ И 珈
8. Proe
¨
Ｌ
rhus the matrixA
I
1gg4l
調刊
ro,=l?
１１０
Now
.'. The inverse
ｌｌｒ
らにｉ
１
〓
ｙ
〓
Ｘ
¨
価
日岡
The given transformatioo may be written ag
1l
4. Veri& thet the following matrir is orthogonal:
隣
xg, !2 =x1
ト
︼
x.2
０機
ll = r, - O,
is regular. Write down tlw inuerx trtnsformatiotrl&emta, 1gg5 S ; Rewa,
!
lz 1 ol [t I
-"]''=Ll 3
^=[l I
masrarmarぁ 2
ｌ
ｏ﹁剤﹁
*
rた
A trarsfornetion from the variableg rr,rr'rs Lo h,!2,!3 ia givan by Y=AX, snd another
trausformalion from ! b !2, )B tb 4, r2, z1 b givet by Z = 8f, where
︶い
ｍ﹁
・
t = Ztt t
1_様
、
8.
￨:￨:1ll
+
llln
tM論
同
か
罰
曝
…
:卜
童
‖臓
ぷ
(2) Eigen
vectors
ca轟 es the colunln vector X into the∞ lumn lettor y by mc
pragEce,lt is onen requlred to ind such vectOrs which transfol
Hence the
matrir is orthogonal.
multiple of themselves.
(i
L€t X be such a vector which transforms into )"J( by meane of the transformation
IX=.AX or .AX-lJX=Oor [,lt-UlX=O
Then
hobleme2.ll
L
Ropreseut each oftlre transformatims
4 = 3y 1+ 2y2, y1 = z1 + ?22 end e = _ + 1y2, y2= 321
by t'he ure of'makice! cnd
t'e composite traurformation
2r'zz'
fifi
"'tiit
This matrir equation represents n homogeneous linear equations
\
which exprcsses
)'
rr,rz in
tcr-ms of
ebrala,19961
f翌]
a+rcoso, rrite the matrir A of transformation and prove that
'If_=i,":"-Jainqr=rsi,,
.a 'eA'. Heace srite the iaverse tmrpf<rmation.
6
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..(jJj)
ｏ
キ′
聾
F=」
MATttMATlcs
UNEAR AtGEBIQA:DETRMiN劇 ¶ゝ MATRICES
aVe a nont五宙 d sdu"On Onし r the∞ 輌 dOnt
… matix L dnguh ia r
Putting λ=-2,we have 3ヒ +γ
性よ
S腱
Hence the eigen lた xtor i3(-1,0,1).Also every non・ zero multiple of this vector is an eigen
″or Jaた Pa■ pec"∴
“
V∝ tor corresponding tO λ=-2
Similarly the eigen vectors coresponding to λ=3 and λ=6 are the arutrary non― zero
mul●ples Ofthe vectOrs(1,-1,1)and(1,2,1)whiCh are obtained from(J).
織職錯
ふFF鶏喘亀鴨臨 1紺鑑帆蹴酔戦胤 t
Lus織
謝力
胤器肥 ltttI帆雛拙蹴
T鮒サ
ポ・
…
Ex3m口 e2払碗臓懃 υ
ご
α
屁
ど
α
は
“
“ れ ¨ぉげ施れ
― f四
F`l“
Hence the three eigen vectors may be taken as(-1,0,1),(1,-1,⊃ ,(1,2,1).
215。
PROPERTIES OF ttGEN VALUES
O)7し ,これ or"・ eな "υ αJ● 2òfα
)
却
驚驚糧胤[二 1り
∴ 董=普 j宙 ngt“ dtt
so
that
￨ム
ー
VI=FIλ
′=0
=*tg
vectOrに ,D.
1黒胤
職
胤
ヽ
痴
千=寺
21λ ][;]=0
Ltぉ
Ifλ l,し ,L
.
鮨物鶴
to the eigen value L then
α α α
__。nding
２２望
〓
Ａ
腱 ―川lX=FIλ
歯 Of′ たノだEcゎ αr di昭ぃ α′
in be ctpable Of壺 y extensicln
P730■
“
òfottr 3,but the method、
COn8ider the square iattnx
+6=o
●f=ルル
es“
■"=jS'た
to mattces ofany Order.I
"u be proved for a ma●
l.
rヱ,y be the_pOnents Ofan eiri vmrc。
′
腱
Inls prOpew口
Ｌ降ト
'た
工+7y+z=0, 議 +y+3レ =0.
the
samO,we
have■Юm the frsttwo
嘔
=ふ =発 Or寺 =昔 =子
琢テ
1島
搬
+32=07
The ttst and third eqtatiOns beiコ
S∬
酬謝翻柵l餞難瑞蹴織∫胤鶴職:蹴」』護朧
X=し
,■ 1′ ,whichお knOvm as the● :“ れは
1,均b.… ¨
ん
ttλ
+ lg(411+
α
鮮
10neXpandb,
oz+a$) -......
be the eigen value3 0fA,then l A―
...(二 f)
VI=← 1)3(■ _λ
l)(λ
―り (■ ―A3)
=_が +′ (λ l+L+L)― ・……
1崚 l
Equating the Hght hand sides of(jJ)and〈
+L=・ 11+a22+● 33・
″α′ねじ
お deた rmjれ a″
c,oFtた。
aztt ofα 鳳力董 Aお
なれυ
`Pm力
Dutttng λ=O in(鰤 ),We get the result.
“
λl+場
ng tte eigen vecね ro,_1)
(2)'■
強竃mple 2・ 35.Fiad tte ear“ υ
.Iz“ α
(3)rFλ
"d etige■
fa"α
JS aれ ●lge″ υαJ“ ●●
力■■´し
'あ
IfXbe the eigen vector corre8pOnding toヽ
¨
・(:Jj)
JjJ)and compa口 ng coerlcients of′ ,we get
HenCe the result.
.
CP3 1/λ jS rんe eligea υα′ e orA 1.
“
then AX=鷹
・¨(j)
Premuldplying both:,ides byA 1,We get A lAX=A lⅨ
lil1111lIIlilliIIli『
盤
￨。 n
ie_ IX=AA IX or X=xA b
as
a+a cA2_9.+lo=o"0+2)a― め0_。 =。
λ=-2)3,6.
yJ z be the cOmponents Ofan eigen vectOr OOrresponding to the eigen valueた
―駆
“
二
d“●●
(4)frλ お an aige2 υ
fo■ οr山
αJ● c.
れ1/λ お aJso frs etgea υ
aJ“ arr軋働●
“
ofA
1.師 pe彎
igan
dgen
valu●
技A,then
1/λ
We knOw that rλ is an eigen value “
ofa ma昴
(2)1.SinceA is an orthOgonal mat転 A lis sane“ Its transpose r.
11:λ
we have
=.1/λ
5-i l…
X
is an eigen value orthe inverse matほ x
A 1.
.
hus the eigen values OfA are
:ち
ie. A lX=(1/わ
Thls being ofthe same fom 88(j),Shows that 1/λ
is an eigen value ofA′
But the matrices A and″
二だ
l響 Ⅲ Same.
で 14′
...(じ )
1[￨=0
Hen∝
「丁
.
have‐ the
same eigen values,since he deteminants i A―
1/λ is il∞ an eigen value ot A.
②
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VI
π
75
UNEAR ALCEttA:DETEPMINttWS.MttRICES
be
then
An
hos the eigen ualues
ツー、
L€tA,
︱
6) If 7\,k,...-.,1n aru the eigcn ualues of a ma*it A,
Lt^ , W, ..,...,ry (m behg a positiae intcger).
tbe eigea varue of.{ andxi tbe corresponding eigeu vector- Trren
dX; = l5X;
We have
2. I
′
″υ
rs● .Arso● P″ ss
￠
=[硯
the characteristic
+
llln -
・
Plこ ←1)竹
″
2 P3=り
e-r=*"-r=*[l ;]-.[l l]=]f
-
4t4
-
?I3
たs8P2υ cr“
IPr= P,
- I * 1OI=(r2 * 4t-
憮
- 2t+
g) +
I+
E
Iby(じ )
d
ル「π
2“
+8=Q
_il=01F―
姜
"λ
,
二
︲
２倒句
計
３■１
・
「
１４４
／／
３
た
た
１
巧一
ｒ
ｒｉ﹁︲︲Ｌ
一
嘲■物
wh"∝ A l=争
３２２
＾
“
フ
一
降ｐ
ｌ一８
一
００１
Hamilton (1g0$-1865) an Insh mathematician who fu l.lrown
il島
Problems 2 12
.-.(iir)
' se€ foptoote or p. 32. william Rowan
lor his work in
１０
O,
firis
１００
- r, ......, A, f respectively and adding, we get
０
〓
knl =
l]
λ
ぬ araCte五顔 Ce叫面 m遍
に
_ 1A +
S)Ga
=λ +5
By C観り項 amilton theo“ m,A3_20A+"=Q
[or the terms on the Ieft caocel in pairs.
(II)
.
５一２
+,t,
l
ぶ二就乱ぶ精 i脚 1留:嚇
批翼ぶぶ∴
APo_1-Pn=io_ rf,
dPn=hJ.
dy[amics.
byA- l, we getA - 4I* 5A- I = 0
,
Now pre.multiplyiqg the eguations by A", A"
(-1)i{o + kgn - t + ..,...
..。
Nowdividingthepolynomiail,t-lf.-ita+rrt2-l-l0Ibythepotynomialt2-41,-5,
+ &nlf.
I"'
API― P2=lll,
1tr)
15
11,
equatim a m tht
０
〓
[(-l)nln
_..(j)
電
=
πA
″α′じ
ァ′0″ ユ。
お
αたご″ ごピ
“
we obtain
MP=IA― Vlx′
―MPlV-1+ざざ -2+.… +島 _lλ +ら
〕
…
Equatiag the coefficiente ofvarious powers of I, we get
`α
珂
用
or
IA―
r +....,
+ &n _
５０
Multiplying
...(ii)
Siuce the product of a matrix by its adjoint determinant
of the matrix x unit matrix.
=
“
αriれ
αtt A=レ
this verifies tbe tbeorem.
Po are all tbe aquare matrices of order n whose
elementa are fuoctions of the
∴ by C)and(:ゴ ),
４３
+p2Ln-2+......+po_rl+po
４孝
I
´
、
¨
一
．
．．
．
．
．
．
-
""Jiilrvrro-irr".
P can be split uD into a number of matrices, containing
teims wittr the sa6e powers of
"' that
i., such
一
″
・
・
´
ヽ
一
＾
¨
¨
“
Now A2
０
︲
■ ︲土 Ч 刊
〓
３
...(r)
`″
∝メ
血
言蕊l血轟
By Cayley‐ Hamilton theorem.A must sat
一
I
311=°
r昴
■郭
if
115λ
Let the adjoint of the matrirA u
crearry the elements ofp wiu be porynomials of the
(n - l)th degree in l-, for the cofacrors ofl"-p.
tbe eremeitsln A i.r wiu be
:hu*
lr'fr: '.-.-.,
etemeDts otA.
"ゎ
ｙ
matrirA is
| =(-l)ztrn +it1ltr-t+....., +in =0
(-l)"Ar +IlAn - I +...... +In =0.
P=p,1a-t
:IIλ ″売。″
A5_4A4_7A3+lLA2_A_1ば cs
一
１２９
８
〓
呵︱
４
ノ
牛
'.hen
;瓦高
The characteristic equation ofA is
5. CANAT.HAMITTON THEOPEM'
I A-U
葛
「
Eaery squan matrk satafue its own characteristic quation,.
i.e.
equatioa for
=0
h, -1Il'
硯
!a
the ath order square
+
E^
I
[(-l)'r.4i' - + [rAn -2 + ...... +
結潔潔震
:認 computatio.'
Asa脚
by-Prcduct ofthe
麓鷲
鑑臨 ∫
l4rye rr*trio,'霊霧
the computaiwr
tlu 鸞
iru,,,"e
if thz L鶴
ル
for 就
馴鰍 of稚
equation and the determinant ofthe matrix are also obtained.
In generai,AmXi = )raX; which ie of the same form as (i).
is an eigea vslue ofAm.
The correspondi.g eigen vector is the same X;.
Hence
'A-l = -
wheuce
A2X;=/,gg1r1=A(\X,)= UIXi) I1ti4)
=
= \2x,
Simiiarly,Asd = \ki
Coa Auotber ucthod of tiadiag tho inverrc.
Multip[yiag (ui) byA-r, we get
?LI"P - 1 + hrAa-2 +..... + lo - 1f + l,Al
proves the theorem-
■Fld∵ 3m andPザ aご隈eigennlu19。fli::
⑫
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1
ぃ
"ご
λ月 r998)
"‐
﹁︲
du‐ ●FI―
:
―
GD`dA躙ら1998)
: _il
2'I7, (I) REDUCNON TO DIAGONAL
■■﹁ fthe C・ m
2 Find theI瀾山餞
・
Fr ,tLP is a diagonal matrit
extension
3 but tlre method will be capable ofeasy
tTbis r€surt wiu be proved for a equare matrir oforder
to matrices of aay order.l
values and
LetA be a Eqqare matrix oforder 3' Let 7'1, 12' )'3 be its eigen
such lhot
0卜
一
珂制珂
＝
lfa**,,o*
",p
4.
ofthe mttc“
:
司
鋼 ●n口″コひ●da 199`)
["]
l"l
rl
i,
)
'
Find thc characteristic roots ald the rharacteristic vectore of tte uatricea :
Ｉ
Ｆ
０００
199'My3● ″
ち 1997′ ん ray● L1996S)
]刊
(o) If tr be aa
Xr
仁SM,2001,',41,gり
(Mに
に)開
i.
(carr.ち
に πa,1鼎
=M“
'
S;7・じ,20000
釉
h鴫 1998:LaP“ ら』
菌
"彎 “"●
P鴫
３１１
・
(,
|
..
UsiagCayley-Ilamiltontheccn,6nd.4,s,n^-[
L
Findthecharaderisticequationoftle-"t irA
lt z -il
-€ -i _{zl
2
- le
byAs-s.e7+zA0-gA5+A.-5As+8a2-2A+I-t
|
I
P'l
=
x2
f" v2
"l'bv P' we
lrr 22 vr
L" "'J
I
have
I3X3]
'9X2'
h*
th2 eisiluatues of Aas its etements'
=
p- IAP f,,
some non-singulot n x n
motrit
P'
p to i is called a slmilarity
This transformation of a matrix A by a uon-singular matrix
transformrtion.
-
- *"
If r
eigen uolues os A'
similar to lrhe metril A' then A has the same
sanre eigen aalue'
rv
is on eigen L'ector of A cornsponding kt the
is an eigen vector of A' then y = F
if
,*
matit A is
(3)Calculotionofpowersofemetrir.Diagonalisationofamatrixisquiteusefulfor
obtaining powers of a matrix.
can be found such that
Let A be the square matrix' Ttren a uon-singular matrirP
ttOn 18“ 由 hed by
D=P-t AP
.
2o00 S)
,2=FIAPllP IAP)=「
Similarly
Ds
lA'P
P.ly'/'P
= ,-rAsP and in general , Dn =
['.'PP-I='I
(二 )
by P-r' Then
To obtain A^, premultiply (i) by P and post-multiply
which gives A".
pyFr
l]
=f3 I
lr I
i"l
r, tu the corresponding eigen vectors'
nif
A
弓 1997S)
Medras,
=
(2)siailarityofnatrices.A squarematrirAofordernisca/ledsimilartoasquarematrix
Aof order
(03翔嘔れ 2α Ю S)
“
(di) I
L'l
idi""tita'
;;;;;;";;;i;""aiu
レ四ら200o Zためi199の
lz -12 1l
-1
-tl(Baipur- t998; panjab, r99Z)
and x3
v
Denoting the square matrix X1X2'Ysl
Cayley-Ilamilto thcorem for the matrirA
“
".d firld itr inverse. “
Lr -r 4
F;l
x,
.'. P -lAP = Fr PD = D , which
ob*TlrematrirPwhiclrdiagonalises,{iscalledthemodglnrtri:ofAaudtheresultinSdiagonal
ii fo""a uy l.*"pi"B the eigen oectors of A
matrix D is known as the epectrol uatrir of A. Tnc matri
synmetrical sqg.are oatrix, ahot tbat the eigen vestor correrpoading to two uuequal eigcu
value areorthogoaal.
efznpir, Igg6;K.;al4 rglglt
l. Using Cayl,ey-Hamiltoo theoteu, fad the inverse of
L Verify
t'r
,-; ,;l,.l ;- io l=ro'whereDisthediagonarmatrix'
fi;l=l;i ',;
=l;;i
-Lill ^;;
;;l L.
;i; #;i-L,i
"l
proves the theorem'
1990
Fo_r a
iISh"that.e"Ш
I
flrrrr )qz l3r3l kr x2 ,rl lrr 0 o0l
eipn value
い
=
AP =A[XrX/s|=lN{1,N{2,AXsl = ti'frr'
of a non-siogulartstri:A, ehow t.hat I A l,zlis an eigen value of the matrir
- tBangalan, lgW S)
"djA.
(b) Show t!!t th" eigeu values of a triaogular maEir A are equal to
the elements of the principal
diagonal ofA.
Anafii, tist)
I. Show that if tr, 1a"...-.., L are thc lahat roots of a matrir .A, thea .A2 has the latent rcotg
h2,122, ... L2.
(Waratgal, 1gg5)
/.
TORM
IfaqugrcnatrisAofor&rnhasnlineorl!independenteigenucc'ors'tlunamatrixPcan&pfound
︲
3.Find the elgen valu● 8 and Oむ簿 ▼醐 hに
77
lJNEAR ALCtBRA:DETRMINAMS.蝉
HIGHER ENCINEER!NC MA■ ■ヨ汎AnCs
=plrrA"p5l =A"
!l*n*"*,findthematixrcprcsentcd
zl
t
*ow,r996l
EranpleZ'SE.FindamatrkPwhichfransprnrsthematrirAofEr'2'SStodiagonalfornt'
,rAr, r
HencecalculateA4.
6 dnd the eigen vectors are I-r' u' "
The eigen values of A (found in Ex' 2 35 ) arl -2, S'
t" tt'" ti"* Jurnns' thi required transformation
(1,
1), (1, 2, 1). writine.i"";"r;";"*t'
￨
￨￨
_￨〇
:
=“
-1,
is
matrix
`
www.Engg-Know.com
3i
(Osmania, 1999; Bangalorc, 1990; I(amla, 19901
?he matrir ofthe given quadratic
■
勾
Is
_l -1 -l1l
_i
Itscharacteristicequation is I A-tI t =0r.".'j':t ,--j
-ll
I t -r 3-rl
I
which gives 1,=2,3,6 as its eigen values. Hencd the given quadratii form reduces to the
canonical form
1rp2 +t,gr2
To
+)qzz
?s2+3y2+6z2
i-".
fitrd the motrix of tmnsformatbn
From [4
(3-
- iJ ] X=
l)r-y+z
0, we obtain the equations
=0
Now corresponding to'
;
lr,
whence
-r+(5-trD -z =0 ;r -y
= 2, we get
+(3
r - y + z = A, - r
- I)z =0.
+ 3y
- z = 0, and r - y + z = O,
i=t=i
.', The eigeu vector is (1, 0,
-
1) snd its normalised form is
11,r
'j2,
O,
-
L/'12\.
Similarly corresponding to l=3, the eigen vector is (1, 1, 1) and its normalised form is
(1,/.J3, L/{3, L/{3).
Finally, corespooding to I = 6, the eigen vector is (1, - 2, 1) and its normalised form is
お ned a
“ “
$/.16,
- 2/16, t/^|il.
Hcnce the matrix of transformation is p =
_m
I tttz L/13 l/'JGl
0 r,/{3 - 2/.J61
I
l- v,lz t/,ls r/'16)
2.I9. NAruRE OF A OUADRATIC TONM
A real quadmtic form X/Jt in n uariables is said to be
(i) poeitive defintte if all the eigen ualues of A > 0.
(ii) negative definite if all the eigen ualues of A < 0.
(dii) positive semidefinite if atl the eigen ualues of A
2 0 and
(iu) negativo senidefinite i;ralJ the eilen values of A<
Xl=[￨￨,X2=[:IX3=[:￨
0
at least one eigen ualue -- 0.
and at least one eigen value = 0.
iadefinito if some of the eigen ualues of A arc + ue and. others Exanple 2.4O. Find the nature of the follouting quadratic forms
(u)
宙
ddけ square
T洲無出11謝密般盤忠譜密憲竜需entおよ
織尾
…=[考』
r"r- ir a =
|
y z〕 ,then
+銘 2+2″ 2+竹工
!刊
sun of squares (i.e. canonicsl form).
ond P is tita matrix of transfornation wbicb ie an orthogonal matrix. That is why the above
metbod of reduction is cslled t}Le ortlugonal transformation.
Frotple 2.90. Red,uce thz quadratic form
+ Sy2 + Sl - Zyz + 2zx - 2ry to the canoniccJ
form. iJso specify the matrir of transformation.
λl,L,L beぬ e eigen values Ofthe mat五 xA and
‰∵
F哺
響
ERMIN“ MATRICES
Lpz +),,2y2 + L1z2
︲
５到可
５
”“２
一
一
．
ｃ引割利月
，
２
噌昨０２
・
餞場Ｌ２，〓
〓
Let
″+げ
LINEAR ALCEBRA:DE「
Hence the gnadratic fgrrn (i) is red,uced to a
degree h any numb"。 f vaHaЫ
For mstance,ifA=[ ￨ :￨,X=[landr=膿
Ы
ぬおac“ d肩おrora.二
１押幸
れ犠¨劉
¨
■ 陶Ⅷ琳
橿＝﹁
〓
〓
。
︲
Ｐ
．
Ｄ
Ｃ
巨
Ｈｒ・ｏ・
が
メ
Ｐ
・
︲
２，中１一
６ｒｌ﹃︱し
Ｉ
〓
ム単響
〓
Ａ
Ｓ “
■ 〓
︲
Ｐ鳳
二午︲
五〇
ｍ
肺・
ｈ
ふ・
Ъ
¨
¨
∞ eXpЮ SsiOn r the“ ∞nd
″ara″ cル需 “
HiCH[R ENCINEERINC MATHEMA■ CS
(il
J
Zry + 2yz + 6zx
+ 5y2 + 322 - 2yz + 2zt - 2ry
(D Tbe matrix of the quadratic form is
+ 5y2 + z2 +
(i\ 3?
'
10)
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ft I
A=lt 5
ls r
3l
1l
1l
ue-
UNEAR ALCEBRA:DETER¨ S
HIGHR ENCINEE':NC MATHI MATICS
10
(1) Coqiugate of a matrir.
The eigen values of A are -2, S, 8. [Erample 2.35]
Two of these eigen valuea being pcitive aad one being negative, the given quadratic form
q"
;;;
[Erample 2.99]. All these eigen values being positive, the
iveu quadratic form io posifrue dzfinite.
ol
I trz ,h/z ol
rt rr1 dshowthatFt,{pisadiagonarmatrix.
Bl""ur=t
*r,"* e-]*'
;$,
"=[f B *3}
rota,D =ECE.
"=|! f]r*"*"*-
changes
"麟
nd the e…
Zyz +
br,
-
SinceF=へ
lQ翼
.
(ui) 8r2 +
?y2 +
k2
-
!2xy
-
Byz +
4zt
So
real.
ExampL 242
h赫
Ⅲ
…直
[J刊 Ⅲ
-i]
=
u
λ
Sttr2‐
獅面餞 ●Lレッ赫
“
“
助面りり●Jbr“の
I― j励巳
コに
'
″嗣
`ψ
(ピ )
R″ λ
.
23arrこ κ
Fig.21.E19m v●
l ν
ari●
:u● ●●
u3hat"●
“
″ 卜∫
A=ト 1121ち
ュ
ぬ
α
α
″
lSわ
場
鵠
t'議 I犠
e
a cano口 cal fom
c+A)1:￨::IF―
IVadras,1998)
pnd transお m轟 on and
i:il::I[￨]:I:[1lF61lf211
∴c― Ar tAp l=[.12 lf司 [112
(Prゅ ac,198'S)
(Moご ras,2“ θS)
・
.20. COMPTEX MAIRICES
ntrix
Itt
Since r X≠ 0,it fouows from(fi)
X=λ This shows that
λis
胤幅露1獄淵:曲試襄:摺躙鯉犠獄 ltt°
『
4i
*'
)
lA″ ごλra,2θ θQ
(j) 3rr2 + 3rz2 + 3132 + 2rrr2 + 2r1:
s-?,...Jit+ ly2 + z2 - ary - 6yz + 4zr
fi1
*'
(じ
(Moご r● 8.2000,ハたごぬra,1998)
4xy to e 'sum of squares'.
reduce the quaむ att fOrln,■ 2+&22+缶 32_2"to
[,
...(正
HencerAX=λ rX
め
But晨om(じ ),7AX=潔 X ..。じ
1:landhencereduCe
and the m嗣
=
′
therefore,X■ =冗 π
and(二 j二 )that
… 卿…ヽ
A'
.
●
´
.A “ヌ =天 X Or「乃′
=冗 r
Ａ
Ｌ
Ｉ劇劇
ЛＴ
口
-
gince
"
∞responding agen vector.Then
―DX=0 0r J【 =Ⅸ …
(Cο J"bar● に,1ま ,θ )
7.Find the eigen V∝bFS Ofthe mat五 XI―
+ 3y! + 322
""*pf*;njrigate
;1'J.
;;il matrix,
(4)Unitarrymtrtr Asquarematritllsuththattf =WLiscalleda"nifirymatrir'This
住 W五健 down the matnx Ofthe qua由にtic formェ
12+ヵじ
22 7ェ 32_4riェ 2+&1ェ3+5x,3(P● ●●α,1990)
6l
;
Hermitian matrix A andX be the
１０４
: 1 llCalttdateA4.
