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8-1-1949
Vector methods in algebra and geometry
Virginia Elizabeth Smith
Atlanta University
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Smith, Virginia Elizabeth, "Vector methods in algebra and geometry" (1949). ETD Collection for AUC Robert W. Woodruff Library.
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VECTOR LiETECOS III ALGEBRA AH) GEOI'ETHY
A THESIS
SUBMITTED TO TIE FACULTY OF ATIAHTA UNIVERSITY
IN PARTIAL FUXFJLBIENI OF TIE EEQUIREHEHTS
FOE THE DEGREE OF MISTER OF ARTS
BY
YIRGIKlft ELIZABETH SMITH
DEflfcRTiEtiT OF lATHEIS-TICS
ATLANTA, GSOiffilA
1//I
AUGUST^ 19k9
J7 W
DEDICATED
TO
URS. RDTH BARMID
■
ACiaiOWIEDGLSMT
For the successful completion of tills thesis, the ivriter is deeply
indebted to Mr. C. B. Dansby, Head of the Mathematics Department at More-
house Gollege and Acting Head of the Department of Mathematics at Atlanta
University,
He has been tireless in Ms efforts to help the nriter mate this
thesis a representative one, and to him goes the credit for tihatever
success has been achieved.
His timely suggestions, constructive criti
cisms, and endless patience have been invaluable to the Ta'iter and
furnished the inspiration necessary to the completion of the task.
The vriter readily acknowledges the extent of her obligation to
I3r» Dansby and -wishes to express her sincere appreciation to him, for
■without his help it would have been impossible for the ttriter to pre
sent this thesis as a partial fulfillment for the Masters Degree in
Mathematics.
iii
iv
TABLE OF CONTENTS
Chapter
Page
I. IMTROD UCTIOH
Vector Algebra.
Vector Geometry......
Vector Calculus...
1
•
••••.*..•••••••••••
••..••.•••••.•••••••••
••
II, FUNDAMENTAL DEFINITIONS
1
1
1
2
A VSGTJiQY...a*.■,••...,«««,,«.«*..««....««•«•««««.•,.«.««
2
DCSUldX..a.....,,,,,,,...•««..*.«..•.««.«•«.....••«.•.,«,
2
Representation of a Vector.....•••••...•••••••••••••••••
Equality of Vectors*
.••*••«,•»•,,•••.«•••««••••«•••
NegatiTe Vector.........
The Unit Vector....
,
Kagnitude of a Vector
.
2
3
3
k
k
A Hull Vector.
Collinear Vectors,,.,..........*.,.»»..«....,...,....„..
h
Ij.
Reciprocal Vectors.,...
111, COMPOSITION OF VECTORS
6
Addition of Vectors,
6
Subtraction of Vectors
Components of a Vector .^
,,
The Unit Vectors T, J, k.......
IV. SCAIAR AMD VECTOR PRODUCTS OF TWO VECTORS
The Scalar or Dot Product
Laws of the Scalar Product...
8
11
,
11
Application of the Scalar Product to Unit Vectors i, 3, k
The Vector or Cross ft-oduct
Application of Gross ft«oduct to Unit Vectors.......
Cartesian Expansion of the Vector ft"oduct
••«••••
V, VECTOR AND SGAlAH HK3DUCTS INVOLVING THiiEE VECTORS
12
13
Hi
1$
16
17
Possible Combinations of Three Vectors
17
The Triple Scalar Product V = a. (F x c)..
Cartesian JSxpansion of Triple Scalar Product V = a.
17
The Triple Vector Iroduct q = a x (b x "e}...,,..........
Expansion axA Eroof of Triplo Vector Product
VI. APPLICATIONS OF THE_VECTOR IffiTIICDS IK ALGEBRA AMD GSOiiETRT...
The Equation (r-.a)oa(r-b)o Represents the Plane
Bisecting at Right Angles the Line A B
The Lines "Which Join one Vertex of a Parallelogram to the
Middle Points of the Opposite Sides Locate the Trisection Points on the Diagonal vMch Does not ^o
Through the Vertex
7
8
"....
The Bisectors of the Angles of any Triangle lieet in a
Poxnt•»•••«».,..•,,.•••••«,,«.,,,....,,..,........,
19
20
22
22
23
26
TABLE 07 CONTENTS (Cont'd)
Chapter
PaSe
The Sum of the Squares of the Diagonals of a Parallelo
gram is Equal to the Sum of the Squares of the
oiu.6S »» ••••••••••• •_•_••• • ••_•_•_•»•• ••»••«••#«•»•• •£•••
jc
Talce Equation of Circle r2 - 2 a.r = 0, Factor With r
siici XnterprGt*•••••••••••••••••••••■•••••••••••••••
Derive an Expression for the Area of a Square of TJhich
33
? = aj i / a2 I is the Semi-Diagonal*••••••••••••••
35
HaJj.
.2U
The Sum of the Squares of the Distances of as^r Point 0
From the Angular Points of the Triangle Exceeds the
Stan of the Squares of its Distances from the Middle
Points of the Sides by the Sum of the Squares of
Ul6
uILUOSa* «•••■••*••••••• a •••••••••••••••••••
Show That ("a - b)» (a / b) = a2 - T>2, and Interpret,••••
BIBLIOGPJ.HIY.•••••••••••••••••••••••••••••••••••••••••••••••••••••••
39
Ul
LIST OF FIGURES
Figures
1* Representation of a Vector
Page
.....•.*»».••••••»•».••.••••
3
2* Equality of Vectors*...§«3a(>9,«*.»....«.«...•*•«.«.•••..**...*
3
3. Reciprocal Vectors,«...•»••
•.*.•»••.•,»•••.*•«••...»
i|
Jl. Addition of Two Vectors.,.....................,.9,..,..,».,,..
6
5>« Addition of Tliree or I lore Vectors..*..
