GROUP (A)- CLASS WORK PROBLEMS
i.e. QR is scalar multiple of PQ
Q-1) If a , b , c are the position vectors of the
points
A, B, C respectively Such that
3a + 5b = 8c , then find the ratio in which
i) C divides AB, ii) A divides BC
∴
PQ & P R belong to same line
∴
points P, Q and R are collinear
Q-3) ABCD is a quadrillateral M and N are
mid points of the diagonal AC and BD
respectively. Show that
Ans. i) We are given that
AB + AD + CB + CD = 4MN .
3a + 5b = 8c
Ans. Let a , b, c , d , m , n be the position vectors of
∴
3a + 5b 5b + 3a
c =
=
8
5+3
∴
C divides AB internally in the ratio 5 : 3
vertices A, B, C, D, M , N respectively.
Since M and N are midpoints of diagonal
AC and BD respectively.
ii) We are given that
3a + 5b = 8c
∴
3a = 8c – 5b
∴
8c – 5b 8c – 5b
a=
=
3
8–3
∴
A divides BC externally in the ratio 8 : 5.
m =
a +c
,
2
n =
b +d
2
2m
= a + c & 2n
= b +d
…(i)
Now AB + AD + CB + CD
Q-2) Show that the points whose position
= b – a +d – a +b – c +d – c
vectors are 5a + 6b + 7c , 7a + 6b + 9c and
= 2b + 2d – 2a – 2c
3a + 6b + 5c are collinear.
Ans. Let
(
p = 5a + 6b + 7c ,
( ) ( )
= 4n – 4m
r = 3a + 6b + 5c ,
(
= 4 n –m
= q–p
=
QR
... (i)
(3a + 6b + 5c ) – ( 7a + 6b + 9c )
= 4a – 4c
(
∴
AB + AD + CB + CD
= 4MN
= r –q
=
)
= 4MN
(7a + 6b + 9c ) – (5a + 6b + 7c )
= 2a + 2c
)
= 2 2n – 2 2m
q = 7a + 6b + 9c ,
PQ
) (
= 2 b + d – 2 a + c (using i)
Q-4) G and G′′are centroids of ∆ ABC and ∆ A′′B′′C′′.
Show AA ′ + BB ′ + CC ′ = 3GG ′ .
Ans. Let a , b, c , a ', b ', c ', g and g ' are the position
)
=
–2 2a + 2c
=
–2PQ [using i]
vectors of A,B,C,A′,B′,C′,G and G′ respectively
with respect to some origin O.
G and G′ are the centroid of ∆ABC and ∆A′B′C′
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2
a +b +c
3
∴
g
=
∴
3g
= a +b +c
=
...[from (i), (ii) and (iii)]
...[i] And
a '+ b '+ c '
3
∴
g'
=
∴
3g '
= a '+ b '+ c '
L.H.S. =
c + a a +b b + c
+
–
2
2
2
...[ii]
AA '+ BB '+ CC '
=
(a '– a ) + (b '– b ) + (c '– c )
=
(a '+ b '+ c ') – (a + b + c )
= 3g '– 3g
= 3GG '
Q-5) D, E, F, are mid - point of sides BC, CA, AB,
respectively of ∆ ABC, show
=
1
c +a +a +b – b – c

2
=
1
2a
2
( )
= OA
= a
= R.H.S.
Hence proved.
(ii) L.H.S.
2
1
BE + CF
3
3
=
AD +
=
(d – a ) + 23 (e – b ) + 13 ( f
–c
)
b +c
 2 c +a
 1 a +b

=  2 – a  + 3  2 – b  + 3  2 – c 






i) OE + OF + DO = OA
ii) AD +
2
1
1
BE + CF = AC
3
3
2
=
Ans. Let a , b, c , d , e and f be position vectors of
A, B , C , D , E and F respectively w. r. t.point
O in the plane Here D,E,F are the mid-points
of BC, CA and AB.
B
∴
E
D
b +c
2
...(i);
e =
c +a
2
...(ii)
f =
a +b
2
...(iii)
(i) L.H.S.
=
3b + 3c – 6a + 2a – 4b + b – 2c
6
=
1
3c – 3a 

6
=
1
c – a

2
=
1
AC
2
C
= OE + OF + DO
= OE + OF – OD
= e+ f –d
Vectors
)
b + c – 2a 2 c + a – 2b
a + b – 2c
+
+
2
6
6
Using mid-point formula,
d =
(
=
A
F
b + c – 2a 2  c + a – 2b 
+ 

2
3 
2

1  a + b – 2c 
+ 

3 
2

= R.H.S. Hence proved
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3
Q-6) AB and CD are two chords of a circle
(
= 2 P – 2P
intersecting at right angles in P. Show that
PA + PB + PC + PD = 2PO where O is centre
of the circle.
=
)
– 2P = 2PO
Hence proved.
Ans. Draw OQ ⊥ AB and OR ⊥ CD
Let a, b, c , d, p, q and r
are the position
vectors of A,B,C,D,P,Q and R respectively with
Q-7) If a , b , c and d are the position vectors of
the points A, B, C and D respectively such
that 2a + 7b = 5c + 4d . Prove that AB and
CD intersect.
respect to centre O. Q and R are the midpoints of AB and CD respectively.
Ans. Given 2a + 7b = 5c + 4d
C
∴
O
2a + 7b
2+7
R
=
A
Q
5c + 4d
5+4
= e (say)
P B
D
Let e is the position vector of E. E .
∴
∴
∴
By section formula,
Using mid-point formula,
E divides AB internally in the ratio 7 : 2 and
CD internally in the ratio 4 : 5
a +b
2
q
=
2q
= a +b
r
=
2r
= c +d
...(i)
c +d
2
∴
E is the point on both AB and CD.
∴
AB and CD intersect in E.
Q-8) In ∆ ABC, O is circumcenter, H is orthocenter
then prove that
...(ii)
i) OA + OB + OC = OH
L.H.S.
(
ii) HA + HB + HC = 2 AO + BO + CO
= PA + PB + PC + PD
=
=
(a – p ) + (b – p ) + (c – p ) + (d – p )
Ans. Let a , b, c , o and h are p.v. of A,B,C,O and H
respectively.
(a + b ) + (c + d ) – 4 p
= 2q + 2r – 4 p
L.H.S.
We know that, if O,G and H are collinear then
G divides OH in the ratio 1:2 where G is
[from (i) and (ii)
(
= 2 q + r – 2p
)
centroid and g is its position vector.
...(i)
By section formula
Now, OQPR is a rectangle, hence it is also
parallelogram
∴
∴
by parallelogram law
OP
= OQ + OR
p
= q +r
)
∴
h + 2o
1+ 2
g
=
3g
= h + 2o
… (i)
i) L.H.S.
...(ii)
Substituting (ii) in (i) we get,
= OA + OB + OC
=
(a – b ) + (b – o ) + (c – o )
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∴
= a + b + c – 3o