1
are compler numbers
is a generalisation ofthe orthogonal matrix in the complex 6eld'
Eremple 2.11. Show thdtlhe eigen ualues of a Hermitian matrix (and thus of o Etmmetric
醐コ″ rd_β ″ nな 2:″
thematrix
１引１刊“
１１
饉
・
)
=L _r--
ル LMユ ,199η
““
Let λ be an eigen
value of a
ｌ
ＦにＰ
3.D■ agon3■ にO the=nabi*
:
A=[a,)
of a motri*
sqtalre matri,r A such that A'= -i is said to be d
that the leading diagoual elements of a skew-Hermitian
matrix are either all zeroo or all purely imaginary.
obt. A Hemition nntriz ic a geturalisation of a rual symnctic mat'ii,. t8 every rsal sJmmetric matfix
matrit
is Hermitia!- Similarly, a sleur- Hirmition t*tri,t is d Ecr'n,r,,li.,t;,n of a tat shew'slmm.tric
2. Show that the linear transforzration
4.IfA=￨―
If the elements
qnd 0E 6eing rcol, tlen tlv mdtri,
(3) Skeo-Eermittan metrir. A
eLew-EerDitiann"tri* ttbimplies
Problams 2.13
(Caric叫」脳
q.
(2)
I s -1 _r[rl
a=l-r
E
rJ
-r
i_r
lhe eigen valuee ofA arc 2, B, 6
j司
p,,,
f
(ii) The matrir of the quadratic form is
0臓
d
= Enl = too-ipJ is catled rrre codugate matrkofA'
Eermittrn rnetrir. A square matri.r A: such that A'=V Is sald to 6e a llermltian
mahir.r lte elemeuts of the leading diagpoal of a Hermitian matrk are evidently real, while
of the element-in the transposed position. For
every other eleaent is th"
t indzfinite.
| -rtz -,ts/z
1.IfA=l-./3/2 rX
+
31
MATRICES
S,° nJuFtetranspose=:[_51・
21「
lf21+6=:12-1 1141…
0
…(jj)
]
(1822-1901). known for his contributions to
-... . N"-"a
the French mathematician cJrarJe s Hermite
algehra and number theory.
far, we have considered matric$ whose elements were real numbers.The elements of a
can, however, be complex numbers also.
"ft*
―」
の
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`■
・
￨■
,￨
３
ｉ
〓
︲
ｏｑ
６０
〓
１一
“
２
司
４
４＋
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・
[211
司ｊ
一
md liil=轟
´
一
¨
一
一
．
．
一
＾
”
´
一
∴
Prod“ tofω
Hellce the result.
Problems &14
I s 7-4t -2+fi1
l-Showthatl 7+{
-Z J+i licaHermitiea}letrir.
2'
Prove that
:
Veetors @nd Solid, Geometryt
5i 3-i
1f
{a) every Hermitian matiir cau be rrritten ar
l-z-
8 is real and skew-s)rmn€tric :
l. Vestor:. 2. Spece coordiaatcr, Direttiolr cosines. s. Section formulae' {-6. Producte of two
vectors. ?. Phyn".l applications. E-fL Produ.ts of tbroe or more vectoE. 12. Reciprocal vector
Eia&. f9. Eqlr.tio6 oii plane. ld Equati,oas of . straiSbt iine. 16. Condition for a line to lie in
a dale. rs. coplamr liaes. l?. s.D. betweea tro linee, lt. Iutersection of three planes- 19.
Eguation of a sphere. 80. Tangeat plaae to a sphcre. il. Cone. &l- Cyliader. 2tl. Quedric
surfaces- 2{. Surfaces of revolution.
A + iB, where A is real and symmetric and
( p.T,ll,, tggg)
(6) ifA is a Hermitiao mstrit rA is a skew-Hermitian
metrir
3' show that e llermitian matrir remains Hermitial when transformed
{.
5'
ll{enrla, 1990)
by an orthogonal matrir.
Show that : (c) the inverse of a uaitary matrir is unitary.
(b) ttre product of two a-nowed unitaqr matricer
is udtary.
Show t}at (o) the eigen values of a skew-Hermitian mstrir (ard thus of
a skew-symrnetric matrix)
3.r. (r) vEcrons
arepurelyimagin&rJrorrero.
(Kemla, lg9}l
(6) the eigen values ofa uaitary
matrix (and thus ofau o,rthqonal matrix) have absolute yalue l. (See
FiR' 2'l)'
il{rrnpur. t99B\
n**,t.r|ff Ii -l1i]o","rr"o-",*
(coticut. leee;Andhm. teeot
一
Ｓ
ｌ一３
，
鮨
¨
″
２
“
…[
山
１
１可
１２
・・
e.
O
A quantity whicb is completely specified by its magnitude ouly is called
a scala.r. Lengt.h, time, mass, volume, temperature, work, electric charge
and numericd data in Statistics are all e-onples ofscalar quantities.
A qrnntity whbh is completely specifizd by i* rnognirude otd d,irection
is calied, o veclot. Weight, displacement, velocity, acceleration and electric
cutrent density are all vector quantities for each involves maguitude and
direction.
ノ
ltus fr represenb
magnitude is the length PQ and direction is from P (starting
point) to Q (eud poiat). We denote a vector by a siugle letter in
capital bold tXpe (or with an armrv oD it) and its magnitude
0eneth) by the correaponding small letter in italicr type' Thus if
V is the velocit5r vector, its magaitude is u, tbe apeed,
ir vector ofunit magnitude is called a unit vector. lbe idea of
rrnil vsctes is oftou used to represent coDcisely the direction of
any- vector. Unit vector corresponding to tbe vectorA is written
a, 3.
A vector of zero magnitude (which cari have no direction
associ,ated with it) is called a zero (or null) vxtor snd ig denoted
A vector ia rcpr.etented by a dirocted liae segrnent.
Fig.31.
a vector whose
The vector●
メ
by 0-a thic& zero.
//
′'A
Fig.32.
員ia‐
郷
行ぷ脇;a潮″
"瞥豊席薄
presmだ the negatveご
“
e
解T
留″
Clearly the vectorsAB,二 r and Po are all equal.(Fig.3・
2).
が響贈管搬鸞帯1琵里鞍蟷
spectlvdy thenあ =c L cdbd the sum or resulね nt of
“
A and B.SpboLcally,we tvrite,
A
P
C=A+B
H9.3■
83
句l
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――
・
―
84
HiCHER ENCINEERING MATHEMA■
CS
O ber"“筋 ′
(3)Subtraction ofv∝ tOre.賀晟 s“ ιttttあれOfa υec“ rBゎれ Aおぬ巌れ′
ぁれ
A αnd“ ゛″だた
in the direction ofox rotatea from oY to*ards oz, therefore'
Notc. As a riclt-handod screw uoviog
-*ldt-nt
aua syde, ltre left'haxled system will' ho,ever' bc
the above system of oro i" ofli
or― Bゎ
I
A+←
B)=A―
35
VECTORS AND SOuD GEOMET鮮
"
B
(4)DEultipucation Ofv4ors by sc」鑽。
We havejust seen that
and
A+A=2A
―A+← ゎ
=_2A.
where both 2A and-2A dentt vectOrs of=naptide tmce that of A;the fomer having the
1醐
same dlrectton as A and the latter the opposite direction.
″
″
ふ勝識脇脇胤 t脇協見
χ
協島
″篤鵠解協島r
ThuS
A=aA
・
α
′
協∫
電
奪
威れ
″〃出競
諄
郡』協揚r嚇
d byぬ
derタ
raprasaれた
οοrル rsゴル
`た
恣糧瞥 1肥掛踊蹴鮒 :
判
Fig.36.
んο
れれ。
Let ABCDEF be the given hexagon,such that
島 =A ttd厳 =B
→
→
→
AC=AB+Bc=A+B
Also AD=2BC=2B
め =島 _晨,=2B― ぃ+Bl=B― A
→
→
Now
DE=―
=_A
∴
I・
“
→
.・
盛灘
ぬ at
AB=and‖
ED〕
nen R=ル =泌 +ル
→
EF=― BC=― B
and
→
a=κ I,品 =yJ,あ =2X
日9.34
1・
.・
→
FA=― CD=― o― → =A,B
[・
.・
ニ
a+鳥,+あ =ユ +晶 +あ
BC=and‖ 属硼
CD=andll期
Hence R=ェ I+yJ+ZK iS Called the pο Siι
relatlve to“ gln O and
3.2. (1) Space co-ordinates. Let XOX andYOY, Z,OZ be three mutually perpendicular
theu O is called the origin.
xox is called, the r-axic,, IzoY t/re y-axiq zoz the z-atie and tahen together these ore
called the co-orrlinato ares.
1...「
Irt
られ
的
and ry-planes. Theo trP, MP, NP are
3.5).
OZ
respectiuely
,'.
the
=d+Yil +zK=r(ll+nrf +nK)
which etpresses a unit vector in terms ofits direction cosines'
ond negdtiue alorq
The three co-ordinate planes divide the apace into eight compartments called octants. The
odant OXYZ in which all the co-ordinates are positive is called the pocitive or first octant.
R
ff,=E=il+mJ+nK
or
ｉゴ
_ The * y, z co-ordinates are
ox,oY,oz'.
If
co-ordinatcs of P.
poeitiue alons OX Oy,
ルπ
IfZ,L π ara由∝rio● cosE2es oFa″ crOr R ι
Prnf.LetDbthefootoftheperpendicularfromP(r'y'z)ouOY'I\en
y = @D = r cos p = rzr' Similarly' --'*' "ad a = lr'
I
F19.35.
.3a
2.
(Dfi,=II+m.I+nX, (ii)lr+m2+n2=1
respectively called the coordinates ofP (Fig. 3.5). Thus the
co-ord.inates of any poiat in space are the perpendicular
distanes fum tle yz, zx and ry plones respectiuely. In
practice, to find the co-ordinotes of a point P d.raw PN
to the ry-plane ; from N draw NA -L to the.t-axis (Fig.
OA=* AIV=y, NP=2, then (r,y,z) are
230P2=ο σ2+cP2=OA2+Aα 2+c7P2
H毎
tie co-ordinate planes.
P be aDy point in space. Draw PL, PM, PN l-s to
zr
Or of P
υ ′
“
慨
ι
%
服
ツ
脱
孝
整機撤轟
Tlw plane YOZ is called, the yz-plane, the plane ZOX
r/re zr-plane, the plane XOY thc xy.plan€ and tahen
the yz,
"几
r=IRI=マ ←2+y2+z2)
lines which intersect at O.
together these are calLed
日g.37.
1= lfl i='J(r2 +*2+o21 ltusl8+n2+#=t
b o, b, c, theu these
obr Directlon retioe. r 0re directioa cosiaes of e lino be proportional
iile'
ofthe
ratioe
ait"Jim cosioes o. direction
;;"
eropoJioJ
0
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are called
﹁
﹁
﹁ :上
__
_
Et tNO旺田 К MAIEMAnCS
Hふ
t
卜嵩 ′=念 μ
=嵩
マ【
CX2 Xl)2+ty2 yl)2+(22 Z■
2・
We have
■We have 礎農 ="1轟
Or
m2(品 ° め〓 "lρ Q一
OP=,11+ylJ+21K
→
οe=・ 21+y2J+Z2K
.・
Or
→
→
→
.
PQ=OO-OP
=
(x2
;1)l + (y2 - y1fl + (22-
-
-+
Thus
d=l {e
z)K / X
1=J(r2-r1)2 +(yz-yr)z+(zz-
-+
and directioa cosines of PQ are proportional
b
x2
-
x 1, y 2
-
- zt
α
Since
ofAi
.'. Its direction cosines are
l.
and
are
-
1+
4, 6
-
9, 6
-
Giveo
B1 + R2 and 281
.
r) ; (2,3,
-
eo弓 CA● 助
Rこ宙颯れ j山 dm
2L競
uero.
zero.
_
?'T\soforcesartgivenbyP=3I-2''andQ=4I+3J-2EdetermiaoaforceBguchthatthethree
forces are in equilibrium.
8. If A anil B are non-collineu vcctors and P = (2r + 3y - 2)A + (ar + 2y + 5) B
IRο r陸晨
ち1990)
and Q=← ″+ター2い +(■ -4y+7澤 ,mdr.y Such that 7P=3Q.
gnd“
proportiona!
9. Five tgrces act ag one veriex A of a rrguiar hurag<ra ir the directions of c'"her
"'ertiees
to tl.e distances of these vertices fromA. Fiud their resultant.
器
禁
B alld
We notethat the mln Ofthe coerlcimtt Ofへ
if
)A+
3&お
wherc I
'凝
1rB + 1C = O,
+ P + 1= 0'
',お
鸞藷山潔
(?i押
淵
1滉品 Other
こ
3甘
Bc.T∬ るT葛夏高訂五
AttB椰
お,″ irわ
"“
"ガ
にα
vettitt be c
Ihe
ヽ
"
」
iH"。A,therefOre,
B_c=あ =減 =凛
f) ; (4, 5, 0) and (2, 6, 2) are thc verticee ofa square.
(Osmania, 1999 S; Battry.alore, lW)l
5. A straight lioe is hcliled to the a:ee ofy end z at angler of 45' and 6Oo. Find the inclioatioa of the
Eag to Ures-r-is.
6. If a line makes aogles o, 9, I with the ares, prove that siaz a + sia2 p + s b"l = 2.
(V.T.U-,2000 ; Osmanio, 1999 ; Mdrcs' 1991 Sl
("1+"2澤 =Q
4 dly｀ R義
veturs A B cnd C an rrlrlitear
it folloss tbat aty thta pointc with rrlsitbr.
Ex―pb
and directioo cosbes of the vectors
ratOnⅢ Ⅲ
and0 0 L伽
巌
お
げ
″ イル …
た
_
昂写
獲
Iれ
B
offii';,;r;;,;;-;;;il'i tu "*YY.y**
i獄 ““c
Heace
1).
l,
ι
2
転 LttP● in30rmcnd
B is
*, - +,a
2. Find a r.nitvcctorparallel to the eugroftbe vectors B1 = 2I +,LI - 5K and82= I+ 2I + $f,.
3. Find the locus of the poiut whoee distalce from the .t-eris is twice its distaocc from tbe point
(r,2, -
型
ガー
01漁 RewritingO o8"2A+"lB―
- 82.
4. Show that the poiats (0,
日9.3■ 0.
R=,I+yJ+ZE
…
6.
Problema 3.1
Rr = 5I - ZI +,tf' and B2 = I + 3J + 7K, frd the magnitu&
0
B='21+yダ +Z2K
Equating coemcients ofL J・ L We get the desired results(J).
ac =./t(o + 1)2 + (? - 6)2 + (10 - 6)21 =3./2
cA =.i (- 4 - 0)z +(9 -712 + (6 - 10)21 = 6
AB2 + BCz = CAz errrdAB =BC, it follows tbat MBC is a right-angled isosceles tri-
angle. The direction ratios
R="lB+“
θ)ルr72α 腱多
`
あ
わ =“ 1(B一詢
ハ
since A=“ 11+ylJ+Zl氏
zt)zl
y1, zz
"2(R―
when∝
n多 3a
Example 3 2.Sあ ″ι
′
議 ι
ル ρoれたA←
"dC(QZ」
`,aO,D(-1,Cの
d`屁 arec琵。
2 cosjnes oFA3.
αng」配 isos“ Zes"cagJa Arsο ″π
We have
rB =./(-1 + 4)2 + (6 - 9)2 + (6 - 6)21 = 3./2
and
ο
二
11ド
Pandol留
d di"cttOn r寝饉 o30fPO a″ X2 Xl,y2 yl,z2 21
and
rtt rcι
お J眸 1,yl,21)● P3d B02,y2,2D:え
“
“滋
=七
器・れ露=“ r Re
)8】
→
酬
po三
i薪
(4)Didance bneen""pOintsR11,yl,Zl)″ ガ QO,y2,22)お
￨
FORMULAE
r陸 ″ :れ ′R← ,y,z)aυ 晟昭 ,たヵ加 Or'陸
:=サ =:=掏鵠 =意
I
■ SECnON
甲卜た
苺
≒
糊弁購蓄甫憲驚載
ぶ
…
鶴
観崎∬
レガ″
3・
A
and AC are Dd3/a
mid-Points of t'be diagonal's OB
andE(A+9/2.
'.
trYom
in=dB-&=|tr"cl- )=:IA― 偶 ―釧
fr_ru
to OA;鮨 Om(j),i
0il, it is clear that D? is parallel
Ilerrce the result.
www.Engg-Know.com
恥
folloWS
3・
11.
¨Kj)
…(￨二 )
﹁■
88
HICH[R ENGINEERING MATH[MAFCS
&Fiodtheratioia.shichthelinejoiaing(2'4'16)and(3'5'-4)isdividedbvtheplane
tM!'ore' 1995; Madumi 1990)
Example 3・ 4。 Sた o″ 働α ル Jれ ′ヵご
πlng o″ ″諭醸 aF a
`′
pposだた stt ι
JcJogramわ ′
をガ‐
rお ι
s"e
Poゴ 2′ oF aル ●
′α臀′
“
ォ
絆″ “ ′
ed′ ル″″
dれgο πご ωガおニ
`法 “
Consiler a parallelo_OABC
Ta■ e O asthe origln and let
2r-3y+a +6=0 .
9. show thatthe tl'ee poiDts
.
N°W SinCe OA is equal to and parallel to CB.
メ
writ*" uryff|S=f
Erample 3.5. The rzsultant
O is R.
=
e
so
If any transoersal cuts their
lines of action ot A"
at
*crrr
ot tn" poiot
called the inoeztre'
ffi'
C
12. UG,
B, C
m
. frr+rz*rs. rr+r-2+ys,4+z:+zrl
-' 3 J
L---T--'-'s
the ce,atriids of the trianglesABC
andA?'C'
13. Show
*.*=*
(*,
=.t &=4, o?=c.
Then the unit vectors aloug OA, OB, OC ate LIOA,
thst the co'ordinetes of the centroid of the
,!, , z,):
lr
I frr
+ 12
r = 1, 2, 3, 4, are
1
1.
+:3 + rr), j(rr +Jz+Jr +rd ; i(zr
+
22*
23 +
{:r'yt' zr)' @z'!z'2il'
prove t'hat
i'r*Bi'*dc=a&''
respectiuely, shou th.at
Letol
"ldt"
triaagle whose vertices are
11. Shorr that the co-ordinates oft'he ccntroid ofthe
(zs, Y3, z$ are
that P trisects DC and oB
of the forces P and, Q acting
\N. Bengal, 1989)
antzid'
(ii) interoal bis€ct rs of th.
→
04=CB,,.θ ,A=3-C
which may be
zt+&r-4E, - 7J+ lOKarccdiaear'
*
The面きphtD ofOAis A/2
∴
gtr,
C ofthe triangleABC' show that thc three
fO. IfA' B, C b€ the Poeition v6ctorB ofthe verticesA' B'
Blopal' 19t)I ; WamngaL 19891
+ B + C)' calhd the
(i) medians c@qr Or noint
$A
the other vettices be■ cAl,3(B)and c(C).
→
r-zr+
tetrahedroE whose vertices are
2t ,.|.
t
tDef. A ictEahedroais a eotid bouaded by fout triangular
B/OB, C/OC respectively. Since their magaitude are P, Q, E,
6;;: iil;il
therefore, the three forces P, Q, B are
tet*,"a'oo ABOD ha' four faces-trrc
ABC, ACD, ADB, BCD. (Fi8' 3 14)'
, forrt o"rtioe B, C, D aatl three paira of
L
itn
=*o+e=#r,*=#"
"
As
oppodte
Now R being the resultalrt of P and Q, we have
goBB
- $ c =o
# "'
But
8,
colinear,
the
the
Hence
*,.& -ft=o
#o.
.4,
#ru
=
C are
therefore,
Fig. t 14'
(Cochin, 1*)9: BttoPol' 19911
…(j)
oA-4A*
coefficients of A, B, C in (i) should be zero.
sum of
prove
ls.
--
Problems 3.2
l.
Prove that the line joining tle oid-poiuts of the two aides of a kiangle is parallel to the third side
and halfOf■
・
●
●do,199θ ′MB￠ agα ι1989)
(M●
2. Prove that (i) the diagonals ofa parallelogram bisect each other; “`λ
(ii) a guadrilataral whose diagonals bisect each other is a parallelogram.
3. In a skew quadrilatcral, prove that:
(■
)thejOins ofthe midl"hts ofopposite sides
4.ABCD is a parュ
iC―
BD=四
Ы
If ABC is a triaugle,
ef
CC
i"
"
Ls脚
parall.-eloeram and Q is a fired point'
:r謝球 :挙導w温
-r
-r!-^ r.ad^,q
H:::H'.,t*:I^tl:::[T:;fiS
I轟 '鍵屁
温
il錦ッ九 x露 ∬
FI鰍
羊
揚
凛rttlttLIIII観
器轟じ
L淵話
i落 ■覆
}lⅢ
lr菅ュ
l宦I竜
ξ
∬躙
群砒ぶ機篤品
雪 ■ ■
∫薯
庶お.1:
"は Lぬ
l∬
FII誼
IFttT,壬
A■ 為 1「
:fぶ
鶴 L√ l燎
滞扁農馬
A.」性:望葛ぶ獣置』:胤逸 ∫卜苺 `L_1_▲
ι^_● ■
●
_ćross
_.● producf'
″′′
the
"力 ^″
latter
the
and
(i) the figure formed by joinirythe mid-points of the adacent sides is a parallelogtam.
(Mcrati“
`do,1990)
former is sometimes call"d i;;;;;uct
another vector
In vector algebra, the division ofa vector by
èct each other_
llelogram and AC, BD are its diagonals. Show that AC+BD=2BC,
‐
is not acEned.
.
5.In a tFapettun,prove that the straight inejoining the mid‐
bry
(m+n)Ol,whereEdividesPQsucht}atn'f8+'EQ' -:f ..a
otlh' pb *^r#;*;:rtit1\
let P be anv poiot on BC' If P-6 is the resultaDt
3.a. PRODUCTS OF ilro vEctoRs
Unlike the product of two sJars or that of a-v,ector
.
t}at their resultant is giveu
points ofthe non‐ parallelも ldes is parallel
ll粥覇鱗盪漱霧翻鯛
to the parallel sld● 3 and halfthelr sum.
6.PrOve that the vectOrs A=31+J-2氏 B=_I+3er+4氏 C=41-2J-6K can form the sides of a
triangle. AIso find the length of the madian biseting the vector C.
じ Maa,I"5s)
7. Find the ratio in whidr xoY plaae divides the join of the points (- 3, 4, 8) and (i, 6, 4) and thqs
write the cmrdinater of the point of divislon.
A and B.
Thus
の
www.Engg-Know.com
A.B=obcos0.
F19.315.
e
VECTORS AND SOuD CEOMErRY
inteqletation A.B is tlu prcdwt
(2) Gconctrical
l.ergth of thz projection of the otlur in thz
-)
OL = l\
I-€t
-+
OM =
B
din*ion
of the leryth of onc veetor aad thg
IfO be the argle betw€etr the lines, then
of the former.
or
then
A . B = ci g6 g = dOM coE 0) = o(OJV) = I .4 I prd. of I B I in the directiou of A
Similarly, A. B= | B I Proj. of I A I i" the direction ofB.
(8)
Hence
Coa l. sins 0= 1-
Conversely when
A.A=′
A. B =0,
aD cos 0
=0 i.e.
cos 0
=0
which is witten as A2.Thus′ ル望
for tlw squan of i.ttmogniludc.
IV. For the mutually prpendhulu unit
I.J=J. f,=E.I
12 =.I2 =82 = 1
which are ofgleat utility.
V. Scdar
prdttct
of
.・
α≠0,b≠ 0),Or O=90°
uectorc
I,
4
cia
aad
or
血
d″ crjoa
perpendiculars to the line of
口Ln we have(A+B)。
(B輌 .ofPR inぬ e direction OfC)
C
C
C W
C 爾 +力 0う
= lC l(Proj.ofPQonC)+ lC l(ProjofQ8onC)
=A.C+Boc
π Sc力 ″ j“
: IA・ BI≦ :A‖ BI
“
"Jが
・ I COS O I≦
IA・ BI=I A ll B llcosOl≦ :A‖ BI
・
wI. sul.ar prduct of ta,o uetors b equdl to the sum of tlu prd,ucts of the.ir orrcspnding
[・
cο
ε
j2● sα
l,
""""●
(Gulturga, 1999) ...(i,
m, n and J', m', z' should be
aCaだ ol b′ ,C'鮨ゴ
reら
π
“ "
0うマ
ぃう
0■
沸 =,lI+ylJ+zlL CO==21+y2J+Z2K
∴
昴 =← 2 '1)I+02 夕 1戸 +● 2 21)K
A180 unit vector U along the glven line is′
..ProiecttOn OfPQ on the giVen line=Pe・
=′ ←2
+ &zJ +
then by the dirtributive law, A . B = a1b1+
Then
and
+ osbs
た。waた crぁ 2c“ [″ sa″
∴
U=II+mJ+nK
31-J+2K
“
υeガ ces a“夕I-2J+2K,
●屯蘭ご力鳩 199`)
BC=1-2・ I+3K
→
CA=-21-J
AB=I+3er-3K
BC=ヾ
14,CA=マ 5,J認 =マ 19.
｀
、
Nび町d.ab ofA3 and AC being
l, m, n ond l',m',n' k
rS h the ttredお ns of the」 Ven Iines are
U・
yl)+4● 2 20.
Leta=1_2J+Ж あ =21+J― K,あ =31-J+2K
1〕
6rK
azb2
″ 1)+″ 02
1+洒 +2K
Lample 3・ 3.Fiaa rた .sf&s aP2d angres oF`晟 rrilmgJe″ あ
21+J― 氏 oだ
a r力″ ″わ se
″ ι凛,2お
′
←2 ″ 1)+"02 yl)+2C22 ZD・
Lt
conporw tts.
"3 101′
vttl房
tltt°
rrl,siltes arc
′ ′ ′
鋼し,a′ お ●
9ビた●ICr ra● +“ +“ =Q(ツ p″dL`″ ●/c′ =b/b′ =c/c′
"`ヵ "ヮ
“
五【 Pい ● ecttOn ofthe u"● JoL.ing● 70p● ints← 1,yl,Zl)cP2d● 2,y2,22)。 7t
Fし 316.
action ofthe vector C。
+nn)
dinction
.