..,•••••••••*•»•
7
......«»...,,»..»».....,
7
6. Subtraction of Vectors..
7• Decomposition of Vectors Into Components•••••*•.•••*•
•■••
8
8. T'i/o Unit Vectors i and j,«••••••••••.*»•••.•••••••••••••,«••«•
8
9. A Variable Vector.•••»,••.••*••••«,,,•»»,•••«,.,,,,,,»•••••••«
9
10. The Projection of One Vector Upon Another.......••••.,,,,•••••
11
11. The Projection of Vectors "c and 3 Upon Any Other Vector ?•••••
12
12. The Resultant of Any Two Vectors.........,....,,,,...,,.,.,.,.
13
13• The Vector E as the Horaal to the Plane of Vectors a and b,,«,
Hi
Ik* Three Unit Vectors I, J, and k*..*.*.............,.
1$
15. A Parallelopiped of Tfliich the Three Conterminous Edges are
Vectors a, S, and c................^..••.....•««.••.,.,,^,*.«
13
16. A Plane Bisecting at Right Angles the Line AB.,,,,,,..,,,,,,,,
22
17. A Parallelogram With lanes Joining One of its Vertices to the
Kiddle Points of the Opposite Side Locating Trisoction Points
on. the Diagonal.*•*•«•*•«•.*.«.«....«...•*...*«.««,»,.....,...
10. The Bisector of an Angle.,,
,
23
.....»..,.
26
19. A Triangle With the Bisectors of its Sides Meeting in a Point,
28
20. A Parallelogram....
32
••••••,»••.
2-L, A Circle.........«•«•■•.•.».*»..»•...,,..,,,.., ...,.... , ......
34
dt-. a bquare..>*.....*............,••.•.,..»......................
35
vii
LIST OF FIGURES (Gont'd)
Figures
.
page
23 • A Triangle and a Point Outside of Triangle From YJhich Vectors
are Dravai to Each Vertex of the Triangle and to the Middle
Point of Each Side of the Triangle •..••••.••.•••.••*•••»••••••
36
£/{.* A farallslograjii*••••••■••••••••••• • • ••• •• •• •«••• »*•««•••• •• •••
38
25. A
CHAPTER I
INTRODUCTION
The word vector is the Latin word for carrier|
and vas defined in this
■eay in 18I|6 by the eminent Scottish Mathematician, Sir William Ro-wan Hamil
ton, "who desired to convey the idea that he was looking upon a -vector as an
operator which carries a particle from the initial point to the terminal
point*
There had been a growing feeling that the processes of analysis irere
in some way artificial and complex.
It is chiefly through the labors of
Gibbs and Heaviside that analysis has been perfected viihich not only does
a-roay with the complexity and artificiality of other analysis but offers a
strictly natural and therefore as direct and simple a substitute as pos
sible, and, at the same time in no wise is at variance, but runs parallel,
to then.
This new, yet old method is Vector Analysis.
Vector Analysis may be considered under three divisions?
(1) Vector Algebra
(2) Vector Geometry
(3) Vector Calculus
The Tffiter makes use of the elementary operations and attempts to re
late and illustrate its power and usage pertaining to some theorems and
formulas -which may be proved or verified by means of Yector Algebra and
Vector Geometry.
Stress is placed on logical aspects more than on tech
nique and details of calculation.
The laws governing operations -with
vectors are then postulated and discussed.
The notations adopted are that of I¥ofessor Willard Gibbs, one of the
great American physicists and mathematicians.
CHAPTER II
FUNDAMENTAL DEFINITIONS
In the study of Mathematical courses that preceded the study in Vector
Analysis, uhen nunfeers, points, and lengths on a line are used to represent
physical and geometrical concepts, it is unimportant -what line is usedj that
is, the line may be drawn in any convenient direction,
of numbers only or numbers and an algebraic sign,
Tfe spoke in terras
We now speak of the
representation of certain physical concepts by line segment where the di
rection of the line segment is of much importance*
1- i Xg£i°£ is a directed segment of a straight line on xAich are
distinguished an initial and a terminal point.
tude and a direction.
A Vector thus has a magni
Any quantity which can be represented by such a
segment nay be called a Vector quantity.
For this reason, velocity,
acceleration, force, motion or displacement are called Vector quantities.
A simple illustration - $0 feet due north.
Fifty feet is the length or
magnitude and due north is the direction.
2« Scalar.
Mien units of measurements have been qhosen, a scalar is
represented by a real number and so is subject to all the laws of ordinary
Algebra.
Thus mass, time, density, coordinates, are scalars.
The number
2 is a scalar.
A Vector, however, involving direction in addition to its numerical
magnitude has an analysis peculiar to itself, the laws of vMch are to be
derived.
3* Representation of a Vector.
A Vector is represented graphically
by an arrow of length equal to its magnitude, pointing in the assigned
-2-
-3-
direction from any point in space.
An example: 3 miles due east.
Figure 1.
The tail of the arrow, 0, is called the origin or the initial pointj the
head, A is called the end or the terminal point.
In figures, ne indicate
the direction in which the Vector is draxm by an arrow.
We shall distinguish a Vector from a Scalar by placing a line or
dash over the Vector symbol.
So that if a denotes a Vector, then a is
its i-component.
Vectors 3re sometimes read with two letters.
In Figure 1, the Vector
Hi denotes the vector beginning at 0, ending at A, and pointing in the
direction from 0 to A,
fr*
/L
Jr
*
O
Figure 2.
h« Equality of Vectors,
Two vectors axe equal if the line segments
defining them are parallel or coincident, and their lengths and directions
are the same.
Thus, in Figure .2, Gil and oTP" are equal.
£• ^fegative Vector,
The symbol - a, in Figure 2, is used to represent
a vector having the length of 0»A« but the opposite direction.
defined as the negative vector of si and is written - a!.