a +b +c
= 3g – 3o ∴ g =
3




M and N are mid points of AB and CD
respectivrly.
m =
a +b
and
2
= h –o
n =
c +d
2
= OH
2m
= a + b and 2n = c + d
= R.H.S. Hence proved
L.H.S.
= h + 2o – 3o
from equation (i)
ii) L.H.S.
=
AD + AC
= d – a +c – b
HA + HB + HC
=
= a – h +b – h +c – h
(c + d ) – (a + b )
= 2n – 2m
= a + b + c – 3h
(
= 3g – 3h
= 2 n –m
= h + 2o – 3h from eq. (i)
= 2MN
= 2o – 2h
(
= 2 o –h
R.H.S.
)
Q-10) If ai + 3 j + bk is unit vector, prove that
)
a2 + b2 + 8 = 0
= 2HO
Ans. Let
(
= 2 AO + BO + CO
)
(
) (
) (
r = ai + 3 j + bk
… (ii)
= 2  o – a + o – b + o – c 
)
⇒
(
)
= 2 3o – 3g 
= 2 3o – h + 2o 
(
)
a 2 + b2 + 9
= 1
∴
a 2 + b2 + 9
= 1
∴
a 2 + b2 + 9
= 1
∴
= 2 3o – h – 2o 
a 2 + b2 + 8
= 0
Q-11) If a , b , c are position vectors of AB and C
= 2 o – h 
= 2HO
r =
But r
= 2 3o – a + b + c 
where A ≡ (0, 2, –1), B ≡ (0, – 1,3),
C≡
≡ (0 , 1,2) F in d x and y su ch that
… (iii)
c = xa + yb
From equations (ii) and (iii)
L.H.S. = R.H.S. Hence proved.
Ans. a = 2 j – k ;
Q-9) ABCD is a quadrilateral M and N are mid points of the diagonal AC and BD
respectively. Show that AD + BC = 2MN .
Ans. Let a , b, c , d , m, n be the position vectors of
points A,B,C,D,M,N w.r.t a fixed point
Vectors
=
…(i)
∴
b =
– j + 3k ;
c =
j + 2k
c =
xa + yb
(
)
(
)
j + 2k
=
x 2 j – k + y – j + 3k
j + 2k
=
( 2x – y ) j + ( – x + 3y ) k
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∴
2x – y
= 1
5
... (i) and
–x + 3y = 2
∴
(
3 3i + j + k
)
... (ii)
(
) (
= 3 3i + j – k + i + 5 j + 2k
Solving equations (i) and (ii) x = 1 and y = 1
(
+ x1i + y1 j + z1k
Q-12) If A(3, 2, -1), B(2, 3, 4), C(0, 5,14) then show
that the points A, B, C are collinear and
find the ratio in which B divides segment
∴
AC.
Ans. AB
=
AC
( 3 +1 – x1 ) i + (1+ 5 + y1 ) j + ( –1+ 2 + z1 ) k
9i + 9 j + 3k
–i + j + 5k
=
= c –a
=
AC
∴
)
9i + 9 j + 3k
=
= b –a
)
By equality of vectors, we have,
–3i + 3 j +15k
= 3AB
( x1 + 4 ) i + (y1 + 6 ) j + ( z1 +1) k
9 = x1 + 4, 9 = y1 + 6 and 3 = z1 + 1
∴
x1 = 5,y1 = 3 and z1 = 2
)
∴
C(5,3,2)
= 3b – 3a ⇒ c + 2a
∴
c = 5j + 3j +k
(
c –a = 3 b –a ⇒ c –a
Let P be the midpoint of BC.
= 3b
Then by midpoint formula, we have,
∴
b
c + 2a
=
3
=
∴
1.c + 2a
⇒ B
1+ 2
divides seg AC internally in the ratio 1:2.
b +c
, where p is the position vector
2
=
p
of P.
∴
p =
(i + 5 j + 2k )(5i + 3 j + 2k )
2
∴ A, B, C are collinear.
=
Q-13) If A(3,1,–1), B(1,5,2) are the vertices and
G(3,3,1) is the centroid of the triangle ABC
then by Vector method, find the mid-point
of the BC.
∴
1
6i + 8 j + 4k 3i + 4 j + 2k
2
(
)
midpoint of the side BC is (3,4,2).
GROUP (A)- HOME WORK PROBLEMS
Ans. Let C be ( x1, y1, z1 ) .
Then the postion vectors a, b, c , g of the points
A, B, C, G are
a = 3i + j – k ,
b = i + 5 j + 2k ,
c = x1i + y1 j + z1k ,
g = 3i + 3 j + k ,
Since G is the centroid of the ∆ ABC, by the
Q-1) A and B are two points with position
vectors a and b . If point C and D have
position vectors 5a − 4b and
prove A, B, C, D are collinear.
Ans. Let c and d respectively be position vectors
of points C and D
c = 5a – 4b
centroid formula,
g
∴
3g
=
a +b +c
3
= a +b +c
3a + 5b
,
8
d =
3a + 5b
8
c =
5a – 4b
5–4
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6
d =
and point Q is common point
3a + 5b
3+5
⇒
point C divides seg. AB externally in the ratio
4:5
⇒
points A, B, C are collinear points ...(i)
⇒
point D divides seg. AB internally in the ratio
5 : 3.
...(ii)
∴
points A, B, C, D are collinear points from
∴
PQ and QR are collinear
∴
points P, Q, R are collinear.
Q-3) If points P, Q, R, S have position vectors
(
)
p, q, r , s such that p – q = 2 s – r . Show
that the lines QS and PR trisect each other.
Ans. p – q
(i, ii)
= 2s – 2r
p + 2r
= q + 2s
p + 2r
3
=
q + 2s
3
p + 2r
1+ 2
=
q + 2s
1+ 2
Q-2) Show that the point P, Q, R with position
vectors a – 3b + c , –2a + 3b + 4c and
–b + 2c are collinear.
Ans. Let p, q ,r respectively be position vectors of
points P, Q, R w.r.t a fixed point.
= m
above is position vector of a point which
p
= a – 3b + c
divies PR and QS internally in the ratio 2 : 1
q
=
–2a + 3b + 4c
i.e. point M (say) divides PR and QS internally
in the ratio 2 : 1
r
=
–b + 2c
PQ
= q–p
(
∴
Q-4) A, B, C are three points with position
) (
= –2a + 3b + 4c – a – 3b + c
PQ
=
–3a + 6b + 3c
PQ
=
–3 a – 2b – c
PQ
–3
= a – 2b – c
QR
= r –q
(
) (
(
)
w .r.t
origin
O.
If
collinear.
)
Ans. 7c
...(i)
)
⇒
∴
= 2a – 4b – 2c
)
= 4a + 3b
c
=
4a + 3b
7
c
=
4a + 3b
4+3
point c divides seg AB internally in the ratio
3:4
= –b + 2c + 2a – 3b – 4c
(
a, b, c
vectors
7c = 4a + 3b , p rove th at A, B, C are
= –b + 2c – –2a + 3b + 4c
= 2 a – 2b – c
PR and QS trisect each other.
points A, B, C are collinear points.
Q-5) P, Q, R are three points with position
vectors p, q,r w.r.t some origin O. If
 PQ 
= 2  −3 


QR
=
2r = 5 p – 3q then prove that P, Q, R are
collinear.
–2
PQ
3
i.e QR is scalar multiple of PQ
∴
Ans. 2r
= 5 p – 3q
r
=
5 p – 3q
2
r
=
5 p – 3q
5–3
QR and PQ parallel or collinear
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7
⇒ point R divides seq. PQ externally in the
ratio 3 : 5.
=
b +c a +c a +b
+
+
2
2
2
=
2a + 2b + 2c
2
⇒ points P, Q, R are collinear.
Q-6) D, E, F are mid-points of the sides BC, CA
and AB respectively of ∆ABC. O is any point
in the plane of ∆ ABC Show that,
(
2 a +b +c
=
i) AD + BE + CF = 0
)
2
= a +b +c
ii) OA + OB + OC = OD + OE + OF
= OA + OB + OC
iii) BC = 2FE
= L.H.S
Ans. Let a , b , c , d , e , f respectively be position
vectors of points A, B, C, D, E, F write a fixed
point .
iii)
(
=2 e – f
A
F
R.H.S = 2 FE
)
 a + c   a + b  
= 2  2  –  2  
 
 

E
a + c – a – b 
= 2

2


B
D
C
= c –b
point D. E. F are mid points of sides BC, AC
= BC
and AB respectvely
b +c
,
d =
2
= L.H.S
Q-7) The position vectors of four points A, B, C
and D are a , b , 2a + 3b , a − 2b
a +c
,
e =
2
f =
a +b
2
i) AD + BE + CF
respectively. Express the vectors AC + DB
and BC in terms of a and b
Ans. Let position vectors of pts. A, B, C, D be
a ,b ,c ,d
then
= d – a +e – b + f – c
c
= 2a + 3b ,
b +c
a +c
a +b
–a +
–b+
–c
=
2
2
2
d
= a – 2b
AC
= c –a
=
2a – 2a + 2b – 2b + 2c – 2c
2
=
0
2
= 2a + 3b – a
= a + 3b
DB
(
= b – a – 2b
= R.H.S
)
= 3b – a
ii) R.H.S = OD + OE + OF
= d +e + f
= b –d
BC
= c –b
= 2a + 3b – b
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8
= 2a + 2b
d
Q-8) A divides seg PQ internally in the ratio
the ratio 5 : 2. Express AB in terms of p, q
and r .
Ans. Let a,b, p, q and r be position vectors of points
A, B, P, Q and R w.r.t a fixed point.
∴
a =
a =
∵
( )
.... (ii)
c (1+ x ) = a + xb
.... (iii)
d (1 – x ) = a – xb
.... (iv)
Adding equation (iii) and (iv)
point A divides seg PQ internally in the ratio
1:3
c (1+ x ) + d (1 – x )
∴
1 q +3p
a + xb
1– x
From (i) and (ii)
1 : 3 and B divides seg QR externally in
∵
=
a =
= 2a
(1 + x ) c + (1 – x ) d
2
1+ 3
q +3p
4
a =
...(i)
point B divided seg QR externally in the
ratio 5 : 2
∴
(1 + x ) c + (1 – x )
(1 + x ) + (1 – x )
A divides OC in ratio
(1 + x) : (1 – x)
or A divides CD internally in ratio
=
5r – 2q
5–2
b
=
5r – 2q
3
AB
= b –a
b
(1 – x) : (1 + x)
i.e.
...(ii)
1 – x 