I
r員 =A and轟 =B thm屁 =A+B.
+コ田
m′
whose
ll'+mrn'+rn'= 0
l=l', 1a=p',a=a'
prallel u
麒 °=マ
=0
In ptti雌 ,A2=。 1+α ′+α 3・
唖 _Angle between two heO“
=t 'lZlm*'-nm')2.
0
nese cod● 。ns easily follow from(j)and(三 :)_
Cor.3.2陸嘔梶 06e"“ 咋r● ●ri26″ ヵο健西腐あ
.
ruo wctors is distributiue i.e.
+ asB,B = 6rI
+mm'+nn'\z
perpendicular is
K, we have the
Draw PL OM and RⅣ
ifA=arl + ad
l-(l!
Cot. 2. Ttrz condition tha, thc litpe
に Orα υ ゎrぉ αscα ′
″ ωλおλsた 22A
“
“
relations
and
(・
tlut ttvir
8=
綱渤
The rccessaty and sufficbnt conditian for two oectorc to be perpndicular is
scalor plnd,uet slwuld, be zero.
When the veetors A ard B are perpendicular, A. B = cD cos g0. = 0
cos2
=112 +m2 +n\112 + ma + r,,zl- {rl'+ mm'+wz)z
=1rn'- nm'12 + (nl'-ln* + (tm' - mt)z
,'.
IL
For
U.[I=@+rzJ+ng.(11+m'll+n3)
1.1.os0=ll'+mm'+nn'
w0=ll'+mtn'+nn'
Propertior and otherrecultr.
I. Scalal pduct of two vectors is commutatirn
ie. A. B = B.Afor A. B =aD coa 0= Da cos (- 0) = B.A
鳳
,
1/マ 19,31ミ19, _3/ヾ 19 and 2/ヾ 5,1/ヾ 5,0,
vre have∞ sA=島
and
L.e.
r
Named after ttre Geroaa mathernaticisn HermannAmcndussctpcrz O843-1921) who is lnora for
iris-wolt ta courolurai mappiuS, caicuiur ofvariarioas anti <iifiereniiai geomerry. ile cucceded Waicrstw
in Berlin University.
\tre
3′
・
0=t5/19
鳥喘+詰・
七十
zA = w- r'JGzr$. fuain d.c.'s of 8C and BA being
l/'h4, -2/114,3,/./14 aDd -t/119, - 3/.i19, 3/{19
i -i .ffi
3 3
-2 -B -da'
have cosa=dI
!i9
ffi
ffi
0
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′
′′
｀
｀
F19.318.
;
=\'(14119)i'e' zB
=aa ^{(14lrs)
―
92
VECTORS AND SOUD GEOMETⅣ
HIGHER ENGINEERING MATHEMATlcs
+→ ;あ ζ
轟Υ
δ
=轟 C― れ
七C+"― →
〒
nndし d.c'sofCA and CB behg-2/V5,-1/も ,o and-1/ヾ 14,2/C-3月 14;
∞
8C=静
∠
・
C=∞ ヽ
青+モ・
台+← 希た
we htte
93
=。
∞s2α +cos2 β+cos2
Example 9.7. Proue that the right-bisectors of thc sidzs of o
trian4le concur at its cistuncentre.
L€t A(4, A@), C(O be the vertices of aay triangle ABC. Ihe
mid-points of the sides BQ CA and AB are
1+"′ δ
=:“ +れ
+“ +ぽ +c― m十 →2+c+"_→ 勺
+が +← ′
..`2+凛 2+″ 2=1]
「
Example 8'9. &oue tlwt the linzs whose direction cosines are giuen by the relations
αJ+b凛 +ca=0● 油′れれ+れ ′+Jπ =O are
(■M、工El,1997Ⅵ ″
rar ra-1+o 1+c 1=0
)″ rpe“
`滋 “
(R● Iva,1990)
0)pa嗣 ″
FAla+も +t=a
=:14Kr2+″
,[of),(+'),[+r)
2+22】
=:.
(′
Let the perpendiculars at D and E to BCandC.4, respectively
irtersec.t at tbe point P(R).
i.e.
and
fn"o.#. ib
=
Or
…(J)
ip.de=o,i.e.t"
or if,
(a-c)=o
..。
Take O, a corner of the cube
three edges througtr it, as the aree.
or it
j.●
(7■ a,2ο οI∫ Os"り πれ 200θ
OB, OC the
cornent are as shovn in Fig. 8.20.
lte four diagonals are
or it
or in
l*'on
and
' , "
ofAA,
uill
be
parallel if 11= 12, mr=
-#+,
nr=
n2-
ffaking necessary signsl
Problem8 33
テ
l=げ
絆―
3.Interpret geomet五
and C=31+J,fnd r such that A+`B Is perpendicular to C
l・
ca■ y(C― A)。
4.Ifl A+BI=IA― BI,shoW
(B‐
C)=0.
that A and B are mutually perpendicular.
5.IfA=I+2J-3K and B381-J+2L show that A+B is perpendicular to A―
angle between 2A + B and A + 28.
‐C(`1,凛 1,■
●on∞
6.Show that the hee conc― ent hes,艶ぬ
-
8u.a'"*,
1sz,
if, (c - c - b\2 = Ao;b
c -a-b=t2{(ob) or i{ c =a + b x2\l@b) = ({o i{6)2
t r/c =.Jc t Jb or if, Ja + Vi + rlc = 0
2 ShOW血
"* +,*, +,
+:+:)=0
l/m1= !r1v1, ir.
1.IfA=I+2J+颯 B=― I+2。「 +K
1 1 I
.r15o1''"' i5' m' G'
cooines
the
‐●
∝ 携=号卜携
・
り
・ y Wmmd朝ル
r rり 2+"1● 2+れ r2=1←
the
o-O a-0 a-0
t*l-"
11
s)
OP,AA', BB' andCC'.
Clearly direction coeiues ofOp are
./(ial' .J(rr1'
α
1/れ 1,′ 2/m2are
:・ :+:=0。
The lines
4/8.
Let OA=OB=OC=a, Then the co-ordinates of
b
"′
with d,ia4onola of a cube, proue that
6
′
2
‐
′
′
α
と
れ
面
壷
甲
“
“
“
“●
)・
"o* =
as origin and O4
(1).
′
1
―
轟
ha
う=Q洒ぬ
ザい
←
...(1)
/11illiflil;￨￨￨:[￨°
れ 1 れ2
(A-R),=(B-R)2ori{A2-2A.n=*-Zn.n
p, y, 6
(′
r∞ts ofthe quadratiё
(jJ)
*" c"t (B-A#),n-r, =^l
=o
Erample 3.8. A line mahes angles a,
co* a + co"Z p + cos2 y +
。
2+“ _α _bソ凛 +われ2=0
If′ 1,"● 1,21;Z2,′ ●2,P32,are the direction cosines oF these lines then′
which shows that .EP is perpendicular to Ag. Hence the recult.
Further PA=PB if I A-B | = | B-B I
orif,
a tom the given relations,we have
ra′
h-E+)
z) (c-B)=o
I
Adding (i) and (ii),
Ehmina血 g
日9_319.
O
日9.320.
1)。 (′
…
`:‐
B.Also cal“ late the
(MaraJLω ●da,199θ )
2,“ 2,● 2),(亀 ,● 3,■ 3)are
ce"*+,*,-+
l*t
L m, n be the direction cosines of tbe given liae which makes angles o, p, 1 6 with
OP,AA',BB', CC respectively. lben
.
cosrr=fr4*-+z);cosp=frt-f +m+n)
9。
Show(by vedor methOds)that the mid‐ pont of the hypotenuse of a五
equ■distant ttom its vertices.
ら
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ght‐
angled tnangle iS
可ト
lO. Prove (by vector mathode) thst the angle ia a semi-cirde is a right angle.
11. show (by vector methods) that tbc diagonak of a rhonb.s interscct at right angles.
12. Show t}rat ttp altitudes of a Eiangle meet in a poiDt (celled tlre oaio.ccnrn).
t3. ABCD is a tetrahcdron haviag the cd&c ac ald AC at right angles to opposite edges A, and gD
respeciively. show tlrat the third pair of opposite edges eB ana cd are asi'at right ingles.
lAndhra, 19!N; Ranchi, 19901
14. Prove by vector m€thods, thet the sum ofthe squares ofdiagonals ofa paraltelogram is equal to the
sum of the squerec of iis si&c15. If A B, C srs coplanar vectors and A is paralel to B, prone that
Hl^.li$ 3fl,}l
"={l8f
Ax
I.
IL
16. Find the angle betwecn the lines whose direclioa cosines arc giv:n by the xluations I + m + z
= 0,
蹴
〕
殺
SWb“
正面
¨
盤
菫
"sa…
(Maご ras,2θ Oθ rりつ
by the equatio■ sI+m tt rl=o,
は
Mi二
『篤鳳尊
驚已
Ls蹴蹴
leequa赫
ar。
いけrf,卜 Q(柵
…■
mi氏属
亀
阜
lil[:h=°
手
,胤
"息
E塁罵躙
uector
are given by 3J+"+詢 =o and
"n
Vector
prduct
zaoea N = %
ois.
of two vectors is not eommutatioe,
Ax B *B xA In facl Ax B =-B xAAxB=aDeiDoN or %OiA
H9.321、
BxA=aDsin(-0)N=-aleia0N or Zt Oil.
The necessary and suffuient cordition for two ,an-zeno uectors to be paratlel is that their
sin 0 = 0, and as such Ax B =
Conversely,
O.
when AxB=O
;a6 sin0=0
8in O=0
="η
1∬
=
pttduct should be zerc.
'when
the voctors AandB are paralle! the anglo 0 between them is 0 or 18o" eo that
12+m2-n2=0.
1&》
uaio elx
\z)
hoperttec and other recul-ts
for
and
rA2B2-1A.svz;
17.Fhd the angle between the lines whOse directiOn cosln“
623n-2■ +口 ″ =o.
r(*
=
(3)
'5
B =cD sin 0N
Or
″I
For rrle cP・ rr・
「
IxI=JxJ=KxK=0
lXJ=聰 JXI=‐ K
JxK=L KxJ=―
I
. KxI=J, Ix K=― J.
どυ ゎrp趙 c亀
:こ
￨
2Q Sho中 that the angle beHeen any節 dlagOna13● fa cube iB cOs 11/3.
コに洒に昴
('.'a+0,b'*0)
0=O or l“・ .In particular,AxA=Q
.rtora.z υecr● ″ヵdLJし】
町 We have the relatiα
几a聡
":
"``".catar aれ
We have
ぃ
“
“
.B)2=aち 2c。80=α 262_α 2ぅ 2 sin2 0=.262_(AxB).(AXB)
6.B)2=A2B2_体
∴
x3)2
2.
cA x B)2=A2B2_ぃ _動
e norm蹴
織通
rFA=a11+α 2J+α 3氏 B=blI+02J+。 3K
・ 6.VECTOR OR CROSS PRoDuCT
(1)DeanitiOn。
=み
力α
cじに
な′
。ぅ。rcmssPttJ“ cr Of′ ″。町
"ぉ
Aαゼ Bお d≠ ″″
αsa υ戯
λ
a y f,
B,
Example 8.f0.
oectorc
.
o*tors
“
NL a untvAOrnorlni tOtheメ aneofAand
Bに L NbrmingaHghthanded system、
lfN be a Ⅲ t Vector normalto the plane ofthe mangle●
3riattθ
A3,then
alld
r.U五 t
“
お勉巌
′
e.ectOr areaげ餞
●
・
︱轟
議F“
Aand,B.
Since
︱
五
υ
躍胤蹴灘躙電
協
- os&, I + (c36r -
IJf
o1Dfl
J
+ (o162
-
a2b{'K
A=4I+3J+8, B=2I-.I+2K, find o unit
ucctor N perpend.icular to
Aand.B such that A, B, Nform o rQht handzd qtstern. Ako find. the anglz betuteen the
ｘ
ＡＡ
縄
︱前
he∫
= (a263
whence follows the required result.
¨
¨
一
fld CJ)JrfOmsが “
読 A αtt B● 4gL′ ‐
■ &ど ,s"“
.
For by u:ing the distrihttve law,we get
"s“
`
: (f)お滉ag2"“ &お ●lsh Qo“ :ag r■ e angre“t“ 曖 Aα だ
(IE)お d面あおPcrp22diCaar r。 ■2′ Jα π
a OrAaJ2d “
B,
AxB=
れ
rゐ θ
Ⅳ on page l∞
３﹄
Ｋα
鷺
1熙 l精嶽 l盤翻離ぶ鯉臆誂ふM、
.
ＪＱＬ
o+coS v=-1.
Ｉα
．﹄
ZL調
∞sO,c¨
V VactOrprd“ cr Orr"υ acrOrsた a輛 ■b“ だυ
`
い +B)XC=AxC+B x CL We postpone its prooftill para
π んじ
αJy,た ●″ Press3● fOr,ル υ
ecrOrprOd“ ′
“
“
ら
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10K),'1t185
(Karnatalw, l99B)
Ｊ■
Also o =.J(42 + 32 + 12; = {Ze and b = 3.
If 0 be the angle between A and B, theo I A x
Thus sin0='/185,'3'/26
whence
B I = cD sin 0'
(Ag.3甥 tttL篇
導:ま ⅧF甲
sin 0 = | Ax B | /ab
ndi“ 為 19“ )
e ABC
ごthe m響」
.Vd●r
=静
area ofΔ
￨
“
」 s。
農 =あ _あ =c_B
and 農 =a_轟 =A… B
.・
Then lル Xあ
￨ル
×あ
卜
Sirnitarly, (ri) foUbws
D〕
屁Xユ l=:【C― DXい ―
('二
)α =ら
‐
印
‐
^=o∞
￨
1弾
郵鱗珊議
Oeゃ SdOntt
P9血 r+P・ Equatlng the絆
“
fron o? ' & =oocos (A +8)'
Thus area
From ttC,we haVe屁
I・
.・
p● れrs
ルれだ‐
C」 2二 Fα″ ′
″αψ 夏〕
VeCtOrially略
た sjと ,BC
担混
OLABj′ ″
23);so that
駈milarlyrmdttplying c■ lV∝ 鯨皿
J
tt result.
13. Proue
(i)
B) = sh A
r.ba cos“ -0+Ca"s“
(i■ )Squ― g(D,WO get
so
(.r{ +
Lま
cos
‐
=a2
8
簾 rar島
ぶ
猟瀾
士
r a 1 0 003 Cキ C10s3.
。
。
「
′=。 2+c2_%cc“ A.
‐
品
Vector pOe… 山
r"Ⅲ A“ d
A and B
鳳器岬
壺
"%9
O b th・
鯛
淵
覇
プ
だ
・
冨
:`励た
暇 ∬
angle bemeen them,shOW tat血 :=ち IA―
職輔鸞鱗紙†ざ
BI
1鶏
/POX=Aand tto=B,
/POe=五十B.
・
B md機
2 rA and B areunit…
1xI=JXJ=0
sO that
￨
Problem8 314
12=′ =1,I.J=0
Let
62+c2+21c coS(■ ―ハ
Fi9.321.
that
and
2
」 )=― ●
(aチ +(鳥 )2+感 .崩 =(屁 )2
B + cos 'tl sin B'
(iD cos (A +8) = cos A coe B - sin A siz B- (Moratlwadz' 1994\
OX OY
Let I, Jdenote unit vectore alongtwo perpendicular lines
sin
1_轟 x屁
￨_i壼文崩
=
+盛
“
。鳥 =_(農 ′
0)Mul“ ungO scal"しけ轟 ,Werta.屁
tlut
Err-ole.8'
yけ CA,潟
●/slnA=C/Sin C,Whence Foll(■
があ =:(滝 X屁 )=::c― A/aXC/a
=静 xA=:識 .Hence
or
Fig.325.
it[舞 :普
(λ )
らcsh(π 一A)=● C in(■ -3)
●/Sinム =0/Sh B
∴
ApEF=AFCE=:AABα
∴
励]翼
lax崩
α
“
Take B as the origm andletthe pition vecto3 0fC anaA be c and A(Fig.3・
the positton VectOrs ofユ鳥 F are
..。
(i)Muldplメ ■ζ(詢
Fig.322.
of′
Za)
+a+崩 L0
a+鳥 =_轟
BXB=Ol
of△ ABC=:IBXC+CXA+AttB I.
mmple 312.Lα
(St2a rOr″
:a)
“
P‐uilacあれ漁r“ ●
〈
CCoSI"fOm“ b)
cosA_
=:lC xA― CXB― BXA+B X Bl
=:IB X C+CXA+Ax BI
we rt C)
COS C+c COS 3.
(置 ).2=。 2+c2_%c
are
sin Bソ
Erarnple b'U. lo any tria ngle ABC' prove that
(f)α /SinA=0/sin 3=C/sin C.
ttBC
andぬ OSeぱ e
Aソ
血
sBIや o
uertizes are A, B, C ls
XBI
COSttI― ●
OP=●
a=62 40"
Eremple $l1..houc tlwt the orea of the triatgtn wh.w
:IBXC+CXA+▲
ir'
p sinめ
then the coordinate8ご P are o 008■ ,―
lfOP=PandOQ=9,
t'hat
Q cc B,q sin B) so
Fig,324.
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(4,_=・
,Ю 9ハ
198
HGHER EI\TGINEEHNG
a la)Pm..thatthe vector area ofthe qundriLtedABCD L:'酪
(6)r31+J_2K andI_3J
en vector8 A=1-"+2XI 4K areぬ
■
e山駆
山ご ap_。
Xj五
gram.Ldお
M I}TEMAIICS
VECIORS AND SOLID GEOMFTTTV
-+
AP xE =
“
lilllililiIIIilli:11llili1lI]Ii:::li「
嗜鳳 hO● ettt J″ ="_
Find the“ ctOr x"d.c」 Fた WhCh
"Q a 3ath,the
pOlnt P遍equation3 A x X=B+4A.Ж
A=I+四 _氏 B=21-J+K"｀
・
12.… by vech品 Ц that
(J)31n fA-3)=8mム “ 3-… ■ 6L B;(r).。。 -3)=…
A“ B+轟 4 8in 3.(C∝
1■
… by.
13.In any tFiangleABC,prOve
(j)ら
=cmA+。
vemr methOds,that
cOs c;(o c2Ec2+。 2_23b。 ∞ “
C
"rrml'jil
Ii198
and (u) the directicn
2 where
Eranple 3'18. Fina the-torquz about
acting thrctgh the pint
IJ+ 2E
L€to&
*igr"
required (FiS. tt
B.30).
″ら 1999)
'
.'. Moment of the
-
"rd
Flg.326.
Besultantforce F=p+e=I_3rI+5E
AB = OB - Oi =(6I+J_3E)-(4I_AJ_2K)=2I+ru*K
.V7bァルdb“ O
.',
FIg.32
1)=-15慟ぼむ
and the fOrce
.
=1曇
■
ng.323.
(3)
^ Moment of a force sbout a point. Suppose the Bometrt
of the force F acting at the pointp atrt
d;dilrii i, required.
日g.329.
F=30(1塑 L〒
e?
dio"frin
iiiii
al
釣
皐曇に―
“
+2●
F19.331.
′
たゎ油θ
″ .fFaら。′ル ′
機 D
“Ofthe mOmentOFF abOut O abng D,
=resOlved I凛武
T(7112響
the line of action of F {Fig. 8.29). If 6 be the angle
and F and N be a unit vector t- to their plane, then
)
￨=器
DrawAMl
between
t5.l.
』堅
→
.Momα lt OfF about O=oA
x F
thua, t/re rak of normal fltu
perunitara-Y.N.
y
+ 169 + 6l) =
+sE
豊F二
.・
Obe We can also apply thir result
to tbe csse ofelectric or maguebc IIu:
{a
sbout e lhe.
Clearly dl = -q+,N
.
∴Nonnal aux ofthe lquid tbrOugh tt in unitttme
A≒
e for.oe
Flg.330.
u-Lt.I.E.,
Let D be the given Iine through the origia O and F the force through
A(_ 4,2,5).
露誡
選鑽翼曹薄職驀懲
i封群
N:APXF
of the moment =
Monent of
i_
=F.4B=(1-3J+5]町 .(21+4J― 聡
=v.転
{ tr)
bt' The mo'n'ient of a force F qbout a rine D is the resoroed. part along D of the moment of
F about any point onD',
Erample a'L7- Fina the moment about a rine thrctqh ttu origin
having directinn of
2l + 2d +8. d,nc to a l0 h,s force acting at o piit
_
i, z, sl ii ttn
_
→
〓1.2-3.4+5.←
yc。。0.ぉ
-+
lr o il
Magritude
(4)
K"*;re; n;;
_-)--.)+
force,& through A.is
Ir ;t rl3l = _2I+ tiU+8f,
= l- 1 -2
and. e = _I+2.I_f,
Dekrmirc tlu worh d,one when tne-paiiae i a*pU"rif*i
A to B, the pasitbn uecton of A
B G;-ii' _5i _ zf a.rd, 6l + itr _ SB,
.
.・
and P be tbe poiat, moment about which of tbe
=pA x AB = (_I_Z.I+gK)x(4I
:
2r +,r -.K of a force represented. by 4r + g,
(Atd.hra, lgg4 ; Marothuoda, Igg4t
forceÎi about p
-+
Erenple 3.15, Constut fortel p
=21_ 6i +6K
act on a pafibk.
and
point
dP =
=F cOs o.J=F.届
respectiuely.
thc
z+J-E
07, = t-s*2g ,oa e? = 4r+8.
Then pi, = A - oa =-r-zr+BK
WOrk dOne=(resO卜 ed part OfF in the赴田市 On OfA3).4B
謝諄
斃:蹴胤器肥
′
moment
rr#x r ie t[e direction of the moment of F about.{.
Ileace the moment (or torque) of F aboutA is.d x f.
鵡柵守鑑
基♂ 咄
(tp . F sin 0) N -fq{p sin 0) N (f ..{M) N
=
the ma$litude ofi} x F
=f . c,which is tbe numerical measure ofthe
are■
_4η .6
=器 71+24r-4η .怖 #埼こ
2+74x2-4× l)=1923.
磐×
“
む
ジ
www.Engg-Know.com
es-
1997t
HCH〔
(6)
■3 MAI EMATCS
'ENG,NEF劇
3.8. PlODt
My
ol a rtgid body
bo rotathg about the axiE
ltr
Hy
O.lt4 reith angt
Let a ri6d
uelocity (l) fadran-s per s€coad (Fig. 3 32). I.€t P be a point of the
that
&
=
= velocity
Ilence V=
J
(|,
YofP
t
Tbs ieader rnust, however,
ia a dhection
J
.'*X#il#'#;*";;
to the plaae MOP.
lil
unit
vector aloas the
l\ N iK.
Anguter velocity o, =
Let A be the point I + 3.,
.1
&
;= ftFT=
directl*
=
?
-
"f
(2t +J
-K
(A x B) x (C
Fitdthz o'locitf of th2 Point
No anbieuiW
(2)
l}rc pobt of alplicatioa of th€ folc€ (-2, {, ?) i6 displ.ced froh tb€ Poitrt (3' '5, 1) to ths Poiot
har th€ diltance
O, i,'n. s"t tlii r";. i" rurldc v harved wheo the ioint of appucatioo nrove!
Mamthwona' 19911
Fhd the work d@e
5.Afor(.F-BI+zr-{Ai!sPPli€dattbsroirt(1,-1,2).FindthsDoEoDtofthsfoEsaboutthePoiDt
tA"an' rW ; $l T U ' 19 St
{2, - 1' 3)
t diltalce 3 units from the origin
lf tft" -"-""i *-tt" fo""i itoit otg'i" ft"" io-po"outr (12' 8, - r4), frnd th€ co'ordinete' of P
?, Fi[d ths momelt of thr f6oo F= 2t +z,- E acting a! lhe pobt (1, - 2,1) ahut z'tri!'
4, 2)
.cts et s PoiDt P which is at
the polat
8. A fotce of 10 Lg acts in a dir€ctioa .qualy ilcliD€d t ttre co'ordurt€ ar'! tlro'ugb
(3, -2, 5). Fild d. r'4aitE te ortrc mdent-or re r"'c. about a lin. thDuah thc
""ffifrl,iB";
9.
directioo ralios arE (2, -3. 6).
A rigid body i! rotatiag !t 25 radirDs Per iecord about an
an! OR' where -R i3 the point
2I-iJ+riehtivetrO.rUatreveloaivorrh€particleofthcbodvattltepoint4I+J+E{Atl
Iendhs ar€ iD
cm).
C
。
by the fotrsi.
,1.
-
')
Then
lAxB l:Cta.Dd lC lcorO=Por-P
r""otainc 6" + S, C f*!r a risht-hs'ded or lefl'harded
triad.
・
︱︱︱︱︱︱︱︱︱︱︱︱ムロＴＩＩＩＪ
8(6, 1, -3).
Also find the v..t r momeot ofth. r€sult{rt of thee folle! acting .t P about the poht Q'
g. Forcc! of rEsritude! 6, 3' 1 urit! act in thc dit6ctioa! 6I + z, + 3S, st - 2J-+ 6I( 2f 3.' - €f
doDG
respectively d r psrticbwhich is disPlaced f r tho point (2, I' J) to (5, -1, 1)' Fird the wo*
(5,
c 15 r{ x {a O would
iT G4 x
l€ -1eY8tT
uolume
The product AxR . C rep**nts nu,rarirall! the
B, C as colzrlninous cdges'
AxB
-J-)
r&lioa.
盤
＋
一
ω
〓
∝
Problen 35
force!
d+J-3X and sI+J-r is ditPI*ed,-ftom the Point
1. A prlticl€ act d oa by co[start
I*'Z.l*fft"tftcp"l"iff+{r+B'Fitrdth€tottlwctdonebvth€forc8.' U(Nnstaha' 1992)
act otr a partict€ P whose position vector is
2. Forces 2I-SI+6X, -I+r-f,ard2l+?.,
4t 3J-2E D€t€rmi.Dc the wo[t dorc by !h6 fo!.er in a disPlic,m€nt ofthe Particl€ to the poin'L
vitb coEpoo.at!