It is thus
-14-
6. The Unit Vector.
A unit vector mil be denoted by adding the
suffix 1 to the symbol representing the vector.
This suffix tells us that
its magnitude is unity.
The vector a then may be considered as one a times as long as its unit
vector "a,} and hence, vje may -write
or
"a - a a *
)
I )
■
a » a0 a^ )
7« Hagnitiide _of a Vector*
(1)
Associated mth a vector 1 is a positive
scalar equal to its length, magnitude, or modulus.
by the notation /SJ,
He shall represent this
If then a is the length of the vector "a,
a ■ y«7
(2)
The magnitude of a vector a vri.ll be sometimes denoted by adding the sub
script 0 to a thus
a0 S a
8» A Null Vector,
(3)
A Hull Vector or a Zero Vector is one yihose modulus
is zero,
if >S7-o,
the vector is a null vector and its direction is undefined*
9» .Cgllinear Vectors,
Parallel vectors regardless of their magnitude,
size or tension are said to be collinear*
10* ^ggigocal Vectors,
The vector parallel to "a, but tdiose length is
the reciprocal of the length of a, is said to be the reciprocal of a.
is illustrated in Figure 3.
Figure 3«
This
-5-
So that if
a
« aa,
then
JLS I
-1
(I*)
CHAPTER III
COMPOSITION OF VECTORS
■'•• Addition of Vectors,
Given any tvro vectors "a and b, rse can con-
struct a vector equal to the sum of vectors a and b Given vectors
To construct a vector equal to their sum*
Figure if..
Since vectors are free and are looked upon as operators
■which move from one place to another, tre can slide them
if they do not have the same origin^ providing Tie do not
alter their direction.
To obtain graphically the sum of the tvro vectors a aid b", draw a
paraJLLel to the given vector aj draw b starting from the end of a
parallel to the given vector bj the vector joining the origin of a with the
end of b is the sum indicated in Figure \\3 by the notation a / b -.
Now
draw "a starting from the end of b~j the vector joining the origin of b" to
the end of a is the sura in question (F / a) •
In other Tiords, it is the
diagonal of the parallelogram of tiiich the two vectors "a and b" are the
adjacent sides.
-7This is known as the parallelogram law of adding vectors.
This is called the commutative law of addition.
Similarly the sum of three or more vectors is obtained by constructing
a polygon having those vectors as consecutive sides*
and drawing a vector
from the initial point of the first vector to the terminal point of the
last.
£
Figure $,
(6)
The sxm. is independent of the way its addends are associated into
groupc.
This is called the associative law of additions
2. Subtraction of Vectors.
To sub-tract one vector from another, add to
the first (a) the second vector reversed (-b).
Figure 6.
If two fectors a and F are drawn from the same origin (Figure 6) the dif
ference a - T> is defined as the vector extending from the end of F to the
end of a.
Figure 6 also shows that
b / (a - b) = a
(7}
3» Components of a Vector»
From addition and subtraction of vectors,
it is clear that any vector q may be considered as the sum of any number of
component vectors, ?Mch rfien joined, end to end, the first one begins at
the origin of q and the last one ends at the terminus of q.
Thus
where a, b, c, d, e, f, and g (Figure 7} arc called component vectors of the
q.
These vectors need not lie in one plane.
Vectors, all of -which lie in
or parallel to the same piano, are said to be coplanar»
Figure 7.
It is often convenient to decompose a vector into tvro or three compo
nents at right angles to each otherj two in case all the vectors under
consideration, are coplanar; three when they are not coplanar.
k» ^£ jfait Vectors i, 3* k.
If vectors are coplanar, the unit vectors
along the x-, y- axes vjith origin at 0, are i and 3 respectively.
n
i
u
Figure o,
In X i and yj, x and y are the i and 3 components
If vectors are non-coplanar, they can be referred to the Cartesian
system of axes.
The three unit vectors along the x-$ j-s S - axes are
called 1, X ¥ respectively.
Let r be any vector equivalent to a vector QA along CK, plus a vector
HE along 0&s plus a veetor
Figure 9<
, S is the magnitude and k the direction! in j3s 7 is the magnitude
and j" the directionj in ;•&, x is the magnitude and i the direction.
If x, j3 5 denote the r-iagnitude of the vectors respectively, vm nay
nvrite for any vector r the magnitude of v/hose components are x, y, e.
r = xi/yj/»k.
.(8)
This vector r is often used to represent the three projections of r along
the three axes respectively.
Let
Then
be the direction angles of any vector parallel to 00.
x = r cos sC )
)
y = r cos /3 ),
)
g = r oos r )
.(9)
-10-
Tliis decomposition of a vector into three rectangular components is the
connecting li P A
bet-ween two or three dimensional Cartesian and Vector
Analysis respectively.
If teo vectors are given
a b a* I / ag ]" / a^ k
and
b = bj I / b2 1 / b3 k
their sum is
.
■
•
(I / F) = (a, / b,,j I / (a2 / b2) I / (a3 / b3) I. • . . . .(10)
This process of finding the
T
sun may be extended to any number of vectors
and shows that the-components of the sum of n vectors are equal to the
stuns of the components of n vectors.
Syrabolically, ne may trrite
To obtain the rectangular components of a unit vector, equations (9) are
used,
so that
r a r (i cosK / 1 eos B / k cos y")
. . . (12)
If Tie divide byr, w get
£ s r» ■ T cos <?C / 3 cos B / k cos K
r .
*
•••••••••(13)
Therefore, the magnitude of the rectangular components of a unit vector
are always its direction cosines.
VI
SCALAR AID VECTOR PRODUCTS OF TY/C VECTORS
•*•• T'ne Scalar or Dot Product.