 1+ x 
Now, equation (iii) - (iv)
∴
=
=
=
5r – 2q q + 3 p
–
3
4
20 r – 8 q – 3 q – 9 p
12
1
20 r – 11 q – 9 p
12
(
)
∴
c (1+ x ) – d (1 – x )
∴
b =
∴
b =
∴
B divides CD externally in ratio
inernally and externally respectively. Show
Ans. Let a , b , c , d be position vectors of point A,
B, C, D respectively C divides AB internally
in ratio x : 1.
∴
By internall division formula,
c
a + xb
=
1+ x
∴
2x
(1 + x ) c – (1 – x ) d
(1 + x ) – (1 – x )
1 – x 
A and B divides CD in ratio ± 

 1+ x 
and
Q-10) If the vectors 2i – q j + 3k
are collinear, then find q.
4i – 5 j + 6k
Ans. Let
.... (i)
(1 + x ) c – (1 – x ) d
1 – x 
(1 – x) : (1 + x) i.e. 

 1+ x 
Q-9) The points C and D divides AB in ratio x : 1
1 – x 
that A and B divides CD in ratio ± 

1+ x 
= 2xb
a = 2i – q j + 3k and
b = 4i – 5 j + 6k
D ddivides AB externally in ratio x : 1
∴
By esternal division formula,
Since the vectors a and b are collinear, the
components of i , j and k in a and b are
proportional.
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∴
∴
2
=
4
–q
–5
=
3
6
1
=
2
q
5
=
1
2
q =
5
2
9
∴
3 : 1.
ii) Putting λ = –3 in equation (2), we get
3(– 3 + 1)
∴
–6 = –3q
∴
q
ii) the values of and q.
∴
Q-12) The position vectors of the points A and B
and –3i + 2 j respectively.
are 2i – j + 5k
Find the position vector of the point which
divides the line segment AB internally in
Then a = 3i + 0.j + pk ,
the ratio 1 : 4
–i + q j + 3k and
Ans. The position vectors a and b of the points A
and B are
c = 3i + 3 j + 0.k
a = 2i – j + 5k and
i) As the points A, B, C, are collinear,
suppose the point C divides line segment AB
b =
in the ratio λ : 1.
∴
∴
divides the line segment AB internally in the
ratio 1 : 4
λ.b +1a
λ. +1
by the section formula for internal division,
–3i + 3 j + 0.k
∴
(
) (
λ –i + q j + 3k + 3i + 0.j + pk
=
)
c =
λ +1
=
∴
(
( λ +1) –3i + 3 j + 0.k
=
∴
=
( –λi + λq j + 3λk ) + (3i + 0.j + pk )
∴
( – λ + 3 ) i + 3qi + ( 3λ + p ) k
By equality of vectors, we have,
–3(λ
λ + 1) = – λ + 3
.... (i)
3(λ
λ + 1) =
.... (ii)
0 = 3λ
λ+p
λ
1.b + 4a
1+ 4
( –3i + 2j ) + 4 ( 2i – j + 5k )
5
)
–3 ( λ +1) i + 3 ( λ +1) i + 0.k
=
–3i + 2 j
Let C be the point, with position vector ,
by the section formula,
c =
p = 9
Hence p = 9 and q = 2
Ans. Let a , b and c be the position vectors of A, B
and c respectively.
b =
= 2
0 = –9 + p
C(–3, 3,0) are collinear, then find
the ratio in which the point C divides
the line segment AB
= –3q
Also, putting λ = –3 in equation (3), we get,
Q-11) If the points A(3, 0, p), B(–1, q, 3) and
i)
C divides segment AB externally in the ratio
.... (iii)
c =
1 5i – 2 j + 20k
5
(
)
2
i – j + 4k
5
Q-13) If P is orthocentre, Q is the circmentre and
G is the centroid of a triangle ABC, then
prove that QP = 3QG
Ans. Let p and q be the position vectors of P and
G w.r.t the circumcentre Q.
From equation (i), 3λ
λ – 3 = –λ
λ+3
i.e., QP = p and QG = g
∴
– 2λ
λ=6
∴
λ= –3
we knoe that Q, G, P are collinear and G divides
segment QP internally in the ratio 1 : 2
Vectors
Mahesh Tutorials Science
10
by section formula for internal division,
1. p + 2q
1+ 2
∴
g
=
∴
p
= 3q
∴
QP
= 3QG
=
p
3
∴
c
=
pi + q j + 7k ;
AB
= b –a
( 3i + 4 j + 5k ) – (i + 2 j + 3k )
=
= 2i + 2 j + 2k
AC
= c –a
Q-14) If A(a, 2, 2), B(a, b, 1) If C(1, 2, –2) are
vertices of ∆ ABC and G(2, 1, –c) is its
centroid find a, b and c.
Ans. Let a , b , c , g be position vectors pf points
A, B, C, G respectively
=
( pi + p j + 7k ) – (i + 2 j + 3k )
=
( p – 1) i + (q – 2) j + 4k
∴
AB and AC are collinear
AB = mAC [m : nonzero scalar]
b = aiˆ + bjˆ + kˆ,
2i + 2 j + 2k
c = iˆ + 2 ˆj – 2kˆ,
∴
g
=
∴
3g
= a +b +c
(
4m = 2 m(p – 1) = 2 p – 1 = 4 m (q – 2) = 2
m = 2/4 = 1/2 4/2 (p – 1) = 2 p = 5
a +b +c
3
3 2i + j – ck
)
= m ( p – 1) i + ( q – 2) j + uk 
Comparing the of i , j , k on both sides we get
g = 2iˆ + ˆj – ckˆ
∵
... (ii)
Given thaty the points are collinear
a = aiˆ + 2 ˆj + 2kˆ,
G is centroid of ∆ ABC
... (i)
1
(q – 2) = 2
2
q =6
Q-16) If the three points A(4,5,p). B(q,2,4) and
C(5,8,0) are collinear then find
= 6i + 3 j – 3ck = ( 2a +1) i + ( 4 + b ) j + k
i) The ratio in which the point C divides
the line AB
Comparing the coefficeients of i , j , k on both
sides
we get 2a + 1 = 6
ii) The values of p and q.
Ans. Let a, b,& c be the position vectors of A, B &
4+b=3
–3c = 1
C respectively.
∴
a = 4iˆ + 5 ˆj + p.kˆ,
5
2
b = qiˆ + 2 ˆj + 4kˆ,
b = –1
c = 5iˆ + 8 ˆj + 0kˆ,
a=
c = –1/3
i) As the point A,B & C are collinear,
Suppose the point C divides line segment
Q-15) If A ≡ (1, 2, 3), B ≡ (, 4, 5) and C ≡ (p, q, 7) are
AB in the ratio λ:1
collinear points. Find p & q.
Ans. Let a , b , c , g be position vectors pf points
A, B, C, G respectively
a
= i + 2 j + 3k ;
b
= 3i + 4 j + 5k ;
Vectors
∴
By the section fromula.
c =
λ.b +1.a
λ +1
(
λ +1 5iˆ + 8 ˆj + 0kˆ
)
Mahesh Tutorials Science
(
11
) (
= λ qiˆ + 2 ˆj + 4kˆ + 4iˆ + 5 ˆj + Pkˆ
)
Then p = i – 2 j + k and q = i + 4 j – 2k
=
internally in the ratio 2 :1, by section formula
( λq + 4) iˆ + ( λ 2 + 5 ) ˆj + ( λ 4 + P ) kˆ
for internal division,
By equality of vectors, we have
5(λ+1) = λq + 4
... (1)
8(λ+1) = λ2 + 5
... (2)
0 = λ4 + P
R (rˆ) di vide s line se gme nt PQ
Sinc e
5 ( λ +1) iˆ + 8 ( λ +1) ˆj + 0.kˆ
r =
=
... (3)
2q +1. p
2 +1
(
) (
2 iˆ + 4 ˆj – 2kˆ + iˆ – 2 ˆj + kˆ
)
3
From (1)
5λ + 5 = λ q + 4
5λ – λ q = – 1
=
2iˆ + 8 ˆj – 4kˆ + iˆ – 2 ˆj + kˆ
3
=
1 ˆ
3i + 6 ˆj – 3kˆ
3
... (5)
From (2)
8λ + 8 = λ 2 + 5
6λ = – 3
x=
–1
2
)
∴
r = iˆ + 2 ˆj – kˆ
∴
coordinates of R are (1,2 – 1).
... (6)
∴ C divides segment AB externally in the
ratio 1 : 2
 –1   –1 
–1
ii) 5   –   q = –1 Substitue λ =
2
2
2
   