Geo;strtcsl htErp
Consirler a paratlelopiped with OA ='A' OB = B'
O? = C as cot€rDitrous edges Oig. 3 33)'
Lt Vbe it5 YoluEe, a the ares of eocb of the two
facer oarallel to the wctors A end B and p t}le perpendi_
culer'distaDce betseeo these faces.
2X) x (3I + 5J + 2K)
lnd its magnitude 9',/(144 + 100 + 49) = 9,/293.
8- A force
of A'xB with
tk scaldr or dat prodtt't (AxB)'c
or
oa
&ritten
is
c.I./
l\4c
tii
inn^.
of in
or llirc iv omittinc tis bt'&tt
hauing l\
parallelopiped hauing
of a Daralelopiped
﹁
-
D) of rny four
tABCl.
+3.I- K)=3I +6'' +2K
= 9(2I + J
Ｋ２２
¨
Ｊｌ５
〓
１２３
o x 4P
t
c;-;A; d*;l';i,**t
] tzr*.r-zrl
- 2D
(lltAr*E)-(I
ploducts of rnore than three vectors Such
bv siDs Buccessivelv the-exptoiion formuls
tvo p'oducts (A x B) ' (c x D) atrd
"l'ti -"tiaer s'Dd a letEr a vecmt
3,9. SCAI.AR PRODIJCT OF IHREE VECIORS
(l) Dsfinition. /f A B , C be abv th'ce ucctot't then
\I(o:tnatata,' i9931
= 9(2I +.r - 2D
K and the point P of the body be (4t + &I + tr) so t-hat
.. VelociEr vector ofP = V =
acrross
;"-;;;a*"d
il:Jix';iilJr]ii"rffiliiiiutioo,
'" being e scaler
vectors, tba former
Flg_332
2rpass,ng throhh the point I +
o.ris pors el to 2l +
+
the bodr u,hase position veio. ts
of tbe form A ' (B ' C)' A x (B ' C) ard
bot that t'he Products
A(B x C) are deaniDgles6.
In Dracticsl aDDticatioa!' we seldo,n come
x n.
i-
MORE VECrOns
u*tots.
Exatrple $18. A riSid bodx it spia iag with anauiat veLcitr 27 tdiaLs Per.s.coad obout
,f
Or
,tte t'hird beirg the v.ctor produd or the
""-'"iGi"g
ottt''o- u'c'o^'
i"a,. iia'a "*t*
;;;;i;;;;5;
ktrowa as tL,f- uector Ptuduct of three
is
u!u''lly
and
vector
a
repreeenL
c,
,
ii
,"a
r*t"a-l
R,nd
roxB=rrrrsiD0.N=oMP.N
= (epeed ofB N
OF IHREI
><
zt op=o. Dxl* pM LoM-+
Norv ifN he a uDit vector I (, aDd R th€E
rch
Cn
With env three vectors A' B, C, we caa form the products (A'B)C'(AXB) C and
C'-lepresents 6 vectot in
f.-g tle proiu"t or a s"alar A ' B aod a vector
re - ir c. fu"-n
product ofveciors A x B aad c' reprcsents a scalar
tl"
H.;;:"#;6. d;"t
Algullr v.focfty
A
Ⅲ
A
3"
席品 lttr]』服 r謂嚇 α
ゎ
θ
軍も
l鑑
協需レ
Ⅲ
H“
h… 缶m社猥ッ盤電ぜ
濯・■
輝
…
′
″
Prod二
励ル″
慨ふ需湖ぶ出:機
F∵ 腹
"あ
`島
υ
m“ お
・
・
ep"meb,ped“ お
e¨ l―Cぱぬ
"",
AB,C are● ●
Y鳳
",thenぬ
r tIP″
copranarぉ
pl錮
e vect¨
田6二酬
￡よ窮溜 [κ 吼鶴盟こ:Ъ 醐露ぷ:驚:.ふ
鍵ぎ
Vdヽ
終
impOrtBnta●
B● ″c滅
Ⅲ
“
響
:麟
0′
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鰭
酔
臨
た PrOdllCtヽ
E“ ″ S″ ″
ゆ
″ ′趨
ÌngCSmJO「
轟 :“
LQl雲
dこ
^B
C`
“
■１可︱︱︱︱
■・ヽ証
HIGHFR〔 NGINEE,:NC MATH[MAT,CS
M呼。Ver_ap-1lelopiped ha宙
,,C,A or C,A・ B tt con範
:
ngへ B,C as cOteminous edges is the same as that ha宙 ng
ed警・
CxA.B=y
AIso,dnce A.B=B.Jし we have
C.AxB=AxBoC=y
A.BxC=Bxc.A=y
BoC xA=CxA.B=V
hS ttl聾
=0
肇]イ
VК】0閤
饉
α
′ 7"お _“ +"+X,31-ガ +低
-13出げ
就胎 "・ P・
脇脇湯釧
Leta-61+3J+2鴫 OB=鋼 -2J+4E10c=51+7J+3K
and δb=_131+17J― K Then43‐ 03-OA=領 _日 +江
Sm‖
・ ,駐鳥 =111+4J+氏 and島 ―-71+14J-3K
悧
∞口anar r崩 ,え局碑。
。
plmtt ia rぬ dr scalar輛ゴe
ぶ階 f蹴 .鸞
“
httd._轟議1■ 霊蹴1讐宵雛驚謬霊
『蹴盟権tVed° "お
玉
蹴∬茫卜
AxC.B=二 IBxA.C=―
iふ
■ CXB.A=― ′
蹴雛
鰍鞣
mate the rliC…
∴the rehtiong(o and(β )can be cOmFctly w五協en as
lABC:=IBC_41‐ ICABl=y
end
IV. Scalot triplt pduct ie distributioe
F・
l+Jd
l・
l・
(r) Volume of the
1
.・
ttu aolurru of tlu tetrahe&tn ABCD
(Mtsotz,
i
=
i
tetrahedmnAsCD
i ofD above tbe planeA8C;
(麗
(2 area
J
of&{SOIr
日 9.334.
ル
"b山
∝sI
=:嬌 ,鳥
,
５一６
〓
tetahedron“ の
１一
６
〓
∴Volumeごぬe
the,電 ult.
２１３
AC=21+J+K and AD=21+2J+3К
０１２
Si7h:larly
１２２
PSition ofthe dot and¨
dF)
血 Ш
04=I+J+X,OB 3 21+J+3Lあ =31+2J+2K and OD=31+3J+4K
鳥 =品 _盛 =I+2K
)Let
Then
―
A
(ares ofaASO x (height
=:嬌,鳥 ,動・
law holds fOr scalar pFOduCts:
St
poit$,
1997
島
「
But because Qお日間ヒ
暉 ,therefore,Pta Hen“
tt
=:mme rthe pannelopi口 withAB,ェ
:
alar triple products are independent oFthe
“
=い +B― ▲―B).CxQ=o
・
.dther P■ Q。 Q=0 0r P=o
=
1
.Q
・dtttHbu血o
3.2g . Sltow tt
find, tlu uolume of tlv tetnhdron furmd b flr
-_ 1,Hence
(1,
1), e, L 3), (3,2,21and (3, s,0.
h" B + c, D -EI I IABDI- TABEI + IACDI - TACEI
"rr the scalar ind vector products oftwo vectorg are both distributive. the tnrth ofdistributive
=fA+DxC.Q― AxC.Q_BXC.Q
=に +B)。 CxQ― A.CxQ_B.CXQ
7142c15■
i':t.A8.
b ' AC- Abl.
i.e.
Lat
P=体 +B)XC― AxC― BxC.
IfQ be any arbitravv‐ ●
F,then P.Q=【 lA+B)xC― ▲XC― BxC】
=ド
FvanpL
′
rw for the fonnar has already been seen and that for the latter we establish as follows
To ptote tF.at
い +B)xC=AxC+BxC.
た
ふ瑶_]キ ‐均
却
中諄 η
Ileuce the pints A, B, C, D are coplensl.
」
￡
…
…
lACBl=IBACi=iCBA:=
AND SOuD eEOMErW
T胤
Thus,ify be the volume …
ofthis parallelopiped,
Ax BoC=IBxC.A=ス
ゝ
V IrA=α ll+α ,「 +● 3EI B=ら 11+う 2J+63K,C=CII+cノ +C3K
η
叫‖言
J
As AxB=o263 α 謹分1+● 36ユ
α lb♪
\rEcrop pnoDucl oF tHnEE VECTOnS
(1)Definttton.Ift\B,Cbcanythrecurr;arirs,tfunthevutororcrossproductofAxBwith
3. 10.
_. ulW
Cb
ptdut
ニ lABC】 =Ica2b3 。 332)I+く。361 ● 16D J+olら 2 α
261)η ・●lI+● 2J+● 3Kl
=ClCa263 ●
(2)
In
361-a103)+C301b2 。 211)WhiCh iS the required resuL.
3● 2)+Cメ ●
r満
pdn
t of threc o*tors
.0y
B. C atd ic writt2n ec (A
;
B) x C.
etl e3lentid ar Ax@xc;*(axB)xc, cryrtldlg.the fact t}lat uator tiple
is tut acrrriaddt*.
IrrFrndor form-la- If A 8, C be any thrtc
uectorE, (A
x B) x C = (C . A)B
word,s
-
(C . ts)
A
(atunrxfijorr,nt)xou&;r=(ot t r . dzmcl@iocett -(outzr .djwnlertrcmc.
db“ ′
臓ヵ嫌"&″ r
_胸 Lご雷霞軍出通究嵩臨認忠需認F鳥乳
FareS」
蹴
the aectot
Here tlrc bractet!
J+cal12 ● 2blメ
Tb€
plaoe
h“
品・″
翼鮮服踏麗tttЪ 燎
脚蝋 :跳慧」
淵尻槻ぷπ″
"
vedo (A x B) x C is pcpeadiarlar to the vettor A x B aod ttre latter is perpendicular to the
we can write
mtainingA ond B. Ilence (Ax B) x C lies in the plane dA end B. As such
there I and ar arrg Somo
￨
」
の
www.Engg-Know.com
(AxB)xC=lA+mB
gcolarg.
...(1)
H,GHER ENGlNEERING MATHEMATCs
lo4
non-coplanar u*tors lt, B, C
is called rlre reciprocal triad ofthz
llultiply both lder ecalarly by C, then C . (A x B) x C = lC' A + raC' B
Yectors are equd'
Tte scalar trir,re product on the left-hand side is zero, sincg,fuo of ita
i', s: c'.*
(2)
.'-?(C.A,*zrtg.3;-g
I
(rr
tlc valuer ofl
and m in (1), we get
(A/ BtxC=n(G.B)A-n(C.AB
"'(2)
Eridently n it *ntenumerical constant. To find it .take the sp€cial case
.fhtn 12) Fver
(I x J) x J =z(J' J) I -n(J' I)J
r..'
*-
11Y.J=zI
漱
A=LB=C =J'
giver n
Cw. Ax
or -I=nI'
1. Hence (2) reduces to
tN, 1".8
(
1・
｀。
C)B― (B・ Cソ｀
=【 α
=く A.C)(B・
L:が通 Lr.″ ●Lave(AxB′
=V…
r,「
I=■ ttLc oだ
tB・
1。
aJ
r=壼
B′
XC・
=謡
…(jjI)
υeCわ r
tP・
)・
Car.
3“ J(I,J,X)お Serf_″ c"「 ●
警 :=LJ′
=J andF=К
.
‐
‐
DcaIり ∞π Cた d・
Cl α″ recip"
ttlc Prod● cお lABCl α″ご lA3′
‐ τみe SCarar ι
“
聾
Q億
′
XCう
B′
・
=″
C■
く
郎
Fo〔
=バ
静
rV on page
V口Lc″ cわ rocaZ
B)x(CXDEれたr"sofA and B.
=喘讐 =品 =品・
おare
υわ
“「 'ri“
“ギ霧 :撤 ∬ 醐 p″ 嗣
bO′ λrなた
たαnded,o「
bO′
'。
qρ ‐
λanル d
λι
雷
轟器 31蹴
e∝
dhgお田°b ⅢⅢ
∞
le the result.
。 a“r∞ "仇 ■dわれ″ ″Й″ へ
“
励銀
∬∬ 明聴憔I義Ⅷ朧
CXDA
We can Wnte
JttFttD卜
lB_IBCDlA
ゴル呻 ″8.rA x 3)x(c xttjれた SOFC α ごQ
'“
“
‐
Aズ B,/rcズ D)=Mx(CXD)where M=AxB
“
R=rA+yB+ZC
B,C。パ
,3,1か
q・
い XB・ C)D
3 tABDIC― IABCID
32 Httipr● cal vectOr"Ⅲ
οns
l Dani働鷹 7■ υ
ect● ′ι
減ガ
“バ,「 ,Ct“ ■ed by'1● た
SubS・ tutinitheseValuFi聯
=0`.DbC-0■・C)D=(AxB.D)C―
F=品 ,7‐ 器 ■ ‐
`α
==器
bttο ′π
vdorS
"rA′
A/BJ/嗽
D:隠
…
…
B=晉聖雲≒ and c=品
For
D
:ごi器
鳥lttI鳳誹乱T彗亀
拗
1路 ∬
T′ 驚灘
翼
組
鳳瀬
ふ
這11灘
:乃句,″
▲
Hence the result fo1loV78 frOm(j)and(西
=A2B21A.B)2 whlch ha3 already been proved in 0 3 6(3■
`duct Of four
}8uCh VeCt耐
=Mパ B′ C′ 1,I.2.λ =1/rB′ C]
1=A.″
Similarly,
(interchanging the dot and crossl
Eり 0・ C)WhenCe fOnows the result.
D)-0饉。
1葦
藝
XC′
Thus
which is known as lagrunge's identity''
L,t B,.{e x D1 - [1Ax B) x C] . D
“
罐
…
HenTA=AB′
xA) + C x (Ax B) =0..
identicallv'
Flr LIts' :'A' c,n - r.r . B)c *-rsllf - ts' oA + (c' BA - (c' A)B which vanishes
(li
Scetrr prcduct of four vectorr
3.11scalar product of
it- A, B, C, D be
vectors, we shall now conlider the
"r;;;;"
..\ r B acd C ,:D, ard prove
that
x (C
′
嗅
ICx助 =ほ ::J
島
嗣
岬
だ
轟
r r=ふ品 =器莉・
ぉ
幣 le"prttdゎ
亀:靴露篤鎮胤lP鷹窟￡:fI鮮撚ぎ、
the required reeult'
Sireiiarly it lan be shown tbat A x (B x C) = (A . C)B - (A' B)C'
ftis
?
tABjl =
. BxC =面
A.″ =A.繭
As
e:T=':*-.n =r' 8aY.
Substituting
also-non-coPlanar'
Propertieo and other renerLs'
Vlhy the ruarlr:r.'tzzci4rocal tria'd'
I.
so that〔 ABCl≠ 0.CleaFly
1獄
'二
…(じ )
￨
:こ
￨[11li]IB+:11ギ
IC
tAssa",1999)
胤織 1翫 1織
[ギ出″
R=(R.パ
上二lα 二
事記
悔
fμ
豊
""
へ二
馴
ZLン
www.Engg-Know.com
k糀1犠
"・
`″
VECTORS AND SOuD GЮ METRY
´d篭貯
ザ“
(r,)
Ax
{B x (C xD))
L Find the
Ａ
〓
Ａ
・
一
メ
Ｂ
ぃ鋤
I....2=A句
=品 _励 =B_予 A=生
塑壼 =皇生
ニ▲・
ザ ♂
￨:キ
日33‐
■ ３．
ι
たヵピα
“ .CXD)+体 xB).(CXD)=a
“
crysora.i"7Sドユ圧■a,1995s」 ]と働ほ
“
- 0) = arn' 0 - sirrz Q.
znd hence show that sin (0 + 0) sra (g
We kaow that
(B x C). (Ax D) =(B.AxC. D)
(C xA). (B x D) =(C. BXA. D)
1994)
-@. DxC.A)
- (C. DXA. B)
(Ax B). (CxD)=(A. cXB. D)-(A. DXB. C)
x C) . (A x D) + (C x A) . (B x D) + (A x B) . (C x Oy = 6
A B, C, D be acting along coplanar lines
OC, OD respectively (Fig. 3.36).
lB
...1;1
Let the vectora
OI,
(C x
-
= -ahcd.eirz
1C
x D) = [a6 ein
0N
=obcd, sir2
0)
...(rr)
Fig.3.36.
atr. (0 + g) eia (0
-
Emmple S".h"r"
g)
-
..{iri)
S
(AxB)x(BxC).
veriry that IABCI tAgC'l = 1.
(D) Prove that Ax.t + B x E + C x C =O.
Prove thot (i) [A x B, C x D, E x r] = [ABI]![CEFI
IABO.
- IABCI[DEfl
6-M.I.E-,
1997;
0.
(Jammu,
1998; Rewa, 1998)
O.{8C cre c, 5, c end the aagler BOC, COA, AOB are
0, {, rg, find its volune.
(l) EOt AION OF A Pt NE
Let P{t,y,z) be any poiDt on the plane through
3.13.
p5*ce follm,s the required result.
4111, y1, z1)
鮨ndλ 昴
CMy30nら 1995」 ′
ら 1992F Bゐ Юヮ
o)AxIB x(CXD)〕 =B.DcA x o― BoC● XD)。
=oXO.I(CXめ XCA x 3)〕
=(BXC).(1(CX幻 .B】 A二【Cxゎ .AIBl
Al
=OXO.(IB.CX詢〕
"ら
1991)
(j)[BXC,CxttAxB〕
=kixc).A1lo xC).Al=lBCA12=lABC12
lten
which ir normal to the vector N
-{+
OP = *I +yrl +zE and OA
Clearly itre vectors
・【CXわ .Al=0
【
・llBCAl=lABCi
1・
-)
lp
= (r
-
x 1)
I
+
AP. N:O
磯 ♪
www.Engg-Know.com
oI
+ 6r, +
cK.
=rrl +yrJ +zrK
+ (y
and N are perpeldicular.to eacb other.
.・
=
-yil t
+ (z
-
z
Nagpul 199k
(Rzwa, 1994)
...(irr)
o6od sin2 0 + abcd, sio2 q = g
(J)IB X C,CXA.AxB】 =IABq2.
(ii)
lr..e L.B L.cl
15. (o) hove that tLMIinhBCl = lu.l
M. B M. Cl
N.B N.cl
lN.A
(D) lbe length of the edges OA" OB, OC of the tetrahedron
qrvl
Substituting the valuee firom (r0, {rtD, (iu) in (i), we get
dcd,
199O')
.
(it-)
0
. Icd sin
IfA.N=0,B.N80,CoN=0,prove that lABq=Q Ineretthsresult geome面山
tAdhn, 1!N3 ; Barryaloe,
0)Pmve」 nt lA+葛 B+C,C,Al=2trABCl.
(r) (B x C) x (A x D) + (C x A) x (B x Ir) + (A x B) x (C x D) = - z{ABClD.
Ax [(F x B) x (C x C)l + B x [(F x C) x (G x A)l+ C x [(F x A) x (G x B)l =
JN
A). (B x D) = [co aiu ( - $)Nl . [Dd sin 0t{
d(Ax B).
1994)
(ii) {(A+B+C)x (B+C)l . C=
14. Showthat
be a udt vector nonnal to the plane ofthese linee, then
(B x C) . (Ax D) = [6c sin (0 . lod sin (0 + o)Nl
ofi
nら
い
(6)SbowtiatvolumcofthetctrahedronhsvingA+B,B+CandC+Aasconcurrentedgesistwice
the volune of ttre tetrahodma hlviEgA B, C ag concurrent edges.
ORし ●●.]998)
0. IfAx (B x C) - (Ax B) x C, show that (Ax C) x B=O.
的・・ n峰』995)
lO. Show that I x (R x I) +., x (R xO + E x (R x K) = 28Wum,. 1999 : Atdhra. 1990 ; Momrhuzda, 1990)
f L If A= I -2I - 3X, B = 2I +.I- E, C = I + &I- A, fad
l&
IfN
(0 + 0) EiD (0
F)(1,1,3),(4,3,2),(5,2,7)and(6,48).
〈
め(2,1,1),(1,-1,2),(0,1,-1)and(1,-2,1).
(』
12.(a)GivenA=2I-rI+3KB= -I+&I+38,C=I+J-2&fndthereciprocaltriad(A'B"C)and
Z4OC=0aBilZAOB=ZCOD=5,
zAOD=0+QaadzBOC=S-Q
=obcd ait
- &I + 4*,
(ii) Do the points (4, - 2, 1), 6, L, il, e,2, -6) aEd (3, 5, o) lie ia a plane.
T€st the liDear depeadency ofthe voctors (1,2, 1), (2, r,4), (4,5,6) and (f, 8,
(i)Ax(BxC)
futding, we get
Tate
,that
volumtr of the parallelolped whose edgBs are rcpreseated by the vestox A = 2I
-5). Ghopl 1991\
Find the volume of the tetrahedron, three of whose coterminus edgcs are
31-J+2町 21+J― Kandl-2J+2K
Cy30nら 1995)
α ttdぬ e volume ofthe卸陸はむ mf― edけ the pointB
OXO.(AXD)+(Cxめ
)8,
(B, D) (A x C) - B . C (A x D).
5.
AlsocomponentofBfA=fi
3・ 22.為
(A x DXB . C) =
B=I+2J-X,(}=3I-.I+2f.
2. Findc soch that the vectorc 2I-J+ E, l+ %I- 3E aodSI+a.I+ SKere coplanar- (Kuuempu,1998)
& (i)Prov€ thattlevectors I-2.I+3f,, -2I+&I-4f.andI-&I+5Eare coplanar. \Punjab,1990)
{.
D口 mple
-
Problemg 3.6
∴ Component ofB along A=Oy(unit Vector alongo
)舎
C- (B. C)Dl
l(B . D)
= (A x CxB . D)
Let OA=Jヽ OB=B and OM beぬ e pttediOn OfB on A(ng.335)
LttA=貯
く
=Ax
i g
日g.337.
/
108
HIむHER
ENCINEERINC MATHEMATICs
-21)珊 .(α I+bJ+cKl=0
yl)+C(Z-21)=0 1
)r
(Jj)
=1)+Ц
rhich is tlrc equation of any plane throtqh the point (rt, Jl, zt).
Ob* Equation (ji) writta as at + by + cz + d. = O is tln general equation of a plane.
Convereely,arery lircarequation in rr!, zrepre*nE aplau andtlecelftcient* of r, y, z are lhe d,irection
'atios of the norml to the plare.
,r
〔
(“ ―κ
l)I+0-yl)J+●
y―
a(スー
++
Cor. l.WritiagOP =
.。
-'. tlle two pinta (x1,!yz1)atdlr2,y2,z2l lie on the sar'le side or on opposile *ides of the plane
ox+by +cz +d.=O,
N=C
It.
where g =6.1t1
is the ueclor equation of a plane through thz point A.
Cot.2. If I, m, n he the diretioa cosinea of the normal to the plane, then
Lt
(fυ
)
aorudfono/r/u cquolianof ,hc p/cwwherep ic tfu pcrpcndiculardistanre from the origin.
(2) Angle between two planer. Det. The ongle hetueen two planes is equal to the angle
etween their normals.
Let the two planes be
oz +by +cz +d=O and c': +b'y +c'z +d! =0.
Now the direction ratio of their normals are a, 6, c and a' , b', c' .
aa' + bb' + cc'
Hence ttre angle o between the planes is given by cr ^
rhich is called tlre
Thc plamt will tuparoIlel (il their normals art parallel), i.e.,
Cot.
pdtdllel to thc ptane
Any planz
+ by +
e
+d=
o
Perpendicular distance of the point (r1,
y1,
*
HL?J;:..
(ii) passec through
b! +cz
z)
+
d
.,,(i)
-+
ofLP are
cta+W+tz+6=0
■ れ=p珂 ∝
饉
Onoftt
on多 =¨
α
+01-g)0+●
←1-の
1-た )c
ヾ。2+b2+c2)
=器
,い
・
↓
″
+d=o
¨.(Jゴ )
IbyⅨ p.91〕
..(1)
fire plane cuts the axeB atA,8, C such thatOA=o,OB=b,OC =c,
it
passes through the pointr A(a, O, 0), A(0, b, 0), C(0, O, c).
ua+6 =0,F6+6=0,p+6=0
s.
--6/a,$=-6tb,t=-6/e
substituting theee velues of G, P, T in trl, (ri)
lbree points foro
It will
I
| " - | r -_Q z + 6 = 0 or
*
oI
1)
iac(r -0) +blg -L)+c(z - 1) =0
(- 1, 2, - 2), if a + c = 0 and -
r-
+
-
-t
.
x + 2(y
-
1) + 1(z
-
l)
r.
3c = 0.
=0
2y
J
lI
--) I 0
xAC=l
of the plane
Kl
1l=-l+2J+K
l-t 1 -sl
→
→
tion ratios are
-1← -0)+2o-1)+l12-1)=0
日9.338.
r
...(ピ
j二
)
(J)pararrer r.ヵ
の
一
ifthe points lie on
-A -z + 3 = 0 which is the required
equation.
Example 3.25 . Firld the equation of the plane whieh passes through the point (3, - 3,
' The sigrr ofthe radical in (ai) ig taken to be positive or negative according
as d is positive or
rgative.
Oba. Ttre perpendiculars to a plane from rwo points are taken to be of0re ssme sigrr
c same gide ard ofdiffenent signr ifihey lie on the opposite sides ofthc plane.
=
...(21
a+b
This vector is nomal to the plane cOntaining AB and AC and its dir∝
(-1,2,1).