The scalar or dot product of two
vectors a and b, denoted by a«b, Sab, a b or ("a b) by various witers,
is a scalar defined by the equation
a* I) = ab cos (ab~) » b"»a. ..*•
•where (a b) is the angle between the two vectors.
The scalar or (tot product is coramutative.
0-
cT~
^~
Figure 10•
a«^ = ab cos (a b") = b»a sho-ws that the scalar product may be looked upon
as the product of the length of one of the tvro vectors multiplied by the
projection of the other upon it,
or
__
0A z 0D - OB x 00.
Evidently, if the two vectors a and b are perpendicular to each other
cos ("a b) « 0 aid their scalar product is zero.
Or, if
a»b ~ o, then aJLb.
If a and b are parallel vectors, cos(a b)= 1 and
a»b~ = ab,
-11-
•••••.. (1$)
-12-
and in particular if "B ■ a,
a-a a a^
The scalar product of a vector into itself is often -written as the square
of the vector, thus,
a. a
In general, to obtain the magnitude of a vectorial expression, it is only
necessary to square it, and the result is the square of its absolute
value, magnitude or size.
■
2# j£ES 2£ £**£ .Scalar Product.
laws of multiplication.
sum (c / <T).
The scalar product obeys the ordinary
Consider the two vectors "c and d as well as their
Consider also their projections upon any other vector B*.
z
Figure 11.
The projection of c on b is CE; the projection of d" on b" is EFj the pro
jection of (c / d") on b" is OF; hence
"cVb / cUb = (c / d)rb =E-(c/d)
(16)
This result is easily extended to the scalar product of the sum of any
number of vectors.
Observe from the above equation, that scalar multiplication or the
dot product is also distributive; that is, it obeys the distributive la%¥
of multiplication with respect to addition.
-233 • Application of the Scalar Product to Unit Vectors I, J, k.
The
application of the scalar product of two vectors to the unit vectors T, 3*
and k is of great importance^ and gives
x.T - 3-7 = k.k a 12 b j2 - k2 = i
j
i.j a j.± a j.k s k«j s bi . i.k a 0 )
Let us consider two vectors a and b given in terras of their components
along the X-, I-, S - a^ces:
a = af I / ag "J / a3 k,
and
F = b^ I / b2 I / b3 lc,
then by (16) and (17)
a.E = (a, T / a2 1 /
k) (b, I / b2 3 / b3 lc)
(18)
Consider any two vectors a and b*.
We vdsh to find the magnitude of
the sum of the two vectors and the angle bet-ween them.
The square of the
magnitude of the sum is found by dotting the sum of the two vectors into
itself.
Figure 12.
In triangle ABC
"c « a / F
Squaring to find its magnitude, we obtain
c«c d c2 s C2 ■ (I / F)» (a / F) = a*a./ 2a»F / T>*h
Or
e2 ss 3?. / 2 lab cos (ab or e) jl b^
.', c2 K the square of the magnitude of the sum, it </• b»
Or
C2 ■ a2 - 2 ab cos (©) / b2
THiere ^ is the' supplement to the angle between a and, F«
This is the Lat?
of Cosines•
^•* ^le Vector o£_ Cross Brxdugt,
The vector product of two vectors
a ard F is a vector^ ?jritten a x b" (in disti.nction from a»b, the dot
product), also V a F or PC Fj bj different authors, and is defined by the
equation.
a x E « 1 ab sin (a F) s - F x a, • • ••••••
(19)
idiere E is a unit vector^ normal to the plane of a end F and so directed
that as you turn the first named vector a into the second one b"5 E points
in the direction that a right-handed screw (cork-sere?/) •would progress if
turned in this sane, manner.
U ab sin (IF) = a x b
»^oi
Figure 13,
-35-
In other words, a x F Is a vector perpendicular to both a and F and whose
magnitude may be represented by the area of a parallelogram of t/nieh a
and F are the adjacent sides*
If the vector F came first instead of the vector a, in the product,
the only difference would be in the reversal of the direction of E, so
that
axFs-bxa
If "a and F be finite vectors and
a x F a 0, then a 11 F
The sine of their included angle must be zero.
This then Is the condition
for parallelism of the two vectors "E and F.
Since any vector is parallel to Itself,
axasQ. .,..
•, • •
£• Application of Cross Broduct to Unit Vectors.
• •(20)
Remeiifoering that the
unit vectors T, 7, and k are mutually perpendicular, It follows from the
definition a x F that (Figure
Figure
"J x k «s I
kxj
k x i s j
i x k
.(21)
.16-
¥e also have* by (20)
TxT=*Jx J«¥x~k = 0.
The cross product of two vectors can be expressed in -terms of their com
ponents.
Thus if
a s a4 I / a2 I / a3 k,
and
b ■ b, i / b2 I / b3 k,
by expanding and using (21), Tie get
a x b = (a, I / aa j / a3 k) x (b, I / \>2 1 / b% k).
a x F = (a, br) (0) / (a, b3) k- (a, b3) I- (a2 b,".- 1 / (a2 b2) (0)
/ (a2 b3) 1/ (a3 b,_) 3- (a3 b2) 1/ (a3 b3) (0}
Grouping terms mth like, unit vectors, rm obtain
(a x b") = (a2 b3 - a3 b2) I / (a3 bf, - a# b3) I / (a, ba- a2 b,)k .(22)
6» C_artg^sian ^pansion of the Ygetcg ^oduct.
can be written
i
1
As a determinant* this
¥
a x b-
Tlie cross product is not commutative.
(23)
CHAPTER V
VECTOR AND SCAIAR ffiODTJCTS INVOLVING THREE VECTORS
■'■• Possible Combinations of Three Vectors,
Fron the three vectors a,
b" and c, the follomng combinations may be derived:
1. "a, (5*g)
(a vector)
2. ii»(Fxc:2
(& scalar)
k» I C£x£)
5» a»tb*^)
(not defined).