Q-18) Find unit vectors along a, b and AB if
A ≡ (1, 2, 4) and B ≡ (2, –1, 5) .
Ans. A ≡ (1, 2, 4)
B ≡ (2, –1, 5)
Let a, b be the position vectors of A and B.
w.r.t fixed point.
in (5)
–5 q
+
2 2
(
a
= i + 2 j + 4k ,
–5+q=–2
b
= 2i – j + 5k
q=3
AB
= b –a
= –1=
0 = λ4 + p
= 2i – j + 5k – i – 2 j – 4k
4λ + p = 0
 –1 
4 
 2
= i – 3j +k
= p
a
p=–2
∴
p = – 2 and q = 3
Q-17) The position vectors P and Q are iˆ – 2 ˆj + kˆ
and
iˆ + 4 ˆj – 2kˆ
res p ectively .
b
=
(1)2 + ( 2)2 + ( 4 )2
=
1+ 4 +16
=
21
=
( 2)2 + ( –1)2 + (5 )2
=
4 +1 + 25
=
30
Fin d
coordinates & position vector of R ( rˆ )
wh ich divid es the line segment PQ
internally in the ratio 2 : 1.
Ans. Let p and q be the position vectors of the
a
unit vectors along a
=
a
points P and Q respectively.
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12
=
unit vectors along b =
AB
Q-20) If a , b , c are position vectors of A, B and C
i + 2 j + 4k
where A (1, 3, 0), B (2, 5, 0), C (4, 2, 0) such
21
2i + j + 5k
30
=
(1)2 + ( –3 )2 + (1)2
=
1+ 9 +1
=
11

b
b





that c = xa + yb then find x and y.
Ans. A ≡ (1, 3, 0)
B ≡ (2, 5, 0)
C ≡ (4, 2, 0)
a = i + 3j
b = 2i + 5 j
c = 4i + 2 j
AB
unit vectors along AB =
=
AB
Now, c
=
xa + yb
i – 3j +k
4i + 2 j
=
x i + 3 j + y 2i + 5 j
4i + 2 j
=
xi + 3xj + 2yi + 5y j
4i + 2 j
=
( x + 2y ) i + ( 3x + 5y ) j
11
Q-19) If ‘C’ divides AB internally in the raio 5 : 6
and if C (1, 5, 0), A (3, 8, 6) find the co-
(
)
(
)
comparing co-efficients of i and j on both
ordinates of B.
Ans. Let a, b, c be the position vectors of points
sides
A, B, C respectively w.r.t a fixed point.
x + 2y = 4
...(i)
A ≡ (3, 8, 6)
3x + 5y = 2
...(ii)
C ≡ (1, 5, 0)
multiplying equation, (i) by 3 and then
subtracting (ii) form (i).
a = 3i + 8 j + 6k
3x + 6y = 12
c = i +5j
3x + 5y = 2
Now, C divides AB internally in the ratio
5:6
c =
( ) ( )
5 b +6 a
i +5j =
5+6
( ) (
5 b + 6 3i + 8 j + 6k
)
11
11i + 55 j
= 5b +18i + 48 j + 36k
5b
=
–7i + 7 j – 36k
b
=
–
7
7
36
i+ j –
k
5
5
5
 –7 7 –36 
, ,
co-ordinates of B ≡ 