Hence the equad● no「 ■le plane through A(0,1,1)iS
¨改め
I
?, =, = i
Substituting these vaiues in (2), we get
AB
*
of lhe equation of the plane.
be pass through (1, 1, 2) and
..
a point on(ハ then
ar+lg+cヵ
points A@, 1, l), B(1, 1, 2) and CCL, 2, -D.
of the equation of the plane. L€t the required equation of the plane be
- z + 3 = 0 which is the required equation
-)J
Otherwise : We haveAB = I + f, andAC = - I + rI - 3K
pkyyp)
a
ら
c
ン2)'th2)・
tFa2)'マィ
ro rg,■ )be
tlu
By cross-multiplication,
Let Pr, be the perpendicular distance ofP(:1, y1, z1) from the plane (i) so ihat the direction
‐
ra
fo
Anyplane through (0, 1,
from the plane
i1a2+b2+c2;
rsines
Intercept
(i )
whence
(I being anv constant)
j)
whictr ue Ore required equations of the bisectiagplaues.
rrqmple ?-21. Find the equation of the plane which
Ql cuts offintercepts d, b, c from the axes,
0
f doc
+ = + = i.
=0
art+hrr+czr+d
o.x +
-
Q
=
r the directioa-ratios of tleir norma*
(3)
u
+ bb. + cd
. (ゴ
ax+b+e+d,.,a't+b'y+c'z+d
.
:',ld;b\c\=t
ur"al6a*"\
i.e.
ifaa,
(f)
“
+ご ″十ご =0
α缶+bレ
I*t P(r, y, a) be any point oa riither ofthe planes bisecting t.Le angles between the planes (i) and (ji).
Then ]. distauce ofP from (i) = l- distaace ofP from (ii),
,8l,=w
i _e.,
saru sign or of opposite $gns.
+by+α +d=0
and
.(■:)
h+uy+nz=p
of the
Con Planec bfuGcting tbe rngles lretweetr two plene..
llris
The planes will fu perpend.i,culcr (il their norma)s are perpendicular),
arr+ byr+ er+d and a4+ byr+e2+ d ore
accord.ing aa
bc the given plaaes.
B eod OA = A(r)becosres(B-A).N=0
109
VECTORS AND SOuD GEOM8RY
(ι
(づ
j)π ●
′
"α
Zゎ
jj)perpendお
`′
rav 2七十3y+52+6=θ
e′ 読
`■
Jα r
“
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Jれ
二
ag,た ●poj几
`J●
rO rle pZα
れes
7ェ
and
:
`s(3,2-1)こ
+y+22=6
1)
αnd
れど(2,― r,5).
3■
+5y-62=8_
lA.MIE.,1990)
(i) Any plane parallel to the given plane ir
?.*, + 3y + fu + h = O which goes through (9, _ g, 1), if
h__2
ltrue the required plaue
is
(ii) Any plaoe through (S,
-
The direction cosires ofthe
toi,3,-6.
ltus
3) + 3(y + 3)
-
- 2 = O.
ie a(.r _ B) + AO + g) + c(. _ l) = 0
linejoining trre pornts (8, 2, - l) and (2,
B,
6(2
1) = 0 or
:
- l, s) are proportional
"it.
The equations of the planes in the two syrtems
二
Eranple 8'?8,. Find thz-equatiorrs to the
- 4y + Sz = 3, S*. + 3y - 1z = g.
equations of the pranecLi-sucting the angres
ilx-4y+52-B t,
=
= p/n.
←1,yl,21)=fP/4J,P/4“ ,P/● り
∴ ★+j:+妻 =ジ C2+m2+め =ジ
2+y 2+2-2=16P 2.
Or
and
ISee p.891
Then,
b, b, c) and
meets the
ti ji"iiinrouen * a,
i
こ
・
abc
-+-+-=ltyz
な +7ッー
…(j)
ar_y+Z-12=0
..。
嗜←V122t′ .c9121ヾ静
tan O=ヾ
0く
e・
(jE)
ABC be any plane through the fixed point
H(a, b, cl such that OA = xv OB - y1,OC = 21.
鵠 =繭
c211)/67 whtchお less than l.
45°
_
pl鳳総管盤Tttebetweenぬ e,Vm
_- Ief
Jm“ ,∞ that
Hence the plane 2+7y-9z=6,bisects the acute angle
Then its equation is
O
Ы sthd atte L辟 eenぬ e
“
“
Probll班n387
ISee Ex.3 24 tt】
.
撫灘撤撃
温∫
鳥:‖ 上胤liffttli舗:盟
l鞘欄四撻柵躙蹴鱚焉鷲拙
:盤
駄 nce″ lies cn it
∴ 貴喘登=上
∞
Ю
ぶa譜 :Teξ 鷺世も11lT猟遣
鳳
...(1)
:篭
11し
=1)・
the given planes are
{ rp;;t;c-;Pr
G{A*'i
92-6=0
轟 +3y― を =9.
3'27. A uariab-re prane passes through the
fued, point
co-ordinate axzs in A, B, c. show that thc locus of the point'comi" a
C parallel to the co-ordinate planes is
1,Jl,・
betwln
_&_9
Let o be thO angle betw∝ n(J)and either ofthe」 ven planes,say:
Itrq6p[s
in P←
5r+3y
whch are the required planes.
Thus the locus Ofc isェ
貴
tuto planes which bisect the angles between the
Ako point out which of ttu ptrrtus bisects the acute angle.
the
∴A=fP/′ ,0,0),3=(0,P/れ ,0),C=(0,0,P/2).
Thus the∞ ordinates Ofthe centroid C Ofthe tetrahedrOn OABC are
+黄 +彙 =1
Ｑ
OC
〓
-
whence follows the desired result'
planas 3r
from the origin.
* ;k, d^*
frA=,
p/m,
＋
--
二
so that OA = p/1, OB
...(J:)
1-t
G-F;5=.t1of4;;n;5
i.e.
""i î
J_
＋
(r) may be rewritten
,..(t)
...(ゴ )
普
Since (i) aad (ri) arc the different equations ofthe same plane.
.'. rL distrhce of (0, O, O) from (i) J_ distance of (0,0, 0) from (id)
=
+.
olBC.
a.ne
+:=1
十
上Ｌ
and
=
=
Hence tlre required equation is 1(r- B)
-3e + 3) _2(z _ 1) =0 or x_By _?z _ LO=e.
Eranple 3'20. A uariabre prane is at d constdnt distance p fiom the irigrn
-'-ibi*""irr, ,n,
arzs at A, B, c- Find the leus of the centroid of the tetrahedmi
isdfil
giveu
As the
pl*" i" str g dirtance p from the origin, therrfore. Its equation is of the foirm
b + my + nz =
where J, rz, n are the d.c.s of the
I
p is
pL
ru: systems of rcctongur.ar a,,as haoe the same origin. If a prane cuts them
q l'^
b, c and ay by clft?/m the origin ; show that
o-2 +b-2 + i-2 =o;2 *b;2 +
-. -
7d+b+2c=}andBc+56-6c=0.
t *
to z in ( 1), the locus of the
aa ct stunces
+ 3y _ 6z + 12 = 0.
Solving these by cross-mulriplication, we get
zl
artd
rlz*b *9 =1.
l)
-
y
to
o
(iii) Any plane through (8, 3, t) ia
d(, - 3\ + bO + 3) + c(z - 1) = 0 which will be J_ to ttre planes
7x +y+22 =6 and Br+!y
-62= g
p
/1
二●
．
if
-
0o .r,
?.t + 3y + 5z
This line is normal to tbe plane (1), ... o, b, c arc pmportional to 1, g, _6,
Substituting these values in (1), the required equation is
1(r
ctrangiDg,rl
｀
Flg.339.
ヽノ
ef
1胃
工 Find a uit vect● r nOmal tO the plane through the poiits(1,2,3),(1,1,1)and(2,-1,3).
www.Engg-Know.com
HIGHER ENGNEERINC MATHEMATICS
12
Find the distaacc ofthe point (3, 2, 1) from the plane passing tlrrough the points
6
md (- 1,0,2).
7. Shos, that the four points (0,
ofthe plane through them.
-
(2, l,
(-/!t,2,0) is the
8. Show ihat th.
(1,2, 1), (- 2, 2, - 1)'
tEiuL Show that the poirt (point
1, 0),
lt,2,0)
- l), (1, 1, 1) and (3, 3, 0) are coplanar.
(r'
1. o). (3.
- I' r)
Find the equation
circumcentre o{the triangle formed by the pgints
(!, l'
vertices.l
Plane!
lMadrar,19968\
10. Find the equation ofthe plane tlrough (1,2, 3) parallel to the plane,lr + 5y - 3r = T.lAndhra, 19911
(rr, yz, zr) at right
11. Find the equatiou of the plane which bisects the join of the points (rr, yt, zt ) and
angles.
12.
- - Find the equation oftlre plaae thmugh the poiots (2,2, f)
2x + 6y +
&
=
9.
ad
(9, 3, 6) and perpendicular to the plane
(Osmaaiz, 1999; Mysore, 1997 Sl
joins
13. A plane contains the points A(- 4, 9, - 9) end B(5, - 9, 6) and is perpendicular to the linc'which
I and C(4, -6, &). Evaluate I and frnd the equation ofthe plaae.
joining
14. Find the equation of the plane through lhe points (2, 2, l), ( 1, - 2. 3i and pamllel to thb line
(Ranchi, 2UN ; Andhru, 1999; A-M-I.E-' 1997 Wl
the points iZ, f , - St, t- f,-S, - Sl.
f5, Fiad the distance betwee! tbe parallcl planes
2-2y+′ +330 and生 -4y+ン +5=0
Let P(r, y, z) be atry point
16. TWo ptane8 are giveo
= 0 and 2r +y +a + 3 = 0, fmd
(i) direction cosines oftheir line ofiDtersection,
(ii) acuta angle between the planes, and
(iii) equation ofthe plane perpendiculsr to both ofthem thmugh the Point (2.2' l)'
1?. Finil the eguation of the plane passing through the line of intersection o{ the planes
lMadras, l990l
x+y +z = t aird 2r +Sy -z +4=0 and perpeidiculartothe planc 2y - &=418. The plane Jr + my = 0 is rotated about its line ofintersection with the plane z = O, thmugh an angle cr.
Prove that the equation ofthe plane is lx + my + z ''1112 + m2\ tan c = 0.
19. Find the equations of the two planes through the points (0, 4, - 3), (6, - 4, 3) other than the plane
(Rourhela, 19901
through thJ origin which cut offfrcm the ues intercepts whoee sum is zero.
20. A plane meets the co.ordiuates axes atA, I, C, such that the centroid ofthe triangleASC is the point
(1) General
r.,
rd
OF
血鳩.Elimlnating′ ,we get
κ―■l y― yl
ι
whtch are theり
whchお the
Cot. 2.
equations of ttlc
for the directioa-ratios ofthe
c9“ こあ
oF′ ル
′
he.
``F″
0
F:9.341.
)
thromA md Parallelto U.
tirc i<rlnir,4 the points
(\,
yr,
2i
dnd (r2, y2,
z2)
ore
liaejoinilg the given points are
a2-tL'!2- ! tz2- z{
lfo reducc tbe geaerrl equetion of r liae to thc qrnnelricrl fora:
for
r
(i) find
and r.
t
poittt on tfu
lia
by prtttiag,z
(ii) find ,tE dir.ctiDn osiItr,s
of
=0
in the given equations
ud
solving the resulting equations
tfu line, from the fact that it is perpeodicular to the normals to the given
planes.
(iii') wrib
ttu cgutian of tlu tiu ia tlu rymnetrbal fon'l
l.'rmrnctrical fonry the equations of th'e line
Eraaple 8'30. Find in
(Osrunia,
r+rt+z+7=0,4r+!-22+2=0-
(Bhopl,!991'l
1999\
$\ To find a point on the line,
Putting z = 0 in the given equations, we have
r+Y+1=0;'k+Y+2=0
Solving,
i=t= *..'
A point on the line is
(- t/s, - 2/s, o\'
(iil
..。
To find the d.irection cosirus l, m, n of the line.
Since tbe line lies ou both the given planes.
proportional to 1, 1,
... It is perpeailiorlar to their normals whose direction cosines are
(:)
4,1,-2.
。..(i:)
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〓
Solving
l+m+n=0;4+m-2n=0'
ユ．
・
.
■●,
〓
ng.340_
二２
straight line which ig the line ofintersection
'the planes (r) and (ii). (F'ig. 3'd0).
確 “酬 rOm oFrた
ι 一．
・
rken together represent a
...(J″ )
″
t“ わr e9“ riO■ OF働
Tlu
..(Jj)
t-rr _ Y-Yr
zz-zt
rr-lr Yl-Yr=r-rl
equations in r, y, z
ar+bY +cz+d,=o
a'x + b'y + c'z + d' =O
(I)
″-21
=
=
“
OL Auy… “ton the
“ h● (fi:)=← 1+J`,yl+潔 ,21+"`).
and沸 =R,then O rV08
Ⅲ RLIfa=▲
一Cυ
■―A=rU oT R=A+tU
A srRAlGHI LINE
forn. T\ro linear
。
蹴鳳肌:T:譜
.鵠龍[鴛譜躙
ご
22. e variable plane is at a constant distance p fmm the origiri and meets the ares at l{' B, C. Find the
(Osmania, 2OO0 S : Calhu,, 109<11
locus of thjcentroid of the trianSle r{rc.
2!1. Find equations ofthe planes bisccting the angle between the planes
r+U+22--9, 4r-3y +l?,2 +12=O
and speci& the one which bisects the acute angle-
Id. EOUAIIONS
parameter.
(最
Jι tt the equa,m Ofthe plme L昔 +首 +:=3_
isgivenbyr-2+i2+z-2'p-2,
given line tbroughA(rl, /1, z1) and parallel to the urdt v€ctor
理
1」 Tttha
1群瑯
"Sh∝ 亀 :ボ 11部
j,1998S)
san,1999;TrIP“ ′
い●
A, 8, C
Through
B,
C.
io
A,
axes
the
p
21, A variable plane at a constant distance from the origin meets
planes are drawn parallel to the co-ordinate planes. Show that the locus oftheir point ofintersection
に ,b,0,ShOW
oD ttre
U=ZI+Й J+れ K.
(″oご ras 1995)
-3r.
byr + 2y -& -2
the point A(r1, y1, z1) and hauing
z-ar
I-tl
tEint. Bequired distsuce = Perpadicular distance of any point on oue of the planes fmm the other.
A point oa the 6rst plane is (0,0,
fom. Equoions of thc linc through
Y-Yr
-l-=-;_=;-
0),
Qfclmla, 1!N0l
lies in the plane of the triangle and is equidistant from its
the equatioo ofthe plane through the point (2,1,0) and perpendicul,ar-to the
9.Find
-'2:-y-z=nandx+2y-3i=5.
(2) Symnetrical
directbn coqines l, n, n ore
1 an<i
...
lte
directioa coshee of the given liae are pmportioual to - 1,2,
form are
(,ii) lbus the equations
-
'arallcl
to
8'31.
findthe
distance of the point (1,
- 2,3\fumthe plane
*-y
+z
=5 measured
Iine tbrougt P(1,
-
2, 3) havine direction ratios (2, 3,
-
direction ra饉
∴∞頭
`あ
Ю
:肩雷電」号
"Pcr●
‖十
LJ",た p誡 ,sln
:::号 )
:メ
e=0
Prove that the pohts (3, 2, {), (4, 5, 2) and (5, 8,
2,r + 2y
-t
+ 15 = O, 1! + z + 29 = 0 and the
lir" +
=5!
=
f
3. FiadthealglebetweeatJrelineofirtersectioBoftheplanes&+!y+z=5andr+r-2'3aadthe
P
d
liaeofiateisectiooftheplerca2x=y+2and
?r+loy=&'
tBatgdon'l99o\
Find the equatim of the liue through the point (-2, 3, 4) atd pamllel ta the planes
and&+3y+52=6.
i. Show that the *" + =t? =|
ル=:(2p+9+「 -6).
b+3y
+ 4z
=5
(Mangalore, 1997\
-parallel to the plaae k+Zy-z=6,and find the
distance
betweeu them-
6. Fiod the cquation ofthe linc throuCt (1,2,
Hence P is
―
ン
→6-η -9+"1
),許 ―
η+η ―
卜均
r v z
i=6=_t
,き
-
1) perpeadicrrlar to each
,x y z
5=.-=-.
ofthe liaes
atr6
7, Fiod the equations of the linea birecting the angle betweeo the lioes
ソ
饒増
洋イT'ff'洲鳳}=lτ せ
therefore.
Fi9,3槙_
O) are colliaear. Fiad tbe equttions ofthe liae through
them.
2. Fiad the angle between the line ofintersection ofthe plauea
Sub8"極 ung these in(`)andsohg,we get
(12-p―
I"1)
Pnoblems 3'8
Fi9.342.
-1.
(Bん opこ
0=
sm e=寺
l.
Also the line PP must be perpendicular to the plane.The
′
lirection ratios ofR「 belng′ ― ′ ,9-g′ ,′ ―〆,we,therefore,
have
=「
2
″
oI+ bn+ca =o
If thz line b pryendicular to tlu pbru, it will be panUel to its aomal
:.
l/a = m/b =n/c.
(a)Let P o′ ,9′ ,〆 )be the image of Po,9,r).Then the
id Point ofPP mustlie on the given plane
whence′ ′=′ ―餞 ,9′ =9-ル ,〆
"｀
j″
Cor.Ifrた ′
[If two poiats P, P be such thst the line PP is bisectod perpeadicularly by a plane, then either of the
■)。 fthe cther i the planel
げ =γ =ギ
"如 ““
y― y1 2-21 “
Hence the required angle e=sln-1〔
in ile s'wne Plane.
=薇 S朝
″―夕l
"ofthe nomalto the plane are c.ac
。r
bisects the acute angle.
xllnts ls the JttF for脅レ
■ 2ザ +γ +ザ =6
′
ルれa
r=施蘭ぼた
“
r o be the angle betteen the une and the plane,then
90° -0 18 the angle between the line andぬ e normal to the
plane(Filg.3“ 腱).
れand the
NOw the direction ratios ofthe line are t鳥
(Kottayam, 19961
lire+ =+ =+
3
and thc planz ar + by + cz + d = 0.
point of iatersection i I Q eh
@ Fina the imqe (reftection) of the
3
′
6) is
2r+1-(3r-2)+3-6r=5 or r=U1.
, - rh , r+11
. rhun the required distance = PQ = {
* #)= ,
1-+
t+2y +22 =9, Lr-\y + l2-z + l2=O and specify the oue which
lte
3
1999)
Aay point ou thie lile is (2r + 1, 3r - 2, 3 - 6r).
This point will lie oa the plane r -y + z = 5
.'.
3
Example 3・ 33.■
(CaJ3● 2ち
T=..T-=;-=r.
f
3
which is the required inage ofthe given he pe(ng.3・ 43).
tle line
二 _2_‐
115
苧 ≒=揺 =長た¥=雫 =雫
3
2 3 -6
lLe
1.
"I[,I;I:T-"*-
-1 - 2 --L'
Eranple
VECTOtt AND SOuD GEOM冊
ENCtNEER:NC MATHEMATICS
HICHE々
14
.
t+2 z-3
x-l _y:! z-3 r-l =-2
=-'-3'
2 -z= 't ; 12
にIE,1997Wり
rA Л
8. Fiud the equatioos of the two straight lines through the origin which intersect the line
1
Q
i(: - -t) =f - 3 = : at argles of 6(P.
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(caた t r99`)
“
VECTOps AND SOuD GEOMErRY
HIGHER[NCINEERINC MATHEMAT CS
6
Erenple &34. Obta;intlu- equ*bn da
●quam● oF」 b line of lnters∝ tlon of the planes R NI=di and R N8=d2｀
N2)=ど 2Nl― ごlN2・
a show that th●
Rx 01×
k
山定lar From the point B to thl lne R=A,lT is given by
lC Show that the equatlon tO the pe…
R=B,た Tx(A― B)xT_
肛・
源譜響野
岬
諏
協
減
1胤
配 I紺
■-2_y+3_′
2
2
e■ne r・
、
i nnd■ e angle be摯
"eenぬ
7r
7ぶ
7.Find
一
=ザ and
i期
+y+z+2 3 0 and is perpendiclllar to the llne 2κ
(7
(7 +
1,-1),iS parallel to the
#
+y=0=ェー=+5.
f
r)
+
+Dy1
lie in the plane (2),
The line 1) will
isfied by all values
.'. the coefEcient
=t+=+
...(1)
+cz1+d)=0
=-O*11=2jontheptane
r+2y+z=
These are the required conditions which state
に Mニユ,1997)
72.
will
-
1)
+8(y +
1) + C(z
-
3) =
I
2A-B+4C=O
be perpendiculer to the given plane,
A+2.8 +C =O
SoMng (fi) and (ai), we
set
…(二 )
if
…(■ )
._〈
″三
)
*=1=g
Substituting tlrese values in (i), we get 9x - 2y - fu + 4 = O
… (:υ )
which cuts t.he giveu plane
υ
r+zot +z=12
along the required line ofprojection.
Oue point on thia line is got by putting z = 0 ia (iu) and (u) and solving, it is (4/5, 2&5, 0).
...(3)
...〈
The direction
= 0.
ratio ofthe
I + 2nt + n= 0 and
...(4)
)
line are fouud, by solving
9J
-
2rz
-
5n = 0
to be 4,-7,10.
...(5)
Hence the required equation8 0fthe line ofproieCtiOn
that
are
$t ihe liru slwuld, be parcllzl to the plane, (iil a point of line should lie in the plane.
v - v| z- zI
Thus the equetion of any plane through the liae-lmn
*
-
.
.A(r
The plane (D
if every point of the line lies in the plane so that (3) is
ofr.
ofr = 0 and the constant term
if tbeir oormalE ale perpeDdicular,
4+
...(21
al+bn+cn=O
er1 +by1 +cz1 +d=O
I
+5=0
...(r)
to the plane
g/6Y + $ 6/5p + (16 + 39/5) = 0
4ix-LLX +2(b + 119 = 0which is the required equation.
t2/6V + F
where
b(mr +yl) + c(nr + z1) +d = 0.
(al+bm+cn)r+(or1
(
2z + 13) = 0
Any plane tbrough the given line is
Anypointontbeline(1)is(Ir+:1, mr+ybnr+211 whichwilllieontheplane(2),if
+
-
3y
mvonple 3.35. Itnd ihc equatbn in tlu sXmmttrical furm of tlu prcje*bn of tlu ibu
l
ylieintheplaneu+by+cz+d=0
a(lr
l(4r+
+&). 1+(-4+3i). (-r)+(?-2t).(-2)=0orif, h=3/5.
5,CONDmONS FOR A LiNETO LiEIN A PLANE
Tofind,theconditbnsthaileti*:f
4y +72 + 16 +
-
(, lyill be perpendianlar
or
and Parallel to the line
plane
4ェ
+ 13 = 0
Subetituting tbis vElue of& in (i), we get
(y■ じ ,2000)
the● tnight hne which passc8 thmugh the point(2,―
=0 ard tlr + 3t -22
+ 76
t -! -b
i.e.ii
the PLne&+y+z=7
拗・
"=:・
of
the■ uatlons
Itt「
(Os"● ●Jα ,2000S,M● dra.● .J"5)
tAndカロ,1993′ B● ■″●I● ‐,1993)
6.Find the eqllation of the plane thmugh the pcints(1,0,-1),(3,2,2)
,-1=;。
説:鳴
The plane
"常
(j)(1,3,4)in tlr phne a― y+z+3=0.
(if)(2,-1,3)ln the plane 3L-2y― ″-9=α
- 4y +72
(7+&y+G1+3k[+(7-?,hP+(16+1&t)-0
or
int
1=者
:LI‖ Td
」凛
thmryhtha lhu of inter*ction of the
and perymdicular to ttte planu x - y - 2z + 5 = 0,
The equation ofany plane througb the line ofintersection ofthe two givea plsnes rs
=鵬
-1・
鮨
°
mぬ eメ me2+"― &=“
.Fhd tte dinanリ
博
●
fわ
ニ魁、
t`L Find the refleunon(unage)ofthe p●
RttI鷲
駕
12 Find the perpendhlar distance oFthe pint(1,1,1)● ●m the me
plalaa por;sirg
plones
ェー4も _y― 零略 _z
4
-7
10・
B
flfhe liac of greeteat alope il a plaac is a line whic]r lies iu
Flg! 3{5'
the plane aad ir pcrrpcodicular to the line of intenectiou of the
plaae with the horizont5l ph.e.
In Fig. 3'45,A4 is t}c linc of intersoction of the given plerc o with the horizontd plane *. 15eu PM
tlram perpendicular toAB, is thc Iiae ofgr,ef,tert slope on the plaae a tbrough the point P.l
"-!-anple 3.36.Assuming tlulinex/4
=y/ - 3 =z/7 os wticol fudthe eo-wtiots of thn line of
gra** slop in tlu pkru 2x + y - 5z = 12 ald passing through tlw point (2, 3, - l). (AM.I.E., 1997\
8(t-tr)+b(Jr-yr)+qz-zr)=Owhercd+bn+cn=0.
Obr fhc egurtion of euy pleno trhrough tbc lioo oflntcraectlon ofthc plrnee
u + by + e, + d, = 0, o'l. + b'y + c'z + d' = O,
ar + \r + cz +d + k(at+ b? + Cz + d) = o.
For (i) it ic an equatioE of tle fint dcgrec ia r, y, z representing a plane and (ii) it is satistled by the
rdiartea ofthi poiats which saiisiy iloih the given pianes, i.?. it contains all the poinLs common to these
res.