(absurd)
3« a x(b"xe) (a vector)
.(2^)
6. a x(E*"c) (absurd)
Of these six expressions, $ and 6 are rgeaningless and absurd, because they
are the scalar product aid vector product, respectively, of a vector 1 and
a scalar (b"»"e) and such products require a vector on each side of the dot
or cross.
As to kj since no definition of the product of two vectors with«
out a dot or a cross has been made, it is yet meaningless.
sider in detail the three remaining triple products.
»
We shall con
The first one of
these, a (b«c) is simply the vector a multiplied by the scalar quantity
(b»c) and is a vector in the same direction as a, but be cos (E ~c) times
longer.
This triple product, then, offers no neT,r difficulties, and means
"a. (E»"c) = a x b c cos (b "c)•
2» ?ne Triple Scalar .Broduct V a it. (b x c).
This product is a scalar
and represents the volume of a parallelepiped of infaich the three contermi
nous edges are a, F, and "c,
This is seen to be the case as F s 0 is the
area of fee base represented by a vector OS 1 to this base; the scalar
product of a and the vector 05 will be this area multiplied hy the projec
tion of the slant height at along it, or in other viordB.$ the volume.
This
volurae, V, nay be obtained by forming the vector products of any two or
three vectors a, b and "o (thus giving the area of one of the faces) and
forming the scalar product of the vector-area with the remaining third
-
-17-
-18vector, it follows that
V =5i(cxa) =ci(axE) = a*(E x c").
If the vectors (c x a), (a x E), and (E x "c) are taken so that they form
an acute angle vdth E, "c, and a, respectively^ then the volume Is to be
considered positive, the cosine term in the scalar product being positive,
(Figure 15). Otherwise, the volume is to be considered negative.
The in
version of the factors in the vector products should change the sign by
0-9^ so that we have
Figure 15.
_ b. (c x a) * (c x a).b = - E«(a x "c) = - (ax c?).E I
~ ~o. (a x E) = (ax b")."c = - "c-(b x a) = - (b x a)»"c )• • • .
= a.(F x "c) s (E x "c)»a = - a- ("c x E) a - ("c x E)*a )
(2li)
By a consideration of these equalities the fQllotdng lavra may be seen to
hold:
1, The sign of the scalar triple product Is unchanged as long as the cycli
cal order of the factors is unchanged.
2* For every change of cyclical order a minus sign is introduced,
3. The dot and cross may be interchanged adlibitum.
The equalities (2it) are called by Heaviside the Parallelepiped Law,.
-19-
3» Cartesian Exgans_ion of Triple Sgalar jfroduct V aa*. (bxc),
The product may be vjritten in terms of the components of its vectors along
the X-, T-, g - axes in the form of a determinant.
If
a = a| i / a2 1 / a3 k,
and
b - b, I / b2 1 / b3 k,
and
<s ■ c, I / c2 1 / 03 Tc,
the triple scalar product is equal to the determinant,
a,
a*» (b x c;) = b~» fc xa) a cs» (a x b)
a3
b,
The triple scalar product can be Tjritten as the minors of a]_,
)
a3, re- )
spectively.
(25)
i a2 [ b3 bi
C3
c3
)
C2
• k» The Triple Vector I¥odimt q = a x (b x "c).
q = I x (F x c) is a vector.
The triple product
In this expression, the parenthesis, or some
separating synibol, is necessary, as ax (Fxc)/"(axE) x "c.
The sign
of this product changes every tine the order of the factors "a and (b x "o)
is chanced in "a x (b" x c), or lAenever the order of the factors b" and "c
is changed in (Fxc),
The vector product being altjays 1 to both of its
components, q is 1 to^'as well as to b x c", hence
q-a a 0 and q*(b xc'JsO
(26)
Equation (26) shoiTS that q lies in the same plane as F and c", either by the
condition that the three non-parallel vectors should like in a plane is
that |[a F qj = 0 or by seeing that it is 1 to a line which is itself 1 to
-20-
b and c.
£• Expansion and jroof of Triple Vector .^gduct.
The -triple vector
product "q a "a x (b x ~c) can be reduced to simpler forms by means of the equa
tion,
a x (b x "c) * b (a. c) - ~c (a. b)
(27)
Proof:
Express vectors "a, b and Is in terms of the unit vectors T? "J, k*j ex
pand both sides and shorr that the two sides are equal.
Given
a ■ a, I /• a2 7 / a3 E
b = b| 1 / b2 j / b3 k
c* = e, I / C2 j / C3 k
F(a.c) = T
| c# / a2 b, C£ / ^3 bj 03)
|
cj / a} bg 02 / a3 b2 03)
/ ¥ (aj b3 c^ / a2 b3 e2 / a3 '03 C3)
H (a H) = ~T (a^ bx ex / &2 ^2 cl / a3 ^3 c
-1 (ax bx c2 / a2 b2 c2 / 33 b3 02)
-k (ax bx c3 / a2 b2 03 / 33 b3 03)
Tlierefore
03) T / (ax b2 cl / 33 b2 03) ~ / (a,
(a."c) a (a2
ci / a2 b3 02) k
and
-c (I.b) = - (a2b2 ci / a3 b3 ex) I / (aX bX 02 / 0.3 b3 c2) I / (ax
C2 ^ a2 b2
Grouping coefficients of like unit «"e"Btors, we obtain
a x (F x c) = E (a.c) -c (a«E) ~!
(b2 c3-b2 c
~ blc3)
As a deterninant^ vre may m*ite
i
■a x ("B" x c) =
al
C2
b3
I
k
a
a3
b2
-21-
Hence,
a x (b x "c) = b (a."c) - "e (a.b)
Q.E.D.