 5 5 5 
y = 10
substitute y = 10 in equation (i)
x + 20
= 4
x
= –16
{x = –16, y = 10}
Q-21) If A (0, 1, 3), B (3, 2, 1) and
A B = xiˆ + yjˆ + zkˆ then prove x + y + z = 0
Ans. Let a and b be position vectors of points A
and B w.r.t a fixed point.
A
= (0, 1, 3)
B
= (3, 2, −1)
a
=
b
= 3i +2 j –k
AB
= b –a
j + 3k
∧
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∧
∧
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∧
∧
∧
∧
∧
∧
∧
∧
∧
= 3 i + 2 j – k – j – 3k
x i +y j +zk
∧
13
∧
∧
∧
∴
3i – k
(m + 2n ) i + (m – n ) j
+ ( –2m + n ) k
=
∧
= 3 i + j – 4k
x i +y j +zk
By equality of vectors,
∧ ∧ ∧
comparing co-efficients of i , j ,k on both side
m + 2n = 3
.... (i)
x = 3, y = 1, z = – 4
m–n=0
.....(ii)
L.H.S
= x+y+2
and –2m + n = – 1
= 3+1–4
To solve equations (i) and (ii),
= 0
Subtracting (ii) from (i), we get,
.....(iii)
3n = 3
= R.H.S.
Q-22) If A (1, 4, 1), B (–2, 3, –5) and C (5, –7, 2) are
∴
Substituting n = 1 in equation (i), we get,
vertices of triangle then find the centroid
of the triangle ABC.
Ans. Let a , b , c , g be position vectors of A, B, C and
n = 1
m + 2(1) = 3
∴
m = 1
Substituting m = 1, n = 1 in equation (iii), we
get,
G respectively.
Let G be the centroid.
LHS = –2(1) + (1)
a = iˆ + 4 ˆj + kˆ
b =
= –2 + 1
–2iˆ + 3 ˆj – 5kˆ
= –1
= RHS.
c = 5iˆ – 7 ˆj + 2kˆ
∵
Hence, m = 1 and n = 1.
G is the centroid of ∆ ABC
=
3g
= a +b +c
=
3g
g
Q-24) If G 1 and G 2 are the centroids of the
a +b +c
3
g
triangles ABC and PQR respectively, then
prove that AP + BQ + CR = 3G1G 2 .
Ans. Let a , b, c, p, q, r , g1 , g 2 be the position vectors
(i + 4 j + k ) + ( –2i + 3 j + 5k )
+ ( 5i – 7 j + 2k )
of the points A, B, C, P, Q , R, G1 and G2
respectively.
Since G1 and G2 are the centroids of ∆ ABC
and ∆ PQR respectively,
= 4i + 0 j – 2k
=
4
2
i + 0j + k
3
3
–2 
4
G  ,0, 
3 
3
∴
a + b + c = 3 g1 and p + q + r = 3 g 2
(
) (
(
) (
= p +q +r – a +b +c
, then find the scalars m and n
c = 3i – k
= 3 g 2 – 3 g1
such that c = mb + nb
(
= 3 g 2 – g1
Ans. c = mb + nb
(
) (
= p –a + q –b + r –c
, b = 2i – j + 2k
and
Q-23) If a = i + j – 2k
3i – k
g1 =
... (i)
Now, AP + BQ + CR
Centroid of ∆ ABC
∴
a+b +c
p+ q + r
and g 2 =
3
3
∴
) (
= m i + j – 2k + n 2i – j + 2k
)
)
)
... [From (i)]
)
= 3G1 – G2
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14
Q-25) If a , b , c are the position vectors of the
(9i + 8 j – 10k ) – (3i + 2j – 4k )
=
points A (1,3,0), B(2,5,0), C(4,2,0)
= 6i + 6 j – 6k
respectively and c = ma + nb then find the
values of m and n.
AC
Ans. The position vectors a , b, c of the points A,
= c –a
( –2i – 3j + k ) – (3i + 2j – 4k )
=
B,C are
= 5i – 5 j + 5k
a = iˆ + 3 ˆj + 0.kˆ, b = 2iˆ + 5 ˆj + 0.kˆ,
c = 4iˆ + 2 ˆj + 0.kˆ,
–
=
Now c = ma + nb
5 6i + 6 j – 6k
6
(
)
....(ii)
From (i) and (ii),
∴ 4iˆ + 2 ˆj + 0.kˆ
(
....(i)
)
(
= m iˆ + 3 ˆj + 0.kˆ + n 2iˆ + 5 ˆj + 0.kˆ
)
AC
 5
=  –  AB
 6
i.e, AC is the scalar multiple of AB .
∴ 4iˆ + 2 ˆj + 0.kˆ
= (m + 2n ) iˆ + ( 3m + 5m ) ˆj + 0.kˆ
By equality of vectors,
m + 2n = 4
∴
they are parallel to each other. But they have
the point A in common.
∴
the vectors AB and AC lie on the same line.
.....(i)
and 3m + 5n = 2
the points a, B and C are collinear.
......(ii)
ii) Refer to the solution of Q.1(i)
Multiplying equation (i) by 3, we get,
Subtracting equation (ii) from this equation,
and 4i – 5 j + 6k
Q-2) If the vectors 2i – q j + 3k
we get, n = 10
are collinear, then find the value of q.
Subtracting n = 10 in equation (i), we get,
m + 2(10) = 4
= (–1) PQ .
PR
3m + 6n = 12
Ans. Let a = 2i – q j + 3k and
∴ m = – 16
b = 4i – 5 j + 6k .
Hence m = – 16 and n = 10.
Since the vectors a and b collinear, the
GROUP (B)- CLASS WORK PROBLEMS
components of i , j and k a and b are
propertional.
Q-1) Show that the following points are collinear :
i) A(3, 2, –4), B(9, 8, –10), C(–2, –3, 1)
∴
2
=
4
–q
–5
=
3
6
∴
1
=
2
q
5
=
1
2
∴
q =
5
.
2
ii) P(4, 5, 2), Q(3, 2, 4), R(5, 8,0).
Ans. i) The position vectors a , b and c of the
point A, B and C are
a = 3i + 2 j – 4k ,
b = 9i + 8 j – 10k ,
Alternative Method.
c = 2i – 3 j + k .
Let
g = 3i + 3 j + k ,
∴
AB
= b –a
a = 2i – q j + 3k and
b = 4i – 5 j + 6k .
Since the vectors a and b are collinear,
a ×b = 0
Vectors
Mahesh Tutorials Science
i
j
15
= i – 2 j + 3k
k
∴
2 –q 3
4 –5 6
∴
i ( –6q +15 ) – j (12 – 12 )
∴
If A,B,C, D are coplanar, then there exist
scalars x,y such that
AB
x.AC + y.AD
( ) (
x 2 j + y i – 2 j + 3k
)
+ k ( –10 + 4q ) = 0
∴
–2i + 2 j =
( –6q +15 ) j – 0 j ( –10 + 4q ) k
∴
–2i + 2 j = yi + ( 2x – 2y ) j – 3yk
By equality of vecotrs,
= 0i + 0 j + 0k
By equalify of vectors,
–6q + 15 = 0 and –10 + 4q = 0
∴
=
15
q =
6
10
q =
4
Hence, q
=
5
and
2
=
5
2
=
5
2
y = –2
....(i)
2x – 2y = 2
....(ii)
3y = 0
....(iii)
From (i), y = –2
From (iii), y = 0
This is not possible.
Hence, the points A, B, C, D are not coplanar.
Q-4) If a , b , c are non-coplanar vectors then
p ro ve
Q-3) Are the four points A(1, –1,1), B(–1,1,1) ,
C(1,1,1) and D(2,–3,4) coplanar? justify
your answer.
that
th e
vectors
2a – 4b + 4c , a – 2b + 4c and –a + 2b + 4c
are collinear
Ans. Let P,Q,R be the points whose position vectors
Ans. The position vectors a , b, c , d , of the points A,
p, q , r are given by
B, C, D are
∴
a
= i – j + k ,
b
=
c
= i + j + k ,
d
= 2i – 3 j + 4k
AB
= b –a
=
=
AC
–i + j + k ,
(
) (
–i + j + k – i – j + k
)
= 2a – 4b + 4c ,
q
= a – 2b + 0.c ,
r
=
PQ
= q–p
PR
–2i + 2 j
(
) (
i + j + k – i – j + k
= 2 j
(
(a – 2b + 4c ) – (2a – 4b + 4c )
=
–a + 2b + 0.c
(a + 2b + 4c ) – (2a – 4b + 4c )
(
)
and
.... (i)
= r –q
= 3 –a + 2b + 0.c
= d –a
=
–a + 2b + 4c .
=
=
= c –a
=
AD
∴
p
∴
PQ
∴
PQ + ( –3 ) PQ
= 3PQ
)
.... [By (i)]
= 0
) (
2i – 3 j + 4k – i – j + k
)
Vectors
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16
∴
PQ and PQ are collinear vectors,
Now,
Hence the points P ( p ),Q ( q ) and R (r ) are
r = x 2a – b + 4c + y –a + 4b – 3c
(
collinear
(
)
5a – 6b +11c
Q-5) Express p = 3i + 2 j + 4k
as the linear
=
combination of the vectors
a = i + j , b = j + k and c = k + i
Ans. To express p as a linear combination of
( 2x – y ) a + ( – x + 4y ) b + ( 4x – 3y ) c
2x – y = 5
...(i)
–x + 4y = –6
...(ii)
4x – 3y = 11
...(iii)
From (i) and (ii) x = 2 & y = – 1
a , b, c
Let p = xa + yb + zc
For these values of x and y
where x , y, z
L.H.S of (iii)
are scalars
∴
)
= 4(2) – 3(–1)
3i + 2 j + 4k
(
)
= 11
(
)
(
= x i + j +y j +k + z k +i
=
)
= R.H.S.
Thus, these values satisfy the (iii) equation
also.
( x + z ) i + ( x + y ) j + (y + z ) k
x + z = 3;
∴
The given vectors are coplanar.
x + y = 2;
y +z = 4
GROUP (B)- HOME WORK PROBLEMS
Solving we get
∴
z =
5
;
2
y =
3
;
2
x =
1
2
p =
1
3
5
a+ b+ c
2
2
2
Q-6) If a , b and c are non-coplanar vectors, show
that the vectors 2a – b + 4c , –a + 4b – 3c
Q -1) a and b are non-collinear vectors. If
c = ( x – 2 ) a + b and d = ( 2x +1) a – b are
collinear, then find the value of x.
Ans. Since c and d are collinear vectors, there
exists scalar t such that
c = td
∴
( x – 2) a + b
= t ( 2x +1) a – b 
∴
( x – 2) a + b
= t ( 2x +1) a – tb
∴
x–2
and 5a – 6b +11c are coplanar.
= t(2x + 1) and
t
= –1
∴
t
= –1
∴
x–2
Ans. p = 2a – b + 4c ,
q =
–a + 4b – 3c , r
= 5a – 6b +11c
In order to show that vectors p, q and r are
coplanar.
= –(2x + 1)
= –2x – 1
∴
3x = 1
∴
x =
To find the scalars x and y such that
r = x p + yq
Vectors
1
3
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17
By equality of vectors,
Q-2) If a , b , c are non-coplanar vectors then
p ro ve th at th e vecto rs 2 a – 4b + 4c,
a – 2b + 4c and –a + 2b + 4c are collinear.
2x + 2y + 3z
= –1
x–y+z
= –3
–4x + 3y – 2z = 4
Ans. Let P, Q and R be the points whose position
Solve these equations by using Cramer’s Rule
vectors are p = 2a – 4b + 4c , q = a – 2b + 4c ,
2
r = –a + 2b + 4c
D =
Now,
(
) (
PQ = q – p = a – 2b + 4c – 2a – 4b + 4c
∴
)
= – 2 – 4 – 3 = –9 ≠ 0
and
(
) (
PR = r – p = – a + 2b + 4c – 2a – 4b + 4c
(
= –3a + 6b + 0c = 3 – a + 2b + 0.c
3
–1 1
3 –2
= 2(2 – 3) – 2(–2 + 4) + 3(3 – 4)
... (i)
PQ = –a + 2b + 0.c
2
1
–4
)
Dx
=
–1 2
–3 –1
4
)
3
1
3 –2
= –1(2– 3) – 2(6 – 4) + 3(–9 + 4)
...[From (i)]
∴
PR = 3PQ
∴
PR + ( –3 ) PQ = 0
∴
∴
PR and PQ are collinear vectors.
P, Q and R are collinear.
∴
p = 2a – 4b + 4c , q = a – 2b + 4c
= 2(6 – 4) + 1(–2 + 4) + 3(4 – 12)
r = –a + 2b + 4c are collinear.
= 4 + 2 – 24 = –18
Q-3) Ex press
–i – 3 j + 4k
combination
of
as
th e
the
= 1 – 4 – 15 = –18
Dy
lin ear
Dz
vectors
, 2i – j + 3k
and 3i + j – 2k
.
2i + j – 4k
b = 2i – j + 3k ,
∴
x=
Dz –18
=
= 2,
D
–9
y=
Dy
D
=
–18
= 2,
–9
Dz 27
z = D = –9 = –3
–i – 3 j + 4k .
Suppose p = xa + yb + ze.
2 –1
–1 –3
3 4
= 10 + 16 + 1 = 27
c = 3i + j – 2k and
p =
=
2
1
–4
= 2(–4 + 9) –2(4 – 12) –1(3 – 4)
a = 2i + j – 4k ,
Ans. Let
=
2 –1 3
1 –3 1
–4 4 –2
∴
p = 2a + 2b – 3c
Then, –i – 3 j + 4k
(
) (
)
+ z ( 3i + j – 2k )
= x 2i + j – 4k + y 2i – j + 3k
∴
–i – 3 j + 4k
=
as the linear
Q-4) Ex press P = 3i + 2 j + 4k
com binatio n
of
the
vectors
, c = k
+ i
a = i + j , b = j + k
Ans. To express p as a linear combination of a , b, c
Let
( 2x + 2y + 3z ) i + ( x – y + z ) j
+ ( –4x + 3y – 2z ) k
p = xa + yb + zc
where x, y, z are scalars
∴
3i + 2 j + 4k
Vectors
Mahesh Tutorials Science
18
∴
) (
(
) (
)
=
x i + j + y j + k + z k + i
=
( x + z ) i + ( x + y ) j + (y + z ) k
x + z = 3;
–
x =
∴
–x – y = 2 gives
∴
y =
x = y = 2;
y+z=4
–
=
3
2
z =
5
,
2
For these values of x and y,
y =
3
,
2
–5x + y =
x =
1
2
p =
1
3
5
a+ b+ c
2
2
2
=
–a + 3b – 5c , –a + b + c and 2a – 3b + c are
collinear
Ans. Let
 1  3
–5  –  +  – 
 2  2
5 3
–
2 2
= 1
Q-5) If a , b , c are non-coplanar vectors, then
sh ow
that
th e
vectors
p
=
–a + 3b – 5c ,
q
=
–a + b + c
r
= 2a – 3b + c
Thus, these values satisfy the third equation
also.
∴
 1
 3
r =  – 2 p +  – 2 q