禽
′
、
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・
…
HIGHER ENG'NEERINC MATH[望
118
A■
鯨 hd
騨欄椰撚鮒職
●
lcs
...(二
κ ■2
)
e hneご rea写
輩嚇現渕
3子
Sd宙 ngttul mdしヽ
:=壬・
ご
・
。
S°
...(jυ
●
iCubr b the‖ ne o“ nters∝ On Oftheメ an“
The line (2) wilt lie in the plane (3), if
on this plane.
al2+ bm2+cn2= 0
.'.
and
)
Ct棚
a猟 :策
]誌
4爺
d… rpenぷ cular"
=ギ =号豊
“
hmunぬ Jm千 =ザ考
樵ぬ田
“
.メ
i[Ъ
蔦
饉 e"飩
6ェ
■nd
lneザ
iC鍼 insれ h¨
価
+4y-52=0
■-2り +3レ
=0
Chれ
憚粧鷲
3
3
=Lよ
)
rPOゎ
“
tiS“
富Ll軍菫鳳翻」
育讐
だ
‐ is is the"“ ●,30れ
=墨
2
A・
"中
Jar2● 弓
′だ r"「 ∞″
“
"""耐
wlich lies on the second line if
Hぬ eeqmtimご the lin∝
=讐 =守
“守
がυ Zhes.
“
皇
∫ザ
響
Jザ =ザ
αだぬe“ ″ ο几げ
“
"『獅
ギ
=¥=ギ
義童=3+仏 Wehtt
Thls value cl― ly satittes the equatton壁
Hence the lines intersd,(:.e. are
COND「 :ON FOR Tlu WO uNEST0 1NTERSECT(ORTO BE COPLANAD
.
B,Dl=0
Any polnt on the ttrst line is(5+47,7+4r, -3-5rb
.
amの be R[11:鋸
iS the c● aι あれOrCOpraれ αrtヶ
e pra2.● .″ raれ ηgrた
Of′ 力
Ex,mメ e33■ 絣 οω 薦″ 議″
∴ 00m
316。
D2 "tl
“
be any pdnt on the plane contaljng AC and BD then AP,AB,CD are coplanar J.a.
14P,AB,CDl=O or OR‐
1,lfbⅢ 午
十&=4 mdお
癬讐
lies
...(6)
anbining ttu lines (I) and (2)'
鳥 ,島,あ卜o
〔
lC― へ】
町 Dl=0.ThiS
.ι
押
z)
These lines哺 l inte"ect rAC,AB,CD are oplanar br whiぬ
じ.
in each of the fOlloWIElg CaSes:
fF幕 i齢
Il難
L』
perpendicub"ぬ eL=星
R=‖
鮒11蝋
J“ planeS IC・
籠 nters・轟
‐
md'』
1t2,
...(5)
l',
ご隈Phe■ hlch contalnsぬ e
S・
is parallel to the plane and its point (x2,
6, c fro;11 (6), (4) and (5), we get
which is tt e eq,&tbn of tltz planc
2 nd ttewau¨
it
Yr-1 ,2-\l
11 l=owhxnisthercquiredcondition'
E1
Dr BzI
It,
y-yr z-41
lt't,
tr1 nt | =o
t1
(5),
get
(3),
(4)
we
and
Aloo eliminatingo, D, c from
I
￨
・
.''(3)
]tl-tr
lll
￨
2‐
cOn血 hg the hne≒ ユ =≒
ザ
-er)=0
a(x2-rr)+b(Y2-Y)+c(22-z)=0
Elirinatingq
￨
ギ =響 =守・
■ ee叫五ぃ ofぬ eメme thr¨ 由 ■eo五鰤n
-yrl+c(2
nI1 + 6m1 + cn1 = 0
where
…(■ )
Hence the equationS Of the■ ine oF greatest slope through(2,3,-1)and having direction
rattos 3,-1,l are
LA“
...(2)
T=m2=下
"'{iii)
thensolving, 4J- 3m+7n=O and 2l+m-6n=0,wehaveI/4=m/17 =n/5
(ii).
plane
the
in
lies
which
greatest
elope
of
the
line
ratios
of
I*t t,, m, ,n. be the direction
2l'+m'-5n'=O
y― y2 2-22
Theequationof anyplanethroughtheline(1)isa(:-11)+D(y
+y-5z=12
性
1■ 11‐
(1,3,2).
‐
立
=ギ =雫
jl,マ
蹴
rttL""9
...(二 )
..(if)
r― L
=-8-5′
●。planar)and tOm (j)their point Of interseCtiOn iS
―
――
...{1)
ギ守｀
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ノ
１︲︲
，
VECiOヽ
HlGHER ENelNEE'INC MATHEMATICS
120
０
〓
５
Problems 8'10
３
l.
and
r-L
3x-2y +z +5=0=2x+3y +42 -4
2 prove dlat撫
Find their point of intersection and, the plane in which they lie.
(Andhru, l9!)8: Kuaempu, 1998: Madras, 1994 Sl
Any point on the first line is P(8r
4, 5r
&-2y+z+5+力 (2r+3y+42-4)=0
-
4)
linee r
-
i:illiliilllililii[「
.闇講
Eヂ =ザ =2-Υ
3y + 2z
-
tu
-
4 = 0 = 23 +y + 42
I = 0 = 2Y + 9c,
ェ
全 ght“
≒ =守就五
・
-0=0
+b+c=0
α―C=0
胤場謝胤躍配器潔1la・
..(j二 )
Hence (iu) and (u!) consr-itute the required
lla“ ,2000)
f鰹∫
だ
洲
on it.
...(jυ
...(υ
●
"IPles Such ines possess a cOmmon
雫 =雫 =警嗣∵ =讐 =7
)
sO that
:
ed sた
A≡ ●1,yl,Zl)and C=← 2,y2,Z'・
Letみ m.π be the direction cosin“ of the shortest diStance
)
EF.
Since EF ttto both AB and CD.
7=0
991)
“
Letthe given skew unes AB and CD be
i =+ =i
3+0-1+H4-5-勾 =Q".力 =:.
Substituting [ = % in (u), we get 17r + !2j + 102 -
`r,コ
“
1。
...(`j二 )
+4y-1+ル (4r+52-2)=0.A130(1,0,-1)lies
“
守 =ザ =子壺ザ =響 =ザ
⑩吻 4″ 鋤
蹴胤出れ,_5お dnwn b htいれ dh“
a鑑 :=田圏霊ま
=謝
ザ =響 =雫狐d響 =ザ =ザ・
圏腱:撚XT∬席織i」 1軍出 ∬S、 :『蝋島a"cttmttShW
血
い
1島 dl
TtttT』 ;ミミ _3鎖認鶴
…(j)
Substituting these values in (i), we get r - 3y + z = 0
Again, the equation ofany plane containing the second line is
缶
力
(Gο
gle3.
'1_。
2●
Solving (ir) and (i,i), we have
end
■と -51‐
1
Also(1,0,-1)lies On(ご )..
I
七
,lC2・
￨ら
lT
3.Find the呻饉mご the sttdght h perpendidar"both the unes
=0
●● -0)+ら 0-0)+C←
where
+
lrr ooCrur'
二_2_二
1
=α
h the plane.
娠Stmtlinelying
=TttRこ
_υ
0
and intemttillg the he≒ 2〓
Substituting this value ofl, (i) becomes 45r - l?y + ?,fu + 53 = 0 which is the required plane.
Example 3'39. I'ind the equations of the line dmwn through the point (1, 0, - 1l and
intersecting the lines
x = 2y = 2z and 3x + 41r = I, 4x + 5z = 2.
The required line will comprise of
(o) the plane containing the first line and the point (f,0, - 1).
(6) the plane containing the second line and the point (1, 0, - 1).
The equation ofany plane containing the 6rst line
2
+“ +d=°
ttd“
蔦
ill:鰍 ∫
equtionsご
...{ゴ )
(SG●Ja譴ち1990)
=響 =¥国手号 =響
3x +2Y +
which wili contain tlle rlrst line iFitt point(-4,-6,1)lieS on it.
+&(- 8 - 18 + 4
“守
coplanar.
are equal, the given lines intersect, i.e. are coplanar.
Putting r = 2 in the co-ordinates ofP, we get (2, 4, - 3) as their point ofintersection.
The equation ofa plane containing the second line ig
i.e.
lAMλ rr8,19",Mysonら 1997S)
=″
& Shor tbat t}e
ofr
72 + L2 + 1 + 5
h -3/lg
t-2 v-S z-1
-I-=-l-=-E-
&FInd the∞ ndltion that the b“
+ tl, which lie in the plane
- 6,
-?t
3r-2y+z+5=0
if
3(3r-4)-2(5r-6)+(-2r+1)+5=0 or r=2,
The point P will also lie in the plane 2x + 3y + 4z - 4 =O
if
2(3r-4)+3(5r-6)+4(-2r+1)-4=0 or r=2.
Since the two values
aaa
are cqtlauar and fi.Bd the equation of the plane contahrng them'
(6) Ure vector method to colvc tfio above problem'
ti“ s
+=#=*
(c) hove that the lines
--'g =-z-3
4
z -Y-2
17ェー47y-2を +172=0.
力e
Exomple 3・田。S■ ow′ ■α′′
are coplonor.
＋一
J.c.
曝
画…… LL陣
３
The equ…
5梅 7
121
ANDSOuD GEOM躙
...(υ
j〕
line.
3b;
〆
www.Engg-Know.com
ll1+mm1+nnr=0
and ll2+mn2+nn2'O'
Fig.346.
■ ■
123
VECTORS AND SOUD CЮ METRY
飾 M電
,流
=山
=洗
.・
- f'y!
ABendCD.
To
t r. rr,
==ff:T".ff
The equatiOn3 0fthe line oFshJttt distance Cn are
″
wbere 0 is the angle behpeen the linesAg and CD.
I
lr1,r", _. r,
wherel, m, n have thevalues as given by(1).
that it is coplanar with both
ia.
生 +y― &=O and 7ェ +y-8z=31.
0価eru30 3 The Vectors B=21-3J+K and D=31-5J+2K
are parallelto AB and CD respdively.
BxD=信
∴
(2)
∴Lngth
..。
ia
【
晟届,〃〕
=0∝ R_ん B,3x珂
and the plane cOntaining CD and EF
ie-
-r-)d
lCP, CD,
=o
-. E-rmple
8'40.
lincs
Itnd
the
r$i#i#=*
i=t-t=
(jj)
λ● Jjrles α″ご
J′
s
,1997S)
,S。 ″
...(ゴ
and any point on the line(jJ)is Fc-3r',-9+2r',2■
and
or
-3(6
+ 3r
jJ)
...(`υ
4〆 )
)
3r, 16 - r-'zr"'2+r -4f
-r-2r')+(2+r-4r'l=0
+3r')
+ 2(16
-r-
2t) + 4(2
+
r
-
4tt) =O
r =- l,r'= L'
+7/ +4= 0, 7r +29r'-22=0,
(iu),
get E = (3, 8, 3) and F = (- 3' (iii)
we
and r' = 1 in
Substituting r = - I in
whence
lJ'r
7' 6) whiclr
are the points on (i) end (rd) neareet to each other.
.'.
I21gthoftheshortestdigtance@O={
lhe equations
of the shortest distance
Obr Thir method il
of the liae of short
日9,347.
=′
..。
Any point on the line(j)is E(6+3r,7-r,4+rl
SinceEFl-boththelines(i)and(ii),.'.3(6+3r+3r')-(16
(Ban alore,I992,S)
dirrctim qinea ofthe shctest distance EF.
`"eF′
.(j)
Then the direction cosines ofEF are proportional to 6 + 3r +
._.(3)
'.' Ef ltobothA8andCD,
:. 2J-3m+n =0,3I.-S:n +?n=O.
so,vi,rs
0●
Ｉ
︼Ｃ
，
n be the
s
_:
nwnitudz and the equztions of the shotesr distance betueen the
;=*=i"* +=*=+.
・
1毛
ザ=ギ =ザ
島=ザ =ザ
.1脇
揚
鶴,1997 W i″
歯
∬ ふ″鶴′
艦1∬ "PSrゎ “ 搬雪
m
Hence (2) and (3) arc the vector eguations of tbe line of shortest
distance.
oba lte conditiou for the given lines to be coplener ir also obtsiDed by equating
the shortest distrtrco
(EI! to zro-
I*tl, q
=1辮￨=￨
'2′
=1獅い
EFI=0orlR-C,D,BxDI=0
К お pad回おぬe'ユ lED・
1 J―
力θ′
■ep● IPI′ .9o■ ′
Example 3・ 41.I弓だ ′
R=A+tB and R=C+/D
so that
品 =へあ =c,燿 b=B and ab=D.
SiTthes.D.ρ t ttO bothAB and cD,it is p_llel to“ x CDia BxD.
鮮鰤忘脚重二
li協劇
三1]
ofS.D.o=pmiectiOn OfAC on EF.
(3)
:Let the vector equatiOn3 0fthe skew linctt AF and cD be
=BxD/IBxDI
121=。
li -l il=Oandl'12 rii
..。
∴″
.
...(1)
thz eauatiow of the line of shortest distonce, we observe
Chh…
十
十
¥資 =轟
Iz,"sth,r
レJ2J力 orSp.0=proieCtiOn oFAC on EF
=(2-0)七 (1-0七 (-2-0七 =七
=漑諄
.-.
.」
Exanple
Jt
8'42.
[(3 +3)2
(En it
#
+(8+7)z +(3 -6)21 =3!30
=
distsnce with the given lines are requirud'
Find the equdtion ofthe plane through
¥=ザ =号
,マ
rie siο rrast aお ra2Ce
www.Engg-Know.com
-
points ofintersection
sometimes very convenieni and is espocially uscful when the
れ響閣ゎル “ ザ =号 =ザー
α
“
んe".
扁ea“ ″れd
=
#
bet曖れ′
the
line
...(i
)
"'( ii
)
le.M.t.E., tggzt
125
v[CTORS AND SOuD GEOMEIRV
The equation ofthe plane containing the line (i) and paraller to (ii) is
(2)烈レ PrcttgS aCy λαυをαcο 滉れ0れ
the thttd.(Fig.348)
F:1∵
71=0
+η +1&=98
G) The planes may form a
to the ttrird but does not lie on
餞
Now the shortest distance between the unes(ピ )and〈麗)
] I盤
Erarple
...(Jjゴ )
1,-2)of(JJ)on the plane(″
北
j)
rthepOint,L
+ b'y
+c'z +d'
da'
is -
- d'c
,]W -a,&;@",
The plane containing the given linb is
(ax + by +cz +/1 + h(a,x
r
-m
(Assc
λ
晟売 ″号
″
“ o働
m, teee\
Thisplaneisparalleltothee-axis(d'c's,0,0,r)ifc+&c,=0
―ct)ェ
+(bc′
tiarqtlor pism,
it.
that the phnes
“
=こ
ヂ“鮨売力だP"2● お
施た4′ ゎ
″′
″O ρra2● ■
“Any plane through
“
the line ofinter80CttOn ofthe planes
11を
...(i)
-1"+1&-28+肛 6L+6y-72-8)=0
or h=-c/c,.Then(i)becomes
―ろt)y+Cc'一ごt)=0
∴ 上 distance ofthe Origin from the plane(ゴ
l[1oz'
-
a'c12 +
(bc'
-
′Zα ″
...(jJ)
"ar議
“
.
∴
ゴ
)
b'cl2l
PrOblem3 3 11
JИ
●agaIonら 199η
翠TLttЪ 十
蕗識
:電 ∬
fお譜、R謡攪議
ど “
the amOunt orclearance between the cables
3,8)_It is required tO detemine
cated such that the co‐ ordinates Or
ギ =響 =ザ a配千 =号 =雫・
(0● 23● II=a,2000S,Myso"=f997)
Find a130 the point where it intersects the lhes.
ic
_
!a'-d"
ar + b,
lc/(。
-b'af (d, -cbl\
d = O = a,x
+
bjr + c,z + d,
+
lGulbarga,
Hめ
=′
2+.2)1/2,cc/cC2+.2)1/2,ab/(● 2+ら 2)1/2.
)are(-2,2,2).
(.b叫
8.tl5.. Prcue
that tle three planes
y + z = 3, (iil x - I + 2z = 4, (iii\ r + z = 2,
form a tri.angutar prixn and find tle area of the normol section of the
(i) 2t,
+
prbm.
I*1 l, m, n be the direction coainee of the line of intensection of the
planes (a) and (iiD, so that I - m +2n = 0, I + n = 0,
-ltnn
whence
,1990)
1=_l=_1.
″
Now
｀
(j).‐
r
in (ii) and (iii)'
・me
ぬ
駐協ぜ脇l出℃メ
eぶ庶誤胤f属梅
To find a point P on this line, put
Any three planes(no“ o Ofwhich are parallel)in""∝ tin one ofthe following ways:
)2・ Pね 3mり mee′ j"● jれ
,ifthe hne ofsectiOn oftwo ofthemお not parJblto the
“
ayD
デ。
1=おエ
.The co‐ ordinates ofthe point P in which(υ J)meets(f″
Erample
19991
8::NTERSECHON OF THREE PLANES
∫
une工
ヽ
e∞u面 o“ し
s轟顔ぬ
λ
=fwЫ ぬ
"』
Hence the poillt P is equidistantÌon the planes(じ )and(j三 ).
+ cz +
ユ
e口 meSDAttMCare
ふ
篭議翻r蹴棚織ι
懲黒F農誡貯鰐蹴説し
rttatぬ
Cil跳
ごa rectanplar panneb●ped Wh¨ edp"e
鵠繊
:峨獄織蹴冦
wladar祠
',lt1G,a'
We haVe
P′
3.Find the magnitude and equaabns ofthe shonest distanco between the lines
s}rcrtest distancs betcre€n:-axis ard t}1g lins
=せ、
」
狙=ゴ湿
200=型生七
Dおた
れ
ギ
発
子
発
昔
告=2● n magnttudo
告
妥ぎ
“OFPル “α
=2 1n magnhd0
ω=堅■ぽ‰攪鉾旦
Dおた
2ceげ Pル
"pra″
.・
υ)
=V=ギ
),
2胸
t}n
=ギ
is (3● +1,-2ri r+3)which lies in the plane(謗
if
2(3r+1)+35(-2r)-39(r+3)+12=0,ム ●.rr=_1.
.
yad● 二」θ98′
Slrow tlrat
...(ゴ
力a sa″ θ
“
Any phtonぬ e
1.Find the magnltude and the equatiOns ofthe ChO● est distance between the lines
{.
Ψ
From早
which is the required S.D.
ザ =ザ =ザ md¥=2子 =ザ
Fig.348.
+(-15+6め y+(16-7λ )z― (28+8λ )=0
“
prar_α ″απごαり ″P″ san′
■ rec P′ α
″ ″″
ZZ Jれ rersec`れ a commoぉ
・
o, (12+6め
A pOint On the 2‐ 8面 s is the o面 gin.
′
と ―d宅
ifthe line of section Oftwo ofthem is Parallel
(See Fig. 3'49)
(j)and(j二 )is
+b,! +c,r+d,)=O
(a+ka)t+(b+hb,)y+(c+hc)z+(d+hd.)=O
lCC′
●OFSec′ あら ifthe line ofs∝ tion oftwo oFthem lies on
“
〔)12■ -15y+162-28=Q (だ )6■ +6y-7z-8=0,
α油ご0だ )2κ +35y_392+92=Q
π
.Proυ O′ ′
憾′め●P● ′ゴ
Jlaυ c a co“ π02どれ●or'■ ″r頭
“
““
1=ち
Example 3'43..9ioa' that the shortest distance between z-anis and the line dt + by
+ cz + d
,0 = a'r.
8-,14. Proue
J二
‐
0
www.Engg-Know.com
t・
.・
=0
2xl+lX←
1)+l× ← 1)=01
127
AND SOLID GEOMEIRY
VECTORS
Also the point P does not lie on the plane〈
jヽ
(8)Theequation*+f+*+2ur+zvy+2rz+d=0rePreeentsaepheirevhoce
-v, -w) end radiur
Hence the glven planes form a tHangular p五 sm.
ccDtro i8 (- u,
Let tt be itS normalsection through P
=.J(u2+f
The equation oFthe plane through P perpendicular tO the line ofintersection ofthe planes
For oa writing it as (r2 +
and(燿 )is,
1←
-0)-10-0)-1● -2)=0
,― y― z+2=0
...(jυ )
e=(峰 ,ヵ ,2).
OR=ギ【И_ンリ2+(輸 _笏 )2+(2-拗)2}=ヾく
ゎ
)
4+1+1)=ギ
JAPeR=:解 X2=:争
譜
:・
=乱
・
odl*
Problems 3・ 12
1.Prove that tt the planes 2-3y-72=0,&-14y-1■
lti,口
MIE,I"3 VD
MI.
commollllne oFinte"ection.
1(I_J+4K)=2
have a
“
+ay
Show thatぬ e planesェ +2y-3=0,■ -4y+′ -4=O and 4レ ■3y-22-24=O forln a trittngular
pnSm
ttaご力ra,19鶉
5.PrOve that the phnes L+3y+と =6,&+4y+i=20,■ ■2y+&=Ofom a p“ m:obtainぬ e
4。
)
cquatiOn oFone oFits edges in the,卿 metHcal fom.
:'SPHERE
(1)De■ A● phe""働 e roc.sora p・
12′
″λJc力 renα ね8α ′α●
●■sた
tlu
sphere
andthe
x2+y2+z2=32
int m tF.esphererぬ Centre
gth議ヂ
ュ
重
ittRIittp°
出"“ ぶ競↓.=°
2「
ich h tヽ
`″
Whe“
ご
=C2_,2
d1-'
S=0
U=0
S+LV 30
S+たび =0
灘 T:ぶ syvSご
m of tbe sPhere
赫。饉
S∝
Fi9.3Sa
the.t!h-"': through the points
+'2 + fuu + 2w + 2uz + d = o
Let the required equation of ttre sphere be xz + y2
(1'
3)'
It passes througb (0,0' 0), (0, 1, - f), (- 1,2, O) and 2'
(0,0,θ )
...(j〕
d=0,
v 'w +l--O
+d=O or
1+4-2u+4u+d=O or -2u+fu+5=O
1+4+9+2u+4u +&a +d=0 or u+2v+3ur +7=0
1+ 1+ 2u -2tt
(du)' we get
Multiplyi:ns (ii) bv (rir) and adding to
u+5u+10=0
αα,ら ,c)=the radius r_
J″
tbpwh
of any cphere througL the circle of
Erample 3'46- Find the equation of
rtdiusb,i- t, 2, 0) and (1, 2, sl' I;@u its cenire atd find the
@, ;
"r disrance""α 解
.
・
dius a" is
pl*rc
ご漁
(2)The● quntion ofthe sphere who3e Centre i3(3.LC)● nd ndiu3 r9
KX-3)2.(y_bF+〈 2-c)2=ノ
-n2).
inbrsutbnof
The Flxed Pomtis called the cマ π′
′
マand the cOnstant distance the mdi“ s ofthe sphere.
For the disan00 0fany pointPKx,y,2)On the Sphere m)m the centre
lm particular rie“ r“ 口 ri。 orri2 sPA2″ ″ヵOse cearre ls rie Orむ
{f
For the equation
=満 =議写
j2■
1r1 =
is
lntersectin a lhe ifo2+。 2+c2+24bc=l and show that the equations oFdhis line are
満
d'
get zTal
(6) The equation
=0,8r-3■ -3&=O Pass through one
2.Prove that the planes R.(7J-5K)+4=0,R (21,5J+3X)=O and彙
3.Prove that the phnes,=qy+こ唸,y=o2+C工、
2=b庄
-
quation of a sphcn b such ttut
$) it k of tlrc uond degree ia 1Y, z,
{ii) tlu cefftcilnt of l.2,trz, z2 an equal,
atd $i\ tt*rc an rw t2rTrllt @nt4ining 7a, * or ry'
plane
is L circle and the saction of a splure bv o
(4) Soction of a
"ph";;';;i;;
great
circle.
its ceflt z is called a
Thur the equations l+ yz + * +fus + 2v1 +btz + d =o [Sphere]
Planel
Ax+By +Cz+D=A
and
and
tr
centre
(Eig'
having
3'50)
taken together represent a circle
R=(拓 ,%,輔 ).
Hen∝ ぬe…
=
*o +w)2 =u2 +u2 +w2 -d'
aa
ir'*,il'*,17'uy
'(r-s'f
+lg
-b)2.+(z -c'f =?'
and comparingwith
it clearly repreeents a sphere whose centre is
(4, 6, c) = (- u, - a, - u) and radius = t/(u2 + uz * -t "
llnla tIE
Solving the equatlons(二 ),(jjJ)and(jυ ),we get
Al∞ Z ttfrom Pon the Jane o=丈
+xl-d).
+ 6/ + }tty) +-(22 + ?tttzl
or
Sol宙 ng the equa● ons(J),(jj)and(Jυ ),we get
∴
!a;,o;1
αPIピ
α
O md radi鴫
「
then th
Solvins (tii) and (u), we Bet u =
From(:i),
″ =υ
-
...('二 )
...(二
j)
じ
(二
υ)
..。
...(υ )
26
15
n, u = - i
+1=f+1三
十
(i)' we get
Substituting these values of u, u, w, d in
-,:
,2*v2*"'-f'-?t-)'=6
which is the required equation ofthe sphere'
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...(υ
ハ
MATHEV,AICS
HTGHER ENGITEERING
123
Its centreお (1ゴ 14,25/14,l1/14)
and the radius =
(-
C25/].q2 + (- 11,/14)2 - OI ='/(gZf)Ztl'
Sirdthe equation of tha sphere which hoAs (,;r,!r,
15,214)2 +
EranpleS-4? .@)
zi
(-u,-u,-ur,
sincotbeplsrepassesthrough
z)
Eeocr the locus ofthe ceotre
and (x2,72,
os the
Eranple
"ir"l"
"i"
plane2x+7-z=3,
(o) Let P(r,y,z) be any poiut on the sphere baving A
A(-tr,rr, z1) and B(x2,y2,2) as ends of diametr (Fig- 3'51)'
Ihen zAP and BP are at rigbt angles.
Now direction ratios ofAP are r - x\ ! - yt, z - 2l
HeDce
r-
rz,
(r - rr) (r - &)
!
- !2,2 -
+ (y
y2) + (z
which is the required equation.