CHAPTER VI
APPLICATIONS OF TIL: VECTOR I.3THGDS III ALGEBRA. AID GECX/ETHI
1, The equation
(r -. a)0 s (r - F)o
represents the plane bisecting at right angles the line A B«
igure 16.
Given the plane M bisecting the line AB at point R making right angles.
To prove: (r - a)o » (r - H)o«
Eroof:
Let 0 be any point in space and P any point on plane I.EI»
Draw QA, OP, and W,
Let
"M. as a,
CB e b.
OP s r, the variable vector.
-22-
-23-
Then,
r - a is the vector constructed from the end of si to the end of
r.
The origin of "a. and r is at point 0,
By similar construction on vectors b and r, tve obtain r — H,
Plane MN is the locus of all points equidistant from A and B.
P is
a point on plane M.
Hence,
<p-a)
The magnitude or length
(F-b)
of these teo vectors (r* - a) and (r - b") may be
represented bys
(r - a)0 ■
(F-i)
and
(r - F)o «
Tie refer to Chapter II, Equa1 don (!)•
Having proved the vectors
equal in magnitude, I conclude that the length, magnitude or size of the
vectors are equal*
That is
(r - a)0 s{p- F}0.
Q.E.B.
2» The lines rjhich join one vertex of a parallelogram to the middle
points of the opposite sides locate the trisection points on the diagonal
•which does not go through the vertex.
Given the parallelogram OBCfl, E the midpoint of CD, F the nidpoint of DC,
OC the diagonals BE meeting 00 at E, and BF meeting OC at S.
To prove BE trisects OC, and BF trisects OC, i.e., R is a trisection
point of OC and S is a trisection point of OC.
This amounts to proving
OH = 1/3 OC, and GS = 1/3 OC.
Proof:
Let CD_= a,
and, OB =f,
Then
Also
and
OE ■ 1/2 a.
..
EB = F - 3/2 a.
__
ER ■ S (b - 1/2 a).
00=
(I^F).
See composition of vectors - Chapter III.
Now
m a m / m = 1/21 / s (F - 1/21),
inhere S is an unknown scalar representing some part or portion of HT.
Also
"0! a t (a /F),
Tdisre t is an unknown scalar representing some part or portion of "OC.
Quantities rMch are equal to the sane quantity or to equal quanti
ties are equal to one another.
Hence,
1/2 I / S (b - 1/2 I) = t (a / F).
Multiplying, me obtain
l/2a/sE-]/2Sasta/tF,
By collecting terms, it follows that
S F / 3/2 (1 - S) a a tT/ t F.
-25The scalar coefficients of a on both sides of the equation are equal.
The scalar coefficient of a on the left-hand side of the equation
is 1/2 (1-S).
The scalar coefficient of a on the right-hand side of the
equation is t«
'•
.
1/2 (1-S) a t.
The scalar coefficients of b" on both sides of the equation are equal.
The scalar coefficient of E on the left-hand side of the equation
is 5.
The scalar coefficient of F on the right-hand side of the equation
is t.
•
•
S "• t«
We now have two linear non-homogeneous equations in two unknowns (S and t).
Solving the equations simultaneously, ire get
S ■ t
1/2 (1 - S>= t
•
Substituting the value of S from equation (1) in equation (2), tie get
1/2 (1 - t) - t.
Multiplying by 2, we obtain
1 - t = 2 t.
By adding t to both sides of the equation3 ne get
1 - 3 t.
Dividing both sides of the equation by 3, we obtain
1/3 - t.
Since
S ■ t,
and
t-3/3
flien
(1)
(2)
-26-
S = 1/3.
Substituting-the value of t in the equation
f«** 3B = 1/3 (a / E).
But
(a / B") is the diagonal OG,
.
,1 conclude that
OR = 1/3 OC.
Similarly, it can be showi that
CS ■ 3/3 OC
3. The bisectors of the angles of any triangle meet in a point.
Be
fore I can prove this I must first prove the Lemma,
The equation of the bisector of the angle formed by tvro vectors a
and B" from the same origin o is r = y (ai / B"i).
Figure 18
Proof of Lemma:
Employ unit vectors along two of the sides as independent vectors.
The bisector is then ai / b"i»
Let OP be the bisector of the angle formed by "a and b v&th P any
point on it»
Through P draw FQ 11 OB meeting OA or OA produced at Q.
-27-
Through A draw a line 11 OB meeting OP at G.
I2U k
mm—m
£\.
cB
Multiplying by | OB], y& get
C[
=
k
=
k
|OR1
CB a k B".
ZtB 0 C =£,C 0 A.
for i,B OA -wss bisected
ABOC =2*0 C A.
If frno parallels are cut by a transversal, the alt, int. IP are =.
t. C 0 A = &A G 0.
Qjiantities inihich are equal to the same quantity or to equal quanti
ties are equal
llGI =
I OS .
If the base l3 of a traingle are = the sides opposite them are equal.
A 0 A C is isosceles.
An isosceles triangle has trro equal sides.
Since AC* = k b,
and
AC = a.
Then
k H :" a.
ies -which are equal to the same quantity or to equal quantities
are equal.
The equation of the bisector OP is
r * x (OG),
r&ere x represents a scalar* portion or part of OP.
P = x (QA / 3[C) .
See composition of vectors - Chapter III.
-28-
Qp,
r == x (a / k Id) ax (aai/k
Refer to Chapter II - Topic 6, Page $,
v «■ x (a ai / a Ei)«
Factoring, -we obtain
-r » x u ^al / Hi),
Hence
•where y represents the product of x a,
Hence the Lemma is proved.
a
Figure 19.
Now let us prove the theorem.