∴
the given vectors are coplanar.
Q-6) Ex p ress
and
–i – 3 j + 4k
com binatio n
coplanar, we should be also to find scalars x
and y such that
c = 3i + j – 2k and
(
)
(
p =
)
=
x –a + 3b – 5c + y –a + b + 5c
=
( – x – y ) a + ( 3x + y ) b + ( –5x + y ) c
–i – 3 j + 4k .
Suppose p =
xa + yb + zc .
Then, –i – 3 j + 4k
r = 2a – 3b + c
=
2a – 3b + c
(
∴
–x – y
= 2
.... (i)
3x + y
= –3
.... (ii)
and –5x + y = 1
Adding (i) and (ii), we get,
2x = –1
.... (iii)
) (
)
+ z ( 3i + j – 2k )
x 2i + j – 4k + y 2i – j + 3k
( – x – y ) a + ( 3x + y ) b + ( –5x + y ) c
By equality of vectors,
Vectors
the
b = 2i – j + 3k ,
Now, x p + yq
=
th e
a = 2i + j – 4k ,
Ans. Let
r = x p + yq
and
of
as
lin ear
vectors
, 2i – j + 3k
and 3i + j – 2k
.
2i + j – 4k
Then in order to prove that these vectors are
∴
1
–y =2
2
1
–2
2
solving, we get
∴
1
2
∴
–i – 3 j + 4k
=
( 2x + 2y + 3z ) i + ( x – y + z ) j
+ ( –4x + 3y – 2z ) k
Mahesh Tutorials Science
19
By equality of vectors,
2x + 2y + 3z
= –1
x– y+z
= –3
p = i + 4 j – 4k .
suppose p =
Then,
–4x + 3y – 2z = 4
Solve these equations by using Cramer’s Rule
D
2
2
3
1
–1
1
–4
3
–2
=
= – 2 – 4 – 3 = –9 ≠ 0
Dx
=
2
3
–3 –1
1
4
i + 4 j – 4k
=
= 2(2 – 3) – 2(–2 + 4) + 3(3 – 4)
–1
xa + yb + zc .
(
) (
)
+ z ( –i – 3 j – 5k )
x 2i – j + 3k + y i – 2 j + 4k
i + 4 j – 4k
= x ( 2x + y – z ) i + y ( – x – 2y + 3z ) j
+ ( 3x + 4y – 5z ) k
By equality of vectors,
3 –2
2x + y – z = 1
= –1(2– 3) – 2(6 – 4) + 3(–9 + 4)
= 1 – 4 – 15 = –18
Dy
=
–x – 2y + 32 = 4
3x + 4y – 5z = –1
2 –1
3
We have to solve these equation by ussing
1 –3
1
cramer’s Rule.
–4
4
–2
2
= 2(6 – 4) + 1(–2 + 4) + 3(4 – 12)
D =
Dz
=
2 –1
1
–1 –3
–4
3
–1 –2
3
= 4 + 2 – 24 = –18
2
1 –1
3
4 –5
= 2(10 – 12) – 1(5 – 9) – (–4 + 6)
= –4 + 4 – 2 = –2 ≠ 0
4
1
= 2(–4 + 9) –2(4 – 12) –1(3 – 4)
Dx =
= 10 + 16 + 1 = 27
∴
∴
x=
Dz –18
=
= 2,
D
–9
y=
Dy
D
=
–18
= 2,
–9
4
–2
3
–4
4
–5
= 1(10 – 12) –1 (20 + 12) – 1(16 – 8)
= –2 + 8 – 8
Dz 27
z = D = –9 = –3
Dx = –2
p = 2a + 2b – 3c
Dy =
2
–1
3
as the linear
Q-7) Express the vector i + 4 j – 4k
combination of the vectors
.
, i – 2 j + 4k
and –i + 3 j – 5k
2i – j + 3k
Ans. Let
a = 2i – j + 3k ,
c =
–i – 3 j – 5k
and
1 –1
4
3
–4 –5
= 2(–20 + 12) – 1(5 – 9) – 1(4 – 12)
= –16 + 4 + 8
Dy = –4
Dz =
b = i – 2 j + 4k ,
1 –1
2
1
1
–1
–2
4
3
4
–4
= 2(8 – 16) – 1(4 – 12) + 1(–4 + 6)
= –16 + 8 + 2
Vectors
Mahesh Tutorials Science
20
Dz = –6
2 0 0
Dx
D
x =
Dy
y =
D
Dz
D
z =
=
–2
–2
= 1
=
–4
–2
= 2
–6
–2
= 3
=
=
0 3 0
0 0 4
= 2 (12) – 0 + 0
= 24 cubic units.
Q-4) Show that vectors
a = i + j + k, b = i – j + k &
∴
p = a + 2b + 3c
c = 2i + 3 j + 2k are coplanar.
ii) ∴ a b c 
3 2 1


= 1 1 2
3 −1 2
Ans. If [a b c ] = 0
then vectors a , b and c are coplanar.
= 1(1) –1(–1)
=2
L.H.S.
=
a b c 