(D) The equation oftbe required sphere is
(: - 2Xn + 2) + C + 1Xv
or
a2*y2+r2-xt
- z]]i(z'
-
2) +(z
*t.ooa
4l{z
ot
*
+ tot -''z - I
=
o
t
:
[,ffhree8
order t'hat (D
- rj'' zt
- G6'i
+2\=a
rrhence (i) be'comes, rz
'*i
+
ti"
0 which is the required
r' * * ^ u * q- & + 4 =
*
,then
ぬ
e areaご嗣er=♂ =醤ェ
S働
●
″
α
れ
″は ″α
… ShA3 a
ι
協管
ttη究淋常焼鯛郷』
繊。
二十二+二 =2.
MIE.,1997Ⅵ
+♂ +め
=ヾ
Srdhs α ♂
猛c“ 43め mぬ試■
e centre dぬ o中
壼ぬ
h:“
め
・
∴The equation ofthe sphere is
2
2+。 _g)2+e― λ
)2=′ +g2+■
ノ+y2+z2_2「 _aヮー2ル =0
_準
Thus the equa●
¨
(:)
・
=零 Similarly OB=を ,OC=%・
on dtheメ aneABC is予 +発 +ふ
;
it"
rao*
on the plaae
ae small as
I + !-:
z -'
jl
;: ;'J ;: I;
6'
Poeible
aud meets the aree
costaot#;ffi";h-,retho origia
6.
-ii-iia orure
-' Arphere
, , 2. ,LD
e(:2
(i) trianeleABC lic6 on t'he sphct
:'l;lr':!l-Orr.
*y- + - =-tr'.-'
oa the rpbe'"
tO t"*l"ar* OA.BC lies
"
p*"* *T*
o. e
,
pr.".
the otigin ou the Plu
inA' B' C' Prove that the
lH":ffil,}
i'i.if::it"
(Assam, 19N\
ftom
of thc foot of thr, psrpeudicular
?i:,
_^
*,, o, ". *""" *. *"
**r*rrr$}rF.ffi+=#-*t*s:;1t, (tuua
;':"":":::#;:;
leeo\
:lr,X"
-
=O in(ゴ ),we have
‐o,ι 認.OA=ェ
J"
of
Hen“
ノ
i"lil t-*J*
iiiU'"
a2ギ _メ =号 ―
幾=警
To ind OA,pttng y=0,″
ffi'ffiHT*.Tfi:::ffi:
radiur'
f"a iti t*tre
;l
equation'
of a diamcter'
aud
ef"o
he mhts A rod B as ends
ofthc gphere wbi& har t-- '
(5. Gujorat, 1989)
s. obtain tbe vector equatiou
(0'
1' o)' (0' 0' 1) €sd
(1'
0)'
0'
pointl
passer through the
)
which
spherc
the
of
4. obtair tlrc oquation
(in magnitude)
r
。
centre
r+Y+z=3
Protrlens 3'13
I-
Iet the given Plane 2, +Y -z - 3 =0
crrt the sphere (1) in the circle PP baving cent;e tr'
(2)
.'. p =perpendicular CL from C on the plane
c_ハ
s ; Mv sore' 1ee4
on t'be Plane
(j)(Kararat 199o S)
andradius ("t={to, t/4+t+14)='l(61/4)-
Ifα be the radius ofthe cittie FP′
1ee4
great circle'
tbe giveo "ircle as it€
Its centre is C(0, 1/2, 1)
F19,352
s; Mdrds,
the given circle is
of any rphere througb
;;'(4
..。
e*L+e--2'
"(')
l-h/2'
o,.!,..1i:ru:1l,.ffi:f;r,.".-'ii=;
its
-t' O"i"
In
Fig.351.
-b-L4=0
+
Li.
zz\ = 0
-
h) ofthe aphere ie'
houing thz circle
tlw equatinn of thz sphere
*0' Fi'd
*,
*,r-ofri.f
- -il"
22,
- yr)$ -
f'g'
*.*.*=t
ertremities of o d.bmeter.
(d Dedwe ttu equatian of the splerc dtscribed on thz lite
ioinins the ooir*s (2, - 1, 4\ and (- 2, 2, - 2l as diameter' Find
in which the sphere is intersected by tlu
oSth"
itn
and those ofBP are
d'' zO*ft*fi=t'
b' b'
*ms.m*g
:"r.
#:*ff i#i#,;,,
; Mv s.re,,ee4,
Shorthattr'cn"istse;-6'0)'(4'-e'o'(6'0'D:[-r,1"'fir}]Yr'-,*-W:;f,;,ia1;
tlle crrErE ' ' ' of thc aphere for wbich
equatioa
t2'
Fioil
the
'great drcte'
11.
=1・
"iTzt'il'i
Qう
www.Engg-Know.com
= o is a
131
VECTOtt A「 4D SOuD G[OMETRY
.g.
Substituting the values ofル
Fiadt.l.eequatiooftblphdGbseirsitoccntreoatheplaae&-5y-z=SandparciDgtnEughthe
(fuipur' 1998;Kttnmpu' 1*6)
:-2y+z=8'
12
.ir.t"lty'*
-2r-?y+1r+8=0,
14provethattheplaaer+2y-z=,lcutsthetphde*+y2+*-r-z-2=oiaacircleofradiuauoity'
of-the aphere which las this chcle a! oae of its great circles.
i.iia-rLo tfru
"if-Uo"the ephere throAL the circle l +y2 + r7 2. * 3y + b - 6 O aad
=
t5. Fiaal the equatior of
(Myna l#)7 S)
′+y2+22_L_夕 +駐 -6=O mdF口 31ng through the pont〈 1,1,2).
es the two required spheres.
I(hhoIErI rybcrcr. I\tto splutrls are said to cut ottlr4onally
if tlz tdngant plo:E; at a point of inhrwtian orc at right anSlzs (F*35s).
llre radii of such epherec tLmruh Oreir poiDt of iatareectioa P'
beiag I to the tangent plaaes at P are also at right angleo.
h
LC Prove thatthe J置
ノ十y2+z2_2,3y+ι -5=0,"● &+1=0
6=0'+η
mdノ +ノ +ノ
―
%Omm極 33me Ⅲ“
嘔賜舞翻貯L蜘
ー
缶―
い
●●
)
Tlutt ttDo spterlus cat onhogonal!1, if tlv squan of tlu dbww
}r;asrzrtluir entrec = sum of ttte squota of tleir rodill
Erauple 9-61. Sltow that thc andition for sphores
20.EQUAT10N OF THE TANCENFPLANE
2● F'ル
ガο
2レ
"2●
￨●
j“ ′
ngear prar α,α り ′。
←1,yl,Zl)● frル響ル″
i +f
ノ+y2122=α 2ぉ =1+"1+露 1=a2・
ard
1,yl,21)tO the rVen Sphere,the
If Ffr,y,z)be any point on the tangent plarle at Pl←
rttion ratios of PIP are■ ―,1,y yl,2-zl.AIso theと田市on ratic8 0f radius OPl are
to
-0,yl-0,21-0.
1←
“
...P101り
【
た raagear′ た″ ″ ←1,yl,Zl)r● めe
ノ +y2+22+2● ェ+2り
ae. 1u - u)z
rO
2.
(Homirpur'
*eC(-u, -u, -w),C(- a', -v', -rl)
* ,2 - d1,
-
19961
aad
-'l1u'2 +u'2 +ua -d.)-
will cut orthogonally, if (Ce)z
u12 + 1w
-
w^)2 = u2 + u2
* -2 -
d,
+
=
f
+
ra
ul * rrt + wfr ' t
2ttu' + 2uu' +?tttw' = d +d'whie.h ie tlre required condition'
.r2 +
y'
+
I
= 9 which are
parallel to the plaoc
6td'hra'
thrul8h (1,
1,
- 3).
t.
2y _22 = 7
at
,
1999)
Nqtw' 1997)
alld
- 1' - l)19911
(3,
(i) kovs
tht
.(Blnpal'
2) aad paecee thiorugL tlte odgito'
(1, 2,
-
(1,
- 1).
point
a,*)e?+y!+22=3'3r-2!+42+3=0atthe
rE Er.6..t
&Fhdtheequatioaof .oetirymtliDGtothe
(kua' l99o\
… (二 )
I
1,
thc liao joiaiug thc points
tlat the ephcre /+/+ ?-*,+ey+14r+3=0 ilivides
retio 1 : 2'
the
in
ederaelly
ead
iataniUy
O)
f,
a)
tf,'S,
[2, - ""d
J + !2 +z'-21'
& Fiad tho shortest
aod the loagort dietalrce ftom trre poiot (1, 2' - 1) to thc ephcrc
+ d2 = e cut orthogonallv if
B
+
2U3'
*'
t
e. Shw that the
?. show
(3,-h/2,1)
2+271+44=0
ユニ=ヽ
ι Jし
l15十三
よ
う
Or」
ヾlC2つ 2_7041二 _!:Or _4
"""*"*l 3;
￨キξ十七
1=-27±
+d'=0
d'
t'he co-ordiaatas
tbe equetioa of the rphera which liee -in the 6rst octaat and touches
planca ir of the form @ + f + zz7 - 21,1r + v + z) + 2* = o'
plaaea.
(ri) Fild the egrreton ortue rpuere p".,i"s o'-"8r, tl, 4, 9) and touclrirrg the co-ordinote
tlre PoiDt
at
+
f
0
+
+
2a
=
- Qy
6. Firrd the egoatioa of tbe spherre whiclt toue:h€r the cpher,e * + Xz a2
{.
+12 +"2
.'. Its centre = ( s, - h/2,1) aail ndius ='l1g + (h2 t +l + 1 - 5l = (5 + &2'l4)'
the sphere G) will touch the plane 8y +&+5=0, if, cistsace ofthe centre
rom the plane = radiua.
3〔
=d+
fiad ihe equatiorr of the trro spheres wbi& pass t'hmugL tho cirdc
* +f + zl'-zo+2f + tu -3=o,2.r+y +z = 4 aod touch the plane & + $r = t{'
pacses
+5=0
-& -22 +6 +h =o
l+f+22-&+b-?z+5=o
,
2W
8. Fiod tho eqgatim of the opfere which is tsDgutial to the plale
The equation ofany sphere through the given ttde is
l
+
3r+4g+fu'=7.
7・
“
+Zu'x.+2u'y +Zw'z
Problems S'14
y+yl,2r● z+″ 1.
l,2 rOi+■ 1,2y ra “
●
rrね eca`″
λo,Pたたたr■ ●
ァ
C ror.ル
″η″ 3dlc● J● アごお
obュ rた coadi餞
"″
"ル
"た
“
嘔 "ro r● “
″′
ο
た PJ● ●●● rLル υ `メ
=′ た
"崚
plo 3蒻. F菫 d ′■● ●9“ α質0■ S ●f rλ ● 3piere3 paSSIag rλ rο gん ′λe rfrcr●
―
“
2+y2+22_缶
“
は Mニユ,19903
鰤 ′
たコb“ 3y+金
暉
+N
Flg.3■3.
1. Find the equationr of thee tangat
planea to the sghere
%--!r-w
`←
_2+5=Qy=0"わ
+ (a
,
or
+滋 +ご =0
“
slg
Now there spheres
sP麟
xxl+yyl十 Z21+ucx+xI)+▼ ● +yl)+■ くZ+Zl)+d=0・
施 oあュ●
■ 力″だた′
Ph″ α 1,yl,Zl)r● ●Spた rel c■ a■ ″′r● _1,y2r● ガ1,′
r`ル ra岬 ′
“
+y'+r2
Zunt
v = .j1u2 + u2
/
1,21)heS On the sphere.
This the desired equation ofthe tangent plane.
S羞五larly′
J
ortlwgonal\ is
their radii
―″D+y10-yl)+ZllZ― ″D=0
=1+〃 1+″ 1=″ 12+y12+′ 12=a2
cut
+rs +2rr+2w +2wz+d.=0
tlhe cenhes of the spheres
Since OPlお nomalto the ngent plane at Pl,OPl上 PIP.
■
in(1),wo get
2+y2+″ 2_餞 _4y-2+5=0
,2+y2+″ 2_麟 _里ノ+″ +5=O andェ
""u
=Ir.I;;
｀Q計
・憲氾盤驚誌it漁4乱ム
巌∫
=°
md=2+ハ z2+ぃ 8y
tと
+1跳
ι
鰯
〇
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ノ
撤簿∵Лl∬聴糧
l轟濶
常減り
憾
lAndたに,199'A
2]+}yz=l,z=t.
irin,1997)
hence:r =3
t'3 Y'l z-2
lm
r-3
v-1 -1
z = I, where -T- =-;
=;
_(11)
ヽ超帆
続
1魔k+"¨ “
掌ずヒ
the req“
鴬稔瓶た″ n
僚
;霊
胤
,.e∞ 枷 h“ い JP ShouH面的
∴ ギtギ +ギ =1嗣 ″ +″
+n″
e…
…鴛
,か
:lξ 露
。
“
=■
鳳慟
轟儀鳳箪
"u話 "
ml%鸞ド
X2+,2+Z2
=1』
′ +,2+″
ル (嘔 +む
謹縣鰹釜轟祠
2_tar+り +″ )は十
;+
requい
醸equatお n
Probl… ●316
&漁轟
e'2_… =Qz=0
…bo3e
tS
1''…
(3,4,5)and
■犠翼鵠胤躙ぷ鷲:aplmettGM、‰。
鳳
4鐵椒雛」i管
+z= l-
い
ulhose wほ
9a"ち
e ttgtter北 S\
｀は20嗣続“
(Nottalo,m.
01編 `麟
mc“ ●
thelr d― ●
l協翻
? +!'+27 =4,r+!
≧3中中哩 1■ P…
魚源識燻
I遍 ::i2+♂ =勧 2 0btaln tbe eqaa● mf,0∞ nt
“
。quRhm“
a:蔦 thatこ
ふ
Ine賞
安響
憎 TWblChp"“
[ふ ″
亀予面」
濃乱器r』:￡諾轟ご惑畠
!篭 l詢言
f
S
:高麟藁 Dyl■ 1圭[ル
ぶ濁c淵
電撃″
島
αnd
″灘
ω驚
ご戸
Ic″ 臓 λ
"懃
:デ
融品虜
脇
`“″
高Ll'」し″
:`み
i品
a解野モ′
"=
rg・
霊脱
=葛脇膨あらむ腸攀摯ヽ
I曽
l堂鸞留
l留 a:
∞
td赫
ま
“け
ぶ翻雛聯ア軍嘔品・
“
:露
Let P← ,y,Z)be any pOlnt cm the co■
、
,OSSible only ifthe equatlon ls homogeneolE
www.Engg-Know.com
F19 354
鷲
e with Vertex 0 8● d
∠
あC=∞ ICセ J50
…iFうも∞Ⅲ
'999)
1996
革
an3 00
Wi蠅 4
報軍綺
li轟椰
農
￨:lililill∫ i器重
yZ← +:)+Fr← +:)+ツ ←+:)=。 WblCh"the
油脇
.hne lll l・
鵞
諜卿
ぶ雅」;lヤ
:摺 ∴
ぼ路1魚
1期驚葛搬 I盤盤織
- l/n ,y = 1- 67r.
期
(i)
:+:+:=・
…(:)
Substituting tle5e values ofr and v ia 2t2 + # = 1'
2 (3 - t'/nt2 + 3 (l - m/ ntz = L
(il i3
ElimiDatingl, ,n,,l fiom (i) aDd (ii)' the locus of the li:le
rノ +ザ
meet路
Or
Any lioe though (3, 1,2) is
DEets
L=tlle pLne
“
囃競 i職額撫轡鰍輌檄孵捌
It
雛霊揺1:脇霊胤競
∬篇
鰍
れ
協
肥
a
2″
Fig ●66
1シ
4
HiCHER ENGINEERINC MATHEMATIcs
Now ttre direction ratios of Op
∴cos 30・
=￢
dfi t, y,z
VECIORS AND SOIJD GEOMERY
and those of OC an
l,
2, 3.
Efr''l2t'/t -l/m
再静号猾濯腎葛
=
z/n
be thc equations ef
.ny
one of the two liues
)
qgt th" plane or+b!+a.=o moets.the
a-r+6-l+c-r=0.
e. Prwo
3Zl. (l)
12
寓 =0
+躍
み
批
服
げ滋載
鷺
赫
璽
中
COr F″
翅昴 ″ Й。
""な
“
∴Vertex ls.A10,o,η .A generatOr ofthe cOne is
・
oru'mirc und,ition ag.,-it raay intdrsJ a
fiv""
cylinder.
、
■1 ‐
b
-.
Pie
=面
α
4*
+ 4Xl
-g2
-
36 = 0 which is the required equation
二[1-づ +[1-等
ofthe cone.
Problemc 8.16
'' XTotattr;S $* XH*ar
arisalougthe line x--?ry=x.
ccoe whose verter iB at the orisia aad semi-verri.ar
aaste ig
2' Fiad the eqrietioo oi-th"-"oo" *hose verticar
Also show
angr e is
*,/Z,wrich has its verrer
t\at
the phne z = 0 orts the conc in tro atraight lines inarin6{
6j sa
3' Find the equation of the circular cone
which pass*
t
tt-rrgr
the ori8:itr aDd ads the line
fta*"r"y1"".^"tbe
:/2 = -y /4 = ,rg.'--ris[t circulr cone gterated
5. Find the equati* orth" .4rrt
origiD- ObtaiD the semi-vertical
"irrotar
aagle
6' Find ttre semi'vertical eagle anti'the
thc point
(rw
S:T#j:i?
", (Eamirpur, lggL
x'I8ht
12J
,
rzs.
o. r,2)"rrd";
and has itr verter at
-
3)
ard the rfircctio cosires
ly)z +
(nu
-
ny)2 = a2m2,
which is the rcquired equation of the.cylinder.
b)t a
from
?he constant diatance is called the nrdi us of tle eflinder.
T-a-plo S'fi.fhz mdius of a twrmal section of a right cirarl,ar eytinder is
(Mysorc, t99s : Andhra, 1994)
by rctatiag tJro t"i ,ri = yi'= ,/s about
the
-.
.
licsa.longthesbaightlinc
passiog through_rhe co-ordiaetc
-:, r-,,
ザ =響 =子 ,μ お"商鷹
2
units ; the a:as
rCrnα たルc1993)
A"ht Ol■ the axi8 0ftheり linder is A(1,二 3,a and itB ditc"on ra饉os are 2,-1,5.
´
∴h actud如菫
are島 ,静鳥・
"cosh“
with semi-ve*icrr angre 3f, vertex at the point
of wUose-aris'aJjin, _ r.
8' Fiadthecqustic!totheliueciDwhichtheplane2r+y_z=0antsthecone
r,
-
cylinder- M. A dsht circular cylinder is a surfau Eene,zted
-qT.,rhr
stmight
line which
is pamllel ta a fi,ud,linz ind i^s at o c,,nst,nt distance
ii.
r"a fr"
Sldfassiaf thro&
-r' rulc
the equatim d the right sirctlar
(2,
日g。 357.
ITず
Hence the 10cu8 0f← 1,yl,21)is
jr',rj
*", *ffrffi
G.ild of tf," eri".
(iiiwmpu, 1996)
orL" ig[t circurar coue having its v€rtex at the origi,
the circley, + z2"qo"tioo
= 2s, z = *
cooe
:
2+″ 2=.2
But these value8 0f=and′ 8a鮨 ″″
Ftg.3.56
a36z
)
雫幸 =雫 .
″
=″ 1_午 and z=zl_等
=I害
鴻
or
.__(″
ne generator c)mts the planey=o,where
ia where
=COS2
II m
Civen g画 dhg d“ L i8=2+″ 2=α 2,y=0
ぬ
昨
ThuS
tjrle guiding curue).
fi:ed liae the arjs of the
"iu
子 ,午 1雫
any polnt On thee cone
sO that
t e arectim ntiOs
cone S°
争tP← ,y,7)be
atぬ
'」
s an angle α with彪 ,we have
嘔
ld
4'r., yr, zD be any point of the cyliader so that the equatiori of the generaror through
8●
JAP冨二
す鑑製ド
ず生
』
職
tcaf
Eernple
0
署借庶田翻露富革乱軌護憲瞥げ ‐
ご
“
2).1__2
=露
―
cos
‐ a= 0.0+3.0+←
ヾ13
"urre
gelrrlrutor aad the
3'87. trind, il:c equatian d a eflhder whrxie gerurrlrting tines hante the
diration
@sin* l" m, n atd whbh pa,,s thtot1gh the chtumference
of tl, fo"a
*
*
,i
o,
in
the zox
=
plona-
=15′
―
‐――
iJ
︰
︰
︰
︰
卜
い
颯
鋼
﹁
嘱
areす
ltr+a,+ult=O in perpendicular linos
CYUNDER
Ihe straight line iu aqy position is called tbe
..
れ
,「
意
cole
A qlindcr is o surfaa gelurzted bx a struigttt rire which is pratter
- -D.!'
to a fired, tinc ard.
sotisfics
tば妊饉
た■Ъ鍛島.楓押"電
・ “… Ⅲheη +れ二0こ【
=普
io which tlre given
plaae ots
tb:giwucoaesouretwegrtz+r, -n=oa,di2
-n,.ar,=o.N"**rr"-t},*"ixroequationsfot
l, m, n.l
cone
i*-yz+Bz2=O.
工rt P← ,y,2)be any point On the 91inder.Dralv Iリ
」
NOw
A■
〇
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f=Pr● edOn OfAP On A■ f(axi8)
イ上to the axis A■f.Then i`P=2.
137
VECTORS AND SOuD CEOMETH
=←
―
⊃畜+。 +め希+・ ―
a
a fUatnsoquationof
︲
．
¨
嶽
Al"oA,P=@
.'. from the
zd
°
r
rt.
LAtrIP,1Att12 + 1Mfi2 = 1142
nレ
3・
53.
た
2ar φ
げ
瞥こ
"動 “
ノ
+y2+′ =llЪ 賀芦
:轟′
宅ぶ霧′
篭‡i競
ぼぬ
e giVen壷 Lけ thrOu由
総
:鵠胤雌轟器
ぼ
露
::零略ふ
警
躍RttCm精
The su.rfacc rcpresented by
.・
...6)
(r) Etripeoid,
genqal equatian ofthe
-
$ # 5=
(, It ia synmetrical about
second dagtc€ in
iB of
t, !, z is called a quadric
the form
t.
each of the co+rdinate planes for only even powers
ofr, y, e occur
(ij) It meets tr[g 5-q*is at A(a, 0, 0), A' (- o, 0, 0) ;
theY-r-ie at B(O' 6' 0)"6, (0' -D' 0) ;
and
thez-"-ig at C(0,0, c), C (0,0, -c).
(rir) Its Bections by the coordinate plaues ar€ elliPses, For the section by the vz-plane
ellipse
Z
Y3*
fi+-"'=L'et*'
(iu) Ttre gurface ie generated by a variable ellipse
*y2-h2,r-
.r,radiu8 0fthe circle=V(OA2_OL2)=マ (9_3)=ヾ 6
"'"3=L-7'z=*;
c to c) and ir liaitod in every
Thus radius Ofthe wlinder← r)=翡
r■ ,y,2)be any point On the cylindeL
OP2=OM2+MP2
(as I varies from
direction-
then
ーギ
-0)∫ +6
ト
:0-0)+ギ :・
Hence
its
-
shaDe is
i CP
a! showu in Fig. 3.60 which is
like that ofanegg.
0)―
2+,2+′ +7+Jz_な _9=O
″
+ ?dy +
iu its equatioo-
(r = 0) ia the
千=士 =子
(1+1+1)=V3_
=2+y2+22=[1デ
\x
+byz +.o2 +4!z+?azs +z[:cy +2w+zuy +bM +d--O
Also O二上tom O(0,0,0)。 n(j)
=ヾ
+ Zflry + try2 +
which can be reducsd to aly of the folloring standard formr ao ureful iu engineering problems.
We now, proce€d to atu{y their shapes.
瓢 sis the reqwed equatlon ofthe nght dコ dar wlinder.
∴盪 8ごぬ e Cylinderお
al
conic
s.2s. ouADPtc s[rPFAcEs
aJ
y+髭 -16)2+4=← -1)2+。 +3)2+c_2)2
2Gピ +29y2+影 2+句 +1。y2-20zr+15け +3レ +71■ 0.
ェーy+z=3
z-pleae,
5. Fiad tho cquatioo to the cylinderr whoso geoeratore intarrsect the
c = O, z = 0 aad are perallel to tlu line r/I =y / n = z / n.
sqr{dceora conlcoid.
thus the getreral equation of t quad.ic surface
百
0(2
or
thecyliaderwhogediFeetingcrEve isJ+22-k-22+4=0, y=0audwhose
(0, 3, 0). Find alco the sror of the aastion ofthe cyliader by a plane parallel to
r-ir e6tai16 lf,s point
町
3駐、
(2)
Exnrboloid of one fieet:
t .#-
5=r.
(i) It is syumetrical about each of the co-ordinate planea for only even powers of t y, z occur
in its equation.
Which is the required equ饉 on.
Probloan 8.f7
It meets fis 1-qrir at.{(a, 0, 0), A' (- a, 0, 0) ;
the y-aris at 8(0, 6, 0), B' (O, - 6, 0) ; snd f,f,s 2-nvis ia imaginary points.
(iD
1. Fiad the equation of the rylinder whose geoeraton are parallel to the lin s s,= y/! z/3 aod whosc
=
" guidirycl,:ryeiatheellipselia.r=l,z=g.
titurha,2NN ;A,A,am, 1999;,{u]d,hm, 19981
2' Fiad the equatioa ofthc right circular cyliader ofradiur 2 whoac arir passea through ( l, 2, 3) and has
orrectroa ratioa (2, _ s, 6).
Maryalort, l99f) ; panjab, 1992 ; Andhrc, 1993\
S'
q" equatioa ofthe right eirarlar cylinder degcribed on the circle through thc points (o,0,0),
I"d
(0, c, 0), (0, 0, c) as guidiag cuwe.
(diD
Ite gectiou by theyz-plane (r = 0) is the UvnerUfa
Its section by the zr-plaoe (y = 0) is the lyperboh
www.Engg-Know.com
$ -t=r,
{o' -lc' = r.