Given BQ, APS
spectively.
and CR, bisectors of the ]^
B, k,
and C of triangle BAG re
To prove BQ> AP and CR meet at point I
By the Lemma:
With reference to vector origin A, equation of AP is r" « x
- b*i)j with reference to vector origin at 0, equation of AP is
r=i/x(ci-Fi)
(1)
-29-
lith reference to vector origin Bs equation of BQ is r = 7 (ai -"ci)j
rath reference to vector origin at 0, equation of BQ is
r - I / y (ai - ci)
(2)
Y/ith reference to vector origin C, equation of OR is r = S (b"i - ai)j
mth reference to vector origin at 0, equation of GR is
r ■ u / 8 (Hi - ai)
(3)
Let I be the intersection of AP and BQ»
Equating the right members of
equations (l) and (2) vie get
S / x (£1 - Fi) m t -/- j (ai - "el).
Or,
"5 - Ti / x Cci - Ei) - y (ai - ci) = 0
(h)
But
3 / c ■ t.
Transposing, we obtain
■g -^a- c ■ c "ci, .................... (5)
Substituting (5) in equation (U):
- c "ci / x ("ci - bi) - y (ai - ci) a 0
(6)
But
a/E/csO,
or
a "ai / b Ei / c *c"i s 0.
Transposing and dividing by b, ive obtain
£, s
1
-a ai - c ci
^^,^
.(7)
By (6) and (7), it follows that
- c ci / x (ci ^ SLJXJL0, .SI - y (ai - ci) = 0.
b
Multiplying by b, -we get
-30-
-bc'ci/xb'ci/xaai/xe'ci-ybai/yb'ci =0
By grouping, we obtain
(- b c/bs/cx/yb) ci/(ax-jrb)Ii = O
By setting coefficients of "ci and a^ equal to zero, we obtain two linear
equations
(
a x -b / = o
}
(8)
(-b.c/bx/cx/ybso)
We shall take the first equation of (8) and solve for ax,
a x - b y = 0.
Transposing, -we get
a x s b j
(9)
By substituting (9) in the second equation of (8) vie obtain
- b e/bx/ex/ax = Q,
Simplifying,
it follows that
- b c / x (b / c / a) = 0.
Or , by transposing,
vre obtain
x (a / b / c) = b c.
OB?,
x - II® _^—__.
(ID)
a / b ^"£-—
In equation (9),
a x = b j
Substituting the value of x from equation (10), we obtain
v =
? k, ,.c , ■ ■ - -
0rs
y s
a
c
»
-31•
be
S
v=ac
.
At the point of intersection of Sq and OR, r from (2) = r from (3),
±» e.
t / 7 («L - "ci) ■ u / S (bi - ai),
Or
T -"u/ 7 (ai --ci) - S (Fi -"ai) = 0
(11)
But
t / a = u".
to
t-ua-as-a ai» ••«••••••••••••«•., (12)
Substituting (12) in equation (11):
- a ai / y (ai - "ci) - S (Fi - ai) a 0,
. (13)
Substituting the value of equation (7) in equation (23), we get
- a ai / 7 (Si -"ci) - sf - a ai - e -01 _ m) m Q
MaltipOying by b3 -we obtain
-ab%/byai-by'ci/a»ai/e«~i/b»i;iii»Q
Grouping like terraSj i« get
(- ab / b y / a * / b s) q / (- b y / e §) 01 s 0
Ey setting coefficients of ai and ^1 equal to zero, rse obtain two linear
equations,
(
-b y/c » = 0
)
(-ab/by/ai/bgaO)
' * * {IU)
Erom the first equd;ion in (Hj.)
b y ■c *• • • •
Substituting (15) in the second equation in (l!+), vm get
-ab/ce/as/bgso
Factoring the last three terras, we obtain.
(1$)
-32-
- ab / a (a / b / g) = 0.
Solving for 8,
we get
8 (a/b/c) «=ab
a *
ab
. * . .
(16)
Erom
b y » c S
Substituting the value of (16) in the above equation, we get
y s_
Or
Hence since the value of y for the point of intersection of BQ and OR is
the same as the value of y for the point of intersection of AP and '*® T
is the intersection of IP, BQ,
AI =
BT
and CR«
b c
a /b |i c
sac
CI 3
a/b;
ab
Q.EJ3,
it. The sum of the squares of the diagonals of a parallelogram is
equal to the sp of the squares of 1jlae sides.
-33Given parallelogram A B G D rath diagonals AC and BD.
To Proves (AC)2
Proofs
/ (BD)2 « (AB)2 / (BC)2 / (CD)2 / (DA)2,
,
Let ffla|,
and
H = F.
But __
BC = a.
Also
DG s F.
Then
Ic ■ a / F,
And
ED ~ a - F.
Hence, to have
IB2 / HJ2 n (a / F)2 / (a - F)2.
Dotting each vector into itself, we obtain
(I/F). (a/F) / (I-F). (I-F)
= (a.a / a,F / F.a / F.F) / (a/£ - a.b » F.a - F.- F)
2 1.F / F2) / (12 - 2 i.F / B2) =
>include that
'Vc)2 / (BD)2 - (AB)2 /(BC)2 / (CD)2
5« Take equation of circle
r2 - 2 a,F a 0,
factor Tidth r ani interpret*
-3U-
Flgare 21»
r2 - 2 aur «= 0.
r. (r - 2 a) = G,
represents the dot product of two vectors, 7 and r - 2 a«
The equation, r » (F - 2 a) » 0, tells us that r 1 (F - 2 a) •
A necessary and sufficient condition for two vectors to be perpendicu
lar is that their scalar or dot product be zero.
Let the point O'lie on the circumference of the circle,
variable vector inhich may move to any point on the circle.
OT or r is a
If r moves and
stops so that point P lies on point 0, -m would have no vector.
That is
if j. F j ■ o,
the vector is null and its direction is undefined..
If point P moves anywhere on the circle except on point 09 the vector
r forms a right angle vdth r - 2 a.