=
GROUP (C)- CLASS WORK PROBLEMS
1 1 1
1 −1 1
2
3
2
1( – 2 – 3) – 1(2 – 2) + 1(3 + 2)
Q-1) Find [a b c ] if
= –5+5=0
= R.H.S.
a = 2i + j + 3k , b = i + 2 j + 4k ,
∴
c = i + j + 2k
are coplanar vectors.
2 1 3
Ans. a b c 
=
∴
[a b c ]
Q-5) Show that A, B, C, D are coplanar, if
1 2 4
1 1 2
=
A ≡ (2, 1, – 3), B ≡ (3, 3, 0), C ≡ (7, –1, 4) and
D ≡ (2, – 5, – 7).
2(4 – 4) – 1(2 – 4) + 3(1 – 2)
Ans. We know that, four points A, B, C, and D are
= –1
coplanar if
[ AB AC AD ] = 0
Q-2) a = i – j + pk , b = i + j – 4k , c = 2i + j + k
find p, if [a b c ] = 0
Ans. a b c 
∴
∴
AB
p
1
2
−4
= 0
1
∴
1(1 + 4) + 1(1 +8) + p(1 – 2)
= 0
∴
p = 14
= i + 2 j + 3k
AC
= c –a
an d
c = 4k
are
th e
coterminus edges of the parallelepiped.
Ans. Volume of parallelepiped = [a b c ]
Vectors
∴
( 7 – 2) i + ( –1 – 1) j + ( 4 + 3) k
AC
= 5i – 2 j + 7k
AD
= d –a
=
( 2 – 2) i + ( –5 – 1) j + ( –7 + 3) k ⇒
=
–6 j – 4k
Q-3) Find the volume of the parallelopiped if
a = 2i , b = 3 j
( 3 – 2) i + ( 3 – 1) j + ( 0 + 3) k
AB
=
14 – p = 0 ⇒
∴
= b –a
=
= 0
1 −1
1
1
a , b, c
L.H.S.
=
 AB AC AD 


Mahesh Tutorials Science
1
=
21
2
3
5 –2
7
1 2 3
∴
2 3 4
= 0
x y z
0 –6 –4
= 1(8 + 42) – 2(– 20) + 3 (– 30)
∴
1(3z –4y) – 2(2z – 4x) + 3(2y – 3x) = 0
= 50 + 40 – 90 = 0 = R. H. S.
∴
3z –4y – 4z + 8x + 6y – 9x = 0
∴
AB , AC , AD are coplanar vectors.
∴
– x + 2y – z = 0
∴
The points A, B, C and D are coplanar.
∴
x – 2y – z = 0
Q-6) Find value of m if points (2, –1, 1),
(4, 0, 3), (m, 1, 1), (2, 4, 3) are coplanar.
Ans. Let A ≡ (2, –1,1), B ≡ (4,0,3), C ≡ (m,1,1) and
Q-8) If A ≡ (1, 1, 1), B ≡ (2, -1, 3), C ≡ (3, –2, –2)
and D ≡ (3, 3, 4), find the volume of
parallelopiped with segments AB, AC and
D ≡ (2,4,3)
AD as concurrent edges.
Given that the points A, B, C, D are
coplanar
AB
Ans. We know that, the volume of parallelopiped
with concurrent edges AB, AC and
= b –a
=
( 4 – 2) i + (1+1) j + ( 3 – 1) k
AB
= 2i + j + 2k
AC
= c –a
(m – 2) i + (1+1) j + (1 – 1) k
AC
=
(m – 2 ) i + 2 j
AD
= d –a
=
=
AB
= b –a
=
=
∴
= i – 2 j + 2k
AC
= c –a
 AB AC AD  = 0


2
1 2
m −2 2 0
0
5 2
∴
∴
∴
= 2i – 3 j – 3k
AD
= d – a = ( 3 – 1) i + ( 3 – 1) j + ( 4 – 1) k
∴
AD
= 2i + 2 j + 3k
∴
volume of parallelopiped
= 0
1 −2
2
2 −3 −3
2 2 3
2(4) – 1(2m – 4) + 2(5m –10) = 0 ⇒
8 – 2m + 4 + 10m – 20 = 0
=
8m = 8 ⇒
m=1
= 1 (– 9 + 6) + 2 (6 + 6) + 2 (4 + 6)
= – 3 + 24 + 20 = 41 cubic units.
Q-7) If O is the origin, A(1,2,3) B(2,3,4) and
P(x,y,z) are coplanar points prove that
x – x + 2y – z = 0. using vector method.
Q-9) Prove that [a + b b + c c + a ] = 2[abc ]
Ans. Let
= [a + b b + c c + a ]
Ans. The points O, A, B and P are coplanar. (given)
=
(a + b ) .[ (b + c ) × (c + a )]
= a = i + 2 j + 3k
=
(a + b ) .[b × c + b × a + c × c + c × a ]
OB
= b = 2i + 3 j + 4k
OP
=
= a . [b × c + b × a + c × a ]
+ b. [b × c + b × a + c × a ]
∴
[OA OB OP ]
∴
OA
∴
( 3 – 1) i + ( –2 – 1) j + ( –2 – 1) k
AC
∴
= 5 j + 2k
∴
( 2 – 1) i + ( –1 – 1) j + ( 3 – 1) k
AB
=
( 2 – 2) i + ( 4 +1) j + ( 3 – 1) k
 AB , AC , AD 


AD
= 0
p = xi + y j + zk
[OA OB OC ]
= 0
Vectors
Mahesh Tutorials Science
22
Q-11) Show that the statement
[∵ c × c = 0]
=
(
(a – b ).[(b – c ) × (c – a )] = 2a .b × c
)
( ) ( )
+ b. (b × c ) + b. (b × a ) + b. (c × a )
a. b × c + a . b × a + a. c × a
is true only when a , b and c are coplanar.
Ans. If a , b , c are coplanar then
(
)
(
a. b × c + b. c × a
=
) +0+0+0+0
[a b c ] = 0
(in box product if two vectors are equal , its
(
i.e., a . b × c
=
[a b c ] + [b c a ]
L.H.S.
=
[a b c ] + [a b c ]
=
(a – b ) . (b – c ) × (c – a )
(∵ a b c  = b c a  )
=
(a – b ) .[b × c – b × a – c × c + c × a ]
2[a b c ] = R.H .S .
=
(a – b ) .[b × c + a × b + c × a ]
=
Q-10) If c = a + 3b , show that , abc  = 0 Also if
=
– b.[b × c + a × b + c × a ]
angle between a and b is π 6
=
(
)
(
∴
a , b, c are coplanar
∴
a b c 


(
)
= 2a.b × c
a + 3b
c
c
c
c
2
2
2
2
2
c2
Vectors
a + 3b
=
(
2
∴
)(
=
a
=
a
2
2
+ 6a .b + 9 b
+ 6 a 6 cos
+6 a b .
)
)
= 0
= (0)
= 0
∴
a
R.H.S.
)
= a.a + 3a .b + 3b . a + ab .b
=
(
if a , b , c are planar than [a b c ] = 0
a + 3b . a + 3b
2
)
= 2[a b c ]
Squaring both sides.
=
(
= [a b c ] – [a b c ]
c =
c
(
)
= [a b c ] – [b c a ]
= 0
R.H.S
c
(
)
= a . b × c – b. c × a
c = a + 3b
2
)
− b. b × c + b. a × b + b. c × a
i.e c is linear combination of a and b
2
(
a. b × c + a. a × b + a . c × a
Ans. c = a + 3b
c
= 0
a.[b × c + a × b + c × a ]
c 2 = a 2 + 3 3ab + ab 2 , P ro ve th at th e
∴
)
value is zero)
L.H.S
= R.H.S
When a , b , c are co planar
2
Q-12) a , b , c are three non-coplanar vectors. If
π
6
2
3
+9b
2
3
+ 9 b2
= a + 6a b .
2
2
( )
+9 b
p=
a b c 