(i
(i.e. DE,
D'E'\
-e. FG, F G"t
i
139
vECTORSAND SOuD CEOMETRY
Its sectioa by the ry-plane (z = 0) is th" etUpu"
{a' + 4
=r
b2- ''
O Coneギ +手
(iu) The surface is geuerated by a variable
ellipae
t
i-#=
ry-plqng.
Hen“
*$,,
=xae r
油shape撻 8
varies fiom *
-
16
-;
and
erten&
16
rnfiritv
―
手=Q
(I)Itis umetrical about each ofthe co・
meets the axo8 0nly at the ttgin.
ordinate planes.
(J〉
on botb sides of ttre
(燿)Ito 3酬出 n by the yz‐ plane● =O IS theI面 r Of…
“
dabrム
・■ぬat i…ヽ
8hOWn h,6161wl,お
y=士
:z("."and動
t lines
,
Its田ぬ m by the“ ‐
plane o=O iS the patr oFstraight hnes
″
zCr.FOC′ and
=士詈
80Ctimけ theり・山
I鮨
r00.
ee=o bthe pi武 d“ 鉾
+手〓0・
手
e dlipseチ
0鴨 daceお辟
,VⅢ 劇
暉γ
priesl
ギギ12=ル ψ
“
興
JⅢ
dS toinmity On both side3 0fthe 7‐
pl呼・Hence
i鮨 lhape is“ shttn in Fig.363
p―bOltti手
olⅢ Ⅲ
ギ=等
(i) It is symmetrical aboutyz- andz:-planes for only even poweri of :andy occur in its
equation.
Oi)
,
It neots the aree at tho origin only and
(iii) Its s€ction by thep-plane (r
= 0)
lz
Flg.361.
0,い
向
聰
b轟議 tWO sh_1手
+手
饉ぬ abOut eachご e00‐ d守
● ・
を臨器
“
(jJ)m meete the z‐ axIB at α o,o,c),σ
(0,0,一 o
_ュ
")L section by the″
Its sectiOn by the 2κ
―
ph,bΨ
ttr…
nけ the
me← =o遍 thehyperbolaf_′
_plane 67=0)iS the品遭
J_plane←
￨
1鍵
脚
=OL遍餞
F蠍
∬
五
i…
is the point
(iu) The surfac€ ie generated by a variable
￨ :
ザ
=上
0.●
ニ
硼
壺
-)
￨'2 0CCur
and the r雌,.3XeSinttarnarypOmむ
.
轟
血
it
extends to
itr
rhape
infiuity
r' =*
"
"tUp""
"ttip""
Hypcrbolic paraboloid:
(O
It
3-6,1
EOB\.
.
-f
*
{ =
z=
n(as
t
valies from 0 to
"L.,
and is like that of a rablc.
=+
1-*
az. bz c
is symmetrical about the yz- and z.r-planes for only even powers of :r and y occur in its
equatiou.
(ii) It mects tbe axer ooly at the origin and touchea the r,y-plane threat.
山詭サメ叫
―
DOD').
{az* {b2.= O
Oril Its section by tbeyz-plane (r = 0) is the parabolay'
一
G.g:
above the ry-plane,
ir ar ahown in Fig.
(o)
.AC3,■ 'CB)
"ef+f=・・
i市
aad
Hence
(J2_DCE,D'CEヽ
ヂ ::=1・
恥価
(z = 0)
日g.363.
―
=三二
手
h認
Its section by the ry-plane
G
302
z, 1iit.
ir the parabola y2 =t
Its sectioo by tbe zr-plane (y = 0) is the parabola
G
,
toudres the ry-plane tbreat.
ItB 6ectio! by the zr-plane
g
= 0) is
Its section by tlre ry-plane
(z
-
_
www.Engg-Know.com
the
0) is tbe
n"rlofa
;r2
1f
part of lines y
=
=-8' ",
z
(ie
.
DOD")
(ie. EOE).
t!r
(not shown in Fig. 3'65.)
:4:
HIGHER ENCNEERINC MATHEMATlcs
pamlblto
the
xoris.
.'. the surfece
which represen ts
gener8td isyP +t2
=tz
r'aris (Fig' 3'68)'
r.'ishr'ci;;lar cliindc' of 'oaius a and ar;is
(2)Blgbtcfrc"ft'*ot"Wf'""f(t)='nr'tbegeneratingcurveisastraightline(y=mr)
as
patsing through tbe origin'
o.'' f +22=Jwla
(Fig' 3'69)'
.'.Theaurfacegeoerated is y2+22=m2l
u*iue*ical anile .-and otis as the x-axis
which rrprvsen te a r*nt+i;;iair*" of
(iu)
Ihe
Flg.366
Fig.365
F:9.364.
surface is gmerated by a variable
uyp"*"ra
{ - $=&,"
=
*
it ertends to infnity on botb sides oftbe t -plane. Ilence its sbape is as shown in Fig- 3'65(7) Cyliaden An equation of the form f(t,yl=0 repreaents a cylinder generated by a
straight iine which is iaraUot to the z-aris and its sectiou by the ry'plane ic the cure
f (*,v)=o (Fig.3'66.)
In particular $) y2 = hrrepresenh a parodlolic eylinder,
and
.-.. ,2
rlt)
t2
j+?=t
represent! ao. etliptb cylind*, fitil
*-#=1
rcpresents
a
F1g.369.
2_ノ th. gene"ating
6)sI=。 w.earO劃
●
.-.
to
r-r-is
so
that
OM = r
MP=′ (0め
As thls curve reVOlv●
or
...(1)
∠yNe=90。 .・ la′
t # "fi = ''
""d;;
=Mo2=ルぃ′+ハ )2=z2+y2.
Now substttuting the values ofMP and MO “
in(1),We
have
*
f
Flg'371'
+ y2
= a2\'
=o2,
reuolution'
which is called an eltipnid of
-
ra2
ヾ。2+z2)=f← )Or y2+z2=Iftx312
F与 367.
which is the equation of the surfoce generated by the
(Fig.
3'67).
revolution of tlw curae r -- f {r) about tfu xtxisSimilarly, thc surface generated. by tle reuohtion of tle cun'n
= tr(r)lr
+ J =lfcr)lr, (ri) r = f (z) about z-axis ia
G) r = f LD ebout y -aris b
The given revolving cune ia called tbe gcruratit4 curue.
tlome cta[dard sur{aces of revolution :
Let the generating curve be y = f (r) in the ry-plane and the axis of rotation be the r'axis
the! the surface generated is y2 + z2 = V lr)12.
+*
Oblr!"fhcroid
of revolution
major axis of the generating ellirr-'i:['sr1[J;f
ir,r:Fil]s
prr
a
called
ie
i"""i;t"a' in thie case'
the aris of revolution and the
r-axis
minor axis of the ellipse liea- along.th3
<
bz,the
a2
$
(Fig' 3'71)'ouUt"
surface thus generated i";t"d;
"pt'"toid
(5) HYPerboloidr of revolutlon'
=l WhiCh representS a
curve iS IF―
手
(i) Whenf (r)= a'16+Jta21, the generating
about the r― a饉 s,the point P
describes a circle unth entre M and radiusル
`P.
Let
っlane
e← ,y,z)ly anyotherpo81● On OfP.Draw oⅣ 上to“
andjoin MⅣ 80that OM=`,MⅣ =z,Ne=y
and
curve
'rh"ol6
is a circle d
.
is an ellipse
b'l\t ' * n2\' the generating curve
(4) Ellipeotd
vEr---- or
r,
- *'oriii-' ii" i ot =
thesurfacegenerated is12+zz -62 11 -t2/o2t
[4-4=t\
"'
[cz'az ]
3.2it. SURFACES OF REVOUTI|ON
I
the surface generated is
i.e. * +YP
!2 +rz =oz -i
ccntre (0' 0' 0\'
and
a
tdius
rc
of
c
spk
is
which
hvpetbolb
cylindcr.
Let P(r, y) be any poini on the curve , = f(r) in the qy'plane. Draw PM
ar.d, MP = y. lbus tbe equation of thig cunre can be written as
),
Ptoltt
- --g'ilo'
> b2, the
*tri*
"Oi驚
*f
:
∬鷲∬IW[算 ∫
T軍 llhts
゛賞:麒朧諄IF」事
滞 li
_
一
一
www.Engg-Know.com
一
︱
¨︲︲
Problems 18 2,pages 570
■ y← 』
=二
68壺
警
観
警
,Whenら
=ィ
6.●
′o・血十
と
0=ギ
OSh告 0-"Sech手
1¨ 0=“ oC"警 °
・
0=半
に
ザ
誰
iげ
4
0=響ユ
に
済げ
Problem●
L“
ey=ギ
7.y← ,の
13・
5,page 535
4塾
2● 60=ザ
由
&“
亜
号堕・
Σ西≒ ≒墜銀
Sh【
=辞
学
ギ嘉藷
・
:導
1‐
4.FCO)=等
璽
ニデ≒Sin7sin塑ギ
0+:由
[述
(22-1)0・
+:“ n50+… .)
“
4詈
キ
ぱげ
・
デ
6鵡ぉ
鮮・
∫
二僣
“
,「
.
&y=ザ
dn警
血
壺
ヽ
普
等
尋
[L発
]
10.Oy← ,の =颯 1-=2-■ 7);oy← ,ハ =号 (1-m8"C082o.
Problems 18・ 3,page 578
・lν 品SinT
←
』=ザニ昔寺鮨
「
2ι ←
n警椰品
め=L二 ¨
繹●
「・
鳥
…絆
蔽
√
“
壼軍監 ……
叩
耐
哺晨
L″
Problemo 18・ 6,page 589
#
&嘲 -0-撃 …→
価
警ャ
,:ρ
C釉パ
4“ m一 &+∞ _撃
平:轟詈
&“
「
ユ“
0-1■ ′
←
め=:― 孝平が等暉り・
√
1
′
超
知
れ‐
●“
ル
にの
=m― ザ
平高齢疇?墜ぞ
薦
(3冽 00s警
・
・8警 ―
・ 0=摯 ″
:「
“ =+:「 "摯轟警 …
.
&υ =型 7■ +等
=…
J=器 +器
1・
て
b
www.Engg-Know.com
;(iil - 8i/25
墾２
一
一
t (2 +'13)l
ｍ
，
it2-'J3 -
＋
=_:2_l=号篭織計
“
静歯・
(')
ｎ
叡
5。
1.
¨
“
nhサ /顔 nh守
←
,yJ― gn警゛
flXl
２
4“
-P"J('O}'
1, Page
２
Problem lo
Problems 18・ 4,page 582
i←
v = Vo cos lP,
,
︱︱ノ
ヽ︱
{.
.
/RCめ
α 一２
`_ヾ
o/鋳澪〕
中 ← π犠
千
にし
￨●
1ソ
"C-2猜
６
騎一
“
ν
ユ
襲L si。レ
a'=70● ・ ′
Sh守
ゴ勾
.
i■■
肝
&“ ←
,r)=3Σ e 品血2‐
=;軋
)・
論
マ
I滋夏鞣,一。
o鰤 Ⅲ
α
″
う
.
[‐
ニ
ユ総
ノ
・｀■■
￢
hHERENCINEttNCMA― B
1176
L
?.
6. :=t1'5,Y=*2.
7ik-i'
6.ェ =±
bblem 19° 9,page8 620
1,y=-4.
6.豊 CCOSh 70+7 coSh 50+21003h30+35∞
A cir,cle : centre (- 1, 1), radius {2.
psge 6(X
1 +r{3, - I -d{3,
Problcm
8. -
19.2,
Problem■ 9・ 10,Page硼
4. -2+0.r, 1-r./3.
-r.J3 ; 4.JS.
5。 ■1■ 3,也 G sin 15° 士じ
oo8 15つ ,tFo03 15・士j3in 15° ).
6.Ellipse宙 thお d atz=± l and maiOr a由 =3.
1
9.士
・拙器譜総霜酬 ∬
品繊網ぶ躙蹂乳蹴躙
reglon bounded by the n■
3and O=ν 2;Cυ いal山
)増
and region abOve it
)コ
"0=υ
between==± 2.
Problen 19・3F page 007
4(:)Right bisectOr ofthe poin鮨
(ゴ
摯‐ 耐
嘲
ザ
Problem 19・ 5,page 612
1.o(2)1/810_98土
。)(り
Jo・
Sin lC藻 :=o)+ι
10。
“[ザ
195)16;o1/2卜 O195±
1/5 cts 473+3.,where■
)‐
a140dulu3=わ dareent=:
)-1,ま 1土 jも ),士 (1+:)/也 ,士
・
5+'0■ +tan-1);(010&3+`0+1)π
F′ 8,ば
Ob&2.
8,0+>-10gヾ 2:い)●
°
C口 20'1′
Lnh".
l i tan l Ctanェ
1■ 1弾 62+COSh η
・
12.o2■ 土i10g o+ヾ 3):(ii)=tbg3+けン
2.Sin
(土
π/5),cis(± 3π /5).
Ю .=3+′
2 32 cos30_24 cos3 0+6 oo8 0
page 013
sin O-3 8in 30+c sin 50.
Problen 19・ 39 page S18
ヽ
“
ν2∞ 80/2,
7.Sin型
軍:0 1)β I SIn π
:
9・
Probleme 19'13, Page
… 629
_2_1=α
3. 397
1'59',
ザ
旦 O Sin c/2r・
ノ
寺斑麗計諾者
lR(1-LCP2)/Lpl
4。
(じ
)1/2,(j二
Problem3 20・ 1,prge 640
AnalytiC.
)Not analytic.0め
4(n and(ι 二
&脚
こ郷 r;
ア』
お:虫ダ
l∫ 1讐
=+。
(υ
)1/120.
)′ ―シ
;
C 8:
リ
―ノ十
瀾ドr"■ ザ撚ェ ′+y2);0):bg♂ +め・
(め
COS Z+C:〔め
(1+￨ソ Z+t;Cめ
〆 ;Cυ )ラし
:0)1+鯰
=3η
12 2tan 10a;2blobg→
.
A=測り /マ LP(1-凛珈ぅ +Lip句 ,0=tan-l
‐
an l子
争
晋オ
・
9。
7.― (2)‐ 1(dn 120-2 8in 100-4 gin 30+10 8in 60+5 8in 40-20 sin 20)
6。
α=0+Dπ
,exCe,when″ =l and
sln:β
2.
■ 土ム士ら± 8:土 j8h司 ,す
土J8h寺 Last four viu`ふ
レ
lmS晋
Problem 19・ 6,page 614
2(二 )が
8.
D・
81n CCOS D Sinh(血
10.
(-1+j)=ミ2,(1-じ )/ヾ 2.
&動 .50=A
α
tt tCOS D COShCmn D-00S
tan 1轟
■
20.
.
r歯
7・
α
6.O COS O「
)ds(22■ 1)″ 6,where a=0,1,2,3,4,5;2 ci3(笏 +1)π /3,where m=0,1,2.
Problem 10・
12 pag0 0B
●cos(o+6恒
30C● ●
oos(α
←1+j)/也
ωザ,ザハ円問
“
0。
ヾ
←in Obl.
4-争 an l CC° S'C0860o
,o‐ 98)】
6.(J)1,-1,o日
).
(Fυ
o―
l■
(め
1。
l.´
=0,1,2,3,4;(鵬め±212.
マ
3土め
3士二
ム訳
,■ マ
(■
bg欧 1+Sin
+"" Ve_
PrOblm 19・ 11,page 625
1(Jυ )州 2 cos rr/9,where r=1,7 or 13.
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9.
Y = a + bX
4.α =0・ 2,b=0・ 0044.
7.α =4・ 1,b=0・ 43.
1. a=223,0=6・ 1879,P=80・ 46.
3. ●=1120,b=551.
Answers to Problerns
Problemg
1-
l.l,
6.`=0・ 115,ら =118.
Proble諄 317,page 25
2.
(i) -2,1+ 3r, 1- 3i.
1a1 2 + {S, 3, -S.
(1,
(2) and
3, - 4).
6. T$o roots between
?. (i) one -ve root between - 3 and - 2 ;
(ii) one +ve root between 1 and 2, one -ve root betrreen -1 and 0.
Problems 1 2,page 5
2.α
:.
6.6,4,-1.
2,6.&:o±
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lα l,:,:・
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七 ::6士 V53
9。
1,4,7.
3. 10r4+970+8ご 2_7■ +1=0.
6rr
yr
+ 4my
-
gr3
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16. (6 - c) (c -a) (o-D) (" - 1) (b - 1)
ca)'
bc
ob
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o6327',ftil-8'
1ど IttI∬I鳥_。
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=2,y=4,z=1,ω =3.
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１４８
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=0,3.
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J)0・ 45
1. c)1.32;Cゴ )229;(ビじ
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6. 2@-bXb -c{c -oflr-Y)1s -21{z-r\'
Problem3 2 4,pages 40
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.
14.-1,-2,-6,-7.
3.4,-2,-1士 J.
18
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o・ュ a→ ″ュ4 o。
呼
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2.y=46・ 05■
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l.
+ my2 - n.
11.y3_30y2+225ター68=0
- 2mnr+
/t'
412.
=146・ 3,ル =-0・
０００
t3 *42-t2
a = 15.8,
- &' +7r - I =e.
7.y3+.3+39)y2+392y+93=。
8n = 0 ; (D) rrrs + m2x2
-
4.
n--1,,*,,
l+151+52r-36=0.
3*3-lb2+9r-2=0.
Y
4. y = 0'485 + 0'397t + 0')J'4r2'
(I)J_"2+2餞 -24=0;(`ビ )ェ `+1&3+60x2+11色 +80=0;(J"),5+7=0.
(a\
c=11.1,6=0.?1
Problen 2'1' Page
2.
12●
1.
t6.4tB.
+27=0.
=
0・
■1. ●=99‐ 86,b=1・ 2.
Problene 1'8, Pege 28
Problems 1 3,page 9
2_36ェ
-r,t
こ=0・ 5012,π
3
,-2,1,4o)1,-2,4,-8
13.c)凛 2_2.,(11)′ ″ ―″ 14_36.
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=0・
7.y=1243,0ヽ 00義 +0・
9977.
こ
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=2,b=1. 3.4,-4,3.
7.― ■
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4 222.30γ
1=監 l》。
1観講。
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r' - 6f + 3:2 + 42.x - 7O = 0.
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2.
3. Yes,9xl-12x2+5x3 5x4=0・
No
Pnoblems 2'0, Pagee 86
1。
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=1,y=31-2,2=ル
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λ
.
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４０■ ２５１１０
・・
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・・
・
Ｌ
■
１１３
．
３
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369/1?5, i2= 241'25. i"= 72/175.
Probleas 2'7, Page gl
1
i
'[; =ii
2,z=3・ 2.
_5μ /3,y=λ -4■ /3,z=ヽ ″ =μ
司ｌ
α
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=1・ 2,y=2・
=λ
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13.
ilns;“
2'1O, Pege 68
23,多 3='・
.
朝到
6. u=1, u=l/2,w=L/3.
r=',le,y-'le,z='le. ll.z=\,v=-L,z=1.
=― ち亀 =:「
=蟄
Problenr
fOr allル
=-5,y=1;ェ =:,y=1・
ｏ４・・
3.
]lO.
1;″
),y=113■ -9),2=ル
ｌ
No solution.
喘
２ｕイ
2. x=2,t=-L,z=L/2.
8. x=Z,y=L-2.
-1/11,y一
・・０５
&rcblens 26, Page 55
l. x=2,Y=l,z=0.
4. t=! =z =e2.
一
〇
3
3
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λ=-1,1,p:″
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9。
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7.
ニ
ー=14-轟
-5,==4/7,y=-9/7,z=0;λ
≠
λ
5,蒻
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2-y+&=9,
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c21+2J+η =5.5.:C21+2+D.
ル +7=0.
■ ― .+2=3. . 1 9。ェー
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18, ル=10・ 2;5″
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L G +}yz +3i-8n-
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7.71-J+2K
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& .t= I,y= -
(め
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2・
15。
l-3:,司
5。
3.325.
7. 6.
し﹁一
′
Problen3 St page 107
1. 7.
4.Not lin― ly
2-4.
dependent.5。
.
&0)Yo8.
C(`)<;(琵 )1:・
4001げ _
1.ェー￨ッー3レ
ttЮ
90° .
18.8・ 57.
11・
14.〈 0(-3,5,2);(琵 )(26/7,-15/7,17/7).
17.2-ェ =37+1=e+1)/3.
13.2繁 -2η -2″ =85。
Z+12‐ 0. ■ 2+3y+し
・
“
)2-20y+292+2=0.
ー
αセ 24=0;(`じ
=0.
4 (`)7L+50y―
;エ
=38
´ 鷹 =霧 =こ土■.● 工
=1
+y+″ =ち =+y+の ″
千
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=響 =雫・
=と絆毛・
篭鮮
・
:≠
Problem3 3・ ■ page 100
217:-241+13J+4K.
21-7J-2K:Vc5η .
&+7=0.
.(D識 ―
■―
―10Z-156=0:Ci)biSn
Proble-3・ 9,page l18
・
珠
:胤 Fl離靖北 T蠅 1.算乳;…
1.40.
7°
+14=0.
l15
=/2-z;=一 y=2/2.
12. 4・ 1.
Problems 3 4,page 97
4. 70・ 5.
‐
19o“ +3y― ル =18;2-3y― e=6.
F・
■2:1:〔 ,手 ) a2:1.
2■
…1(シ
1・
oFthe hexagon.
3.A,3,C form a△ ,rt.∠ ″ at C
32=5。
amJ海 3ngle.
ギ =≒2=篭手・ 2げ 3。
=号 =守・
cギ
+2=ザ =守・ La
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・ ¥=曙 =響 ;雫 =雫毛島
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Problems S・ 2,page 88
1. 5
10・生 +り
+夕髭 =9・
1412-1聴
+lη +0と -78=0;(ガ )=+3秒
"。
・
3_OP-3
'+y 3+z‐
。‐
i : lll=0
6.碍
0)83・
16.0鳥 ,予番,番・
17.28r-lη +19=0・
Problens 3 1,page 86
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; piltori."yrta"r.
i il,ui*."r*cyiinrlerwitbariralongz'aris
shoetr'
ュ R3_R・ cA+3)+A・ B=o.
2ノ +y2+22_2+"_と =0;(1,-1,2):お
4.0)″ 2+y8+z2_と _"_■ +3=0,(b)ヨ貯2+ノ +め _雛 +y+→ -1=0・
.
(j)=2+y2+ノ _“ _"T″ =0,(“ )ノ +y2+28-“
t#
+
Prcblenl3'lt,
Problen3 3・ 13,pq"129
9。
u2
z.
.
126
"t ttd semnd Phnes cut」
gvt o'
+B* +gy. -t'zt+&,-241 -1:&+2A=o'
- tt'.- tt?r" ) tttu - =
]
1$f +'"*
+#
.G il li*'
i *l.il-Yz-n-o=o''
ilfiy,;{#il',yi-x;,ifl,:xl,rffih:#Irrn(nv-,,'l2\+cn'=o
. '
77;静
Problen3 3・ ■2,par
3'17, PagP 136
Problor
Pmbleng 3・ 11,page」腱
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S. EYferboloiil
tso
i'
?. Rigbtcirculart*ilt'
j
―
=O and昔 +昔 +:=11
け―
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origtn'
if,Uffit't*,:;mlJ:'i;!Ti'rffJ-
-u
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鳳
13.13← 2+y2+め _3臓 -2“ +Ⅲ +176=0. 「
10。
そ
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L-8y+z+10=0. 15.ェ
ノ +y2+22_2_η _2-6=0.
14.31′ +ノ +ノ )―
16。
3・
・券∞
SeC O・ SeC4 Q
2+ノ +ノ _11=-7y+:F2-lF=0.
koblemr{3,P88e
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149 page 131
1. 3“ +4γ +5″
e.-f,ro'uz- ■
場.1● 8L″,COS4■
Problem8 4 1,page 144
0・
ro I t# (3s - 2)'
&:1008← +“ ″ 勾 ―r
±15マ 2=0.
+&-11=0・
2.2+ノ +′ _2+勾 +″ -3=0;ノ +ノ +′ +2+匈
4(麗 )●2+ノ +22_1そ +y
■ c),2+y2+22_10y_1し -31=0
(2+y2+22p+10r_均 _2=0・
工0・ 6=,2・ 4〓 .&3t,V6.n缶 +y+z+6=0.
舌
11.ノ +y2+z2+性 ,10J― L+12=0. "は (12/5,4,9/5).
2島僣
需 +訓・
静―
147
tt Z)+98=0.
ooOく臓
幼r
4.士 p
coS Cr t″
6。
L金
C"e+P3Va+α
ぱ
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COg“ +P372p● 五
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・千←り
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(2)"'r
hblem33・ 15,page 188
1.cpz_ッメ=4●Kat― アルー●
・
7.
252r+36げ
+7(メー6209-264/+353a+700+1352-436=0.
z,聴 /zD=0
■ Oy2+Z3=(レ /CDa cう ′はノ
43ピ +げ +ノー勾 ―け ―極 +“ +む +1し ― =0.
Lノ +y2_認 B_2-η +e-1,o.
“
*e
(rq""
cos
一・ｏ︼
Problem●
+ノ +′ +2← +ッ +z+1)‐
.
(3r + n tan- 3) +
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ノ
m80+″
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Problen3 3・ 10,pag0 134
1・
ノ十y2=22 tan2∝
2ノ +ザ +ノ +8り
■ ″ +40y2.1922_“ り _72レ +鮨 =Q
_レ +睦 =0■
ー
L.ウ (-1ソ・れ_1)"in
yz士露土
り =0,‐8 10レ13);,=y/1t l=2/1L l.
● ‐ 14/t41);%♂ -1げ二ピ=0。
7.2ピ +ず +37z2_480■ 1レ
“+12+130y+1睦 +203=0.
Sh「
αWhere α=COt
Problem3 43,page 150
3‐
& =/-1=y/4=2/2;=/-1=y/2=z/0.
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