The diameter 2 a divides the circle into a' serai-circle*
The radius
vector 2 a and the variable vector F are drawn from the same origin 0«
Hence, the vector that begins from the end of vector 2 a and ends at
the terminus of vector F represents the difference vector r - 2 a»
-35This proves the theorems
An angle inscribed in a semi-circle is a right angle*
6* Derive an expression for the area of a square of vjhich
r ■ ai x / a2 3
<D_
is the semi-diagonal*
Figure 22.
If the semi-diagonal is r » ai T / a2 3"* then the diagonal is
By definitions
A square is a quadrilateral Tvhose sides are equal, and tiiose angles
are equal*
Let /a/ represent the length of each side.
Let AB s X - Axis,
and
A D » T - Axis.
The diagonal 2 (a^ 1 / a£ 3) is expressed in the components of a on the
X_5 Y-, axes using the X, 3 unit vectors respectively*
I / I = I . (ai T / a2 3) / (ai 1 / a2 3)..
By coEtoiningj we get
a / a « 1 » 2 (ai i / a2 3)*
The figure, triangle ADC brings to crar mind lythagorean Theorem -
q2 a a2 / b2s
Let us apply this theorem by dotting each side into itself«
-36-
a. a / a. a « (2(ai i / &2 I) »2 (ai i / a2 I) )
a2 / a?^- l| (a2 / &2)»
Combining the sum of the right-hand member, we get
2 a2a It (a2 /af).
Dividing by 2, we obtain: 1? »• 2 (a2 / a2).
Hence,
The area of the square equals to twice the sun of squares of
and
7. The sum of the squares of the distances of any point 0 from the
angular points of the triangle exceeds the sum of the squares of its dis
tances from the m.ddle points of the sides by the sum of the squares of
half the sides.
Figure 23.
Given triangle ABC mth B0fclie mdpoint of AB, R the midpoint of AC and Q
the midpoint of BO.
Also given any point 0.
To proves
(OA)2 / (0B}2 / (0C)2 - /"(OR)2 / (0±-)2 / (0Q)2 J = (CR)2 / (1P)2
i.e.
""
-37-
(0A)2 / (0B)2 / (00)2 - (cR)2
(0P)2 / (0Q)2 / (CR}2 / (AP}2 / (BQ2)9
VtooSi
Let GfC = ji*
OB - b.
and
OC
Then
AB = F - a.
BC = c - b5
and
s H - c.
Also let
and
AC = "c CA = a - c»
and
GE ■ a / 1/2 (c - a) a H!
OQ ■ c / 1/2 (b - c) =
"OP ="b / 1/2 (a"
IPs 1/2 ( F -a).
BQ ■ 1/2 (c - b)•
GR = 1/2 (a - "c}.
Therefore;
(0Q)2 / (o?)2
•
I2
=2 a.F /
(AP)2 / (BQ)2 / (CR)2
IL^Ji / 1 - "c ^
I-c.
2 I.e / c2 / b2 ^ 2 b.c / c2 / 12 / 2I.b / b2 /
2 / c2 . 2 e.b" / F2 / 12 «2 aTc""/ e2).
/UE2/U c2).
-38-
a? / F2 / c2
(OC)2.
I conclude that,
(CH)2 / (OQ)2 / (0P}2
(AP)2 / (BQ)2 / (CR)2 E (0A)2 / (0G)2 / (0B)2.
8. Show that
(a - b) x (a / F) a 2 a x F
and give its geometric interpretation.
(a-l?) x (a »4 E) =axa/axF-bxa»Fxbs
But
a x a = o,
arid
F x F s o.
A necessary and sufficient condition for the cross product to vaiish 1e
that their vectors "be parallel*
By the definition of the cross product^
Eenee,
(a x F) / (a" x H) = 2 a x F,
Therefore^
(a - F) x (a / F) = 2 a x E.
Or,
2(1 - b) x (a / b) 1 m 2 1 a x F L
i
<k
L
-39-
a 2c b a E ab sin (ab) s - b x a*
Given tiro vectors "a and b" ike. , magnitude is the area of the parallelogram QA.CE of i/vhich a arid b are the adjacent sides*
OA is s. side of the parallelogram GACEj CD is the diagonal of parallelogran OBDC,
Hence, the parallelogram OBDC is turo times the area of the parallelo-
gren OHC3S,
the area, of a parallelogram Ofl.CE #1030 adjacent sides are l/2
of the diagonals of the parallelogram OBDC, is 1/2 of the area of the
parallelogram. C3DG.
9, Show that
(a - 5) . (a / E) s a2 - F2.
and interpret.
(a-fc) « (a/'b) =a«a/a.b"-b" . a - b . b"
(a - b") . (a / B") ■> 12
B2
Figure 25.
(a - b) and (a / b) represent the difference and the cum of tr:o r;iven
vectors a and b" respectively.
The difference and the sum of'the two given vectors see the sides of
the parallelogram OBDG.
The dot product of the sum and difference of the two vectors ("a - b")
\
and (a / b)' equal the difference of the squares of the magnitudes of the
two vectors,
QTj the dot product of the adjacent sides of a parallelogram equals
to the difference of the squares of the semi-diagonals of the parallelo
gram*
BIBLIOGRAPHY
Books
Coffin, Joseph George,
Vector Analysis; An Introduction to Vector Methods
and Their Various Applications to Physics and Mathematics.
Second Edition, Tnird Printing, July l^lj.6.
New York:
John
Wiley and Sons, Inc.
Phillips, He B»
Torks
Vector Analysis*
Tenth Minting, January 19h®«
New
John Wiley and Sons, Inc.
Taylor, Jarces Henry,
Analysis.
Hall, Inc.,
Vgjgtar Analjygis: With an Introduction to Tensor
Fourth Printing, August 19h7»
70 Fifth Avenue.
4a-
Net? York:
Frentiee-