,q =
c×a
a b c 


,r =
a×b
a b c 


then prove that
i)
a . p + b.q + c .r = 3
ii)
( a + b ) .p + ( b + c ) .q + ( c + a ) .r = 3
2
= a 2 + 3 3 ab + 9 b 2
b×c
Mahesh Tutorials Science
23
Ans. i) L.H.S.
= a. p + b.q + c .r
(
a. b × c
=
[a b c ]
= a.
iii) a . p.
) + b. (c × a ) + c. (a × b )
[a b c ]
[a b c ]
c .q
= c.
(b × c ) ⇒
(a b c )
(c × q )
(a b c )
[a b c ] + [b c a ] + [c a b ]
=
[a b c ]
=
[a b c ] + [a b c ] + [a b c ]
=
b .p =
[a b c ]
= 3
)
(
)
(
)
ii) L.H.S. a + b . p + b + c .q + c + a .r
=
(b × c ) (b + c ) . (c × a )
(a + b ) . [a b c ] + [a b c ]
=
(
b . b ×c
3[a b c ]
(
=
(c c a )
(a b c )
= 1
[a b c ]
=
(a b c ) ⇒
(a b c )
(
+ c +a
)
(a b c )
(b b c )
(a b c )
⇒
(
)
c . a ×b
c .v
=
[a b c ]
=
0
=
(a b c )
=
(c a b )
(a b c )
=
(a b c )
(a b c )
(a b c )
(a × b )
.
Since a b c  is a scalar.
)⇒
0
(a b c )
⇒
=0
Let a b c  = k
(a + b ) . p
(b + c )
=
(a + b ) .
=
1
a b c  + a b c 
 

k
=
1
a b c  + 0

k 
=
=
⇒
=
⇒ b .p
=
k
{
(
b . c ×a
}
=
1
a b c 

k
1
0×k
k
(
)
(
)
= 1
and c + a .r
(b c a )
(a b c ) ⇒
(
a .r
=
)
(a b c )
=
(a b c )
(a b c )
=
(
= 1
)
(a b c )
a . c ×a
= 1
Similarly b + c .q
(a b c )
(a b c )
0
abc
)
= 0
(a + b ) .p + (b + c ) .q + (c + a ) .r
= 1+1+1=3
Vectors
Mahesh Tutorials Science
24
Q-13) If the vectors
ai + j + k ; i + b j + k ; i + ck a ≠ b ≠ c ≠ 1
are coplanar then show that
1
1
1
+
+
= 1.
1–a 1– b 1– c
Ans. Let
p = ai + j + k ,
c 
1
1

+
1+
+
1
–
c
1
–
b
1
–
a


= 1 ⇒
1– c +c
1
1
+
+
1–c
1–b 1–a
= 1
1
1
1
+
+
1–c 1–b 1–a
= 1 ⇒
1
1
1
+
+
1–a 1–b 1–c
= 1
q = i + bj + k ,
Q-14) Show that if four points A, B, C D with
r = i + j + ck
position vectors a , b , c , d are coplanar then
and p, q , r are coplanar.
∴
bcd  + cad  + abd  = abc 

 
 
 

[ pqr ] = 0
Ans. ∵ four points A, B, C, D are coplanar
a 1 1
∴
[ pqr ] =
1 b 1
1 1 c
Applying C1 – C2 and C2 – C3
a −1
0
1−c
∴
∴
0
1
= 0
b −1 1
1−c
c
AB AC AD are also coplanar
∴
 AB AC AD 


∴
AB . AC × AD
(
= 0
)
= 0
(b – a ) . (c – a ) × (d – a )
= 0
(b – a ) . c × d – c × a – a × d + a × a 
by taking (a – 1), (b – 1), (1 – c) common
(a – 1)(b – 1)(1 – c)
=
1
1 0
a –1
1
0 1
b –1
c
1 1
1–c
= 0 ⇒
∴
a b ≠ c ≠1
∴
(a –1) (b – 1) (c – 1) ≠ 0 ⇒
1
1 0
a –1
1
0 1
b –1
c
1 1
1–c
∴
= 0
(
= 0
) ( ) ( )
=
– a . (c × d ) + a . (c × a ) – a . (a × d )
b. c × d – b . c × a – b . a × d
0
a × a = 0 


b c d  – b c a  – b a d  – a c d 

 
 
 

+ a c a  + a a d 
= 0
b c d  – a b c  – a b d  – c a d  + 0 + 0

 
 
 

= 0
b c d  + a b d  + c a d  = a b c 

 
 
 

GROUP (C)- HOME WORK PROBLEMS
= 0
Q-1) Find a b c 
c
1
1
–
–
1– c b –1 a –1
= 0⇒
i)
; b = 10i + 3k
;
a = 2i – 2 j + 3k
c = i – j + 2k
c
1
1
+
+
1 – c b –1 1 – a
= 0
ii)
a = i + j , b = j – k and c = k + i
Vectors
Mahesh Tutorials Science
2i – 2 j + 3k
Ans. i) a =
25
Ans. i) a =
j – 2 j + k
b = 10i + 3k
b = 3 j – 2 j + k
c = i + j + 2k
c =
2
a b c 


ii) ∴ a b c 
j + j + 5k
−2 3
1 –2 –1
10 0 3
=
1 −1 2
a b c 


=
3
1
2
1
1
5
= 2 (0 + 3) + 2 (20 – 3) + 3 (–10 – 0)
= 1 (10 – 1) + 2 (15 – 1) + 1 (3 – 2)
= 6 + 34 – 30
= 9 + 28 + 1
= 10
= 38
3 2 1


= 1 1 2
3 −1 2
a b c 


ii)
=
1 0 2
2 1 1
2 3 3
= 1(1) –1(–1)
= 1 (3 – 2) – 0 (6 – 2) + 2 (4 – 2)
=2
=1–0+4
=5
Q-2) Find ‘p’ if a b c  = 0 where,
Q-4) If A ≡ (1, 1, 1), B ≡ (2, 1, 3), C ≡ (3, 2, 2) and
, b = i – j + k
and
a = i + j + k
D ≡ (3, 2, 4), find the volume of the
parallelopiped with AB, AC and AD as
c = 2i + 3 j + pk
Ans. i) a = i + j + k
concurrent edges.
Ans. A ≡ (1, 1, 1), B = (2, 1, 3), C ≡ (3, 2, 2),
D ≡ (3, 2, 4)
b = i – j + k
let a , b, c , d be position vectors of points A,
c = 2i + 3 j + pk
a b c 


B, C, D respectively with respect to a fixed
point.
= 0
∴
1
1
1
1 –1 1
2 3 p
= 0
1 (–P – 3) – 1 (P – 2) + 1 (3 + 2)
–P–3–P+2+5=0
– 2 P = –4
a
= i + j + k
b
= 2i + j + 3k
c
= 3i + 2k + 2k
d
= 3i + 2 j – 4k
AB
= b –a
P=2
=
= i + 2k
Q-3) Find the volume of parallelopiped whose
co-terminus edges are
i)
,
, b = 3i + 2 j + k
a = i – 2 j – k
c = i + j + 5k
, b = 2i + j + k
,
ii) a = i + 2k
c = 2i + 2 j + 3k
(2i + j + 3k ) – (i + j + k )
AC
= c –a
=
(3i + 2j – 2k ) – (i + j + k )
= 2i + j + k
vol. of parallelopiped
=
 AB AC AD 


Vectors
Mahesh Tutorials Science
26
(a b c )
1 0 2
=
2 1 1
= 0
so,
2 1 3
vectors are co-planar
= 1 (3 – 1) – 0 (6 – 2) + 2 (2 - 2)
ii) To show that, vectors are co-planar we
=2–0+0
have to prove that
=2
(a b c )
Q-5) If the volume of the parallelopiped whose
coterminus edges are
(a b c )
, 2i + j – k
, – i + 3 j + 2k
–3i + 2 j + nk
a
= 0
b
= 2i + j – k
(a b c )
c
=
–3 2
2
–1 3
–i + 3k – 2k
= 7
= 0
vectors are co-planar
∴
, b = i – j + k
Q-7) If a = i + j + k
(
find a . b × c
and c = 2i + 3 j + 2k
n
1 –1
=
= 3 (12 – 12) – 1 (4 – 4) + (6 – 6)
= –3i + 2 j + nk
(a b c )
3 1 3
1 3 2
2 6 4
is 7 find n.
Ans. Let
= 0
=
7
= i + j + k
Ans. a
2
–3 (2 + 3) – 2 (4 – 1) + n (6 + 1) = 7
–5 – 6 + 6n + n = 7
)
b
= i – j + k
c
= 2i + 3 j + 2k
7n = 7 + 21
i j k
1 –1 1
2 3 2
7n = 28
b ×c =
n=4
Q -6) Show that the following vectors are
coplanar.
i)
, i + 3 j – 4k
, – i + 4 j – 3k
3i – 5 j + 2k
ii)
, i + 3 j + 2k
, 2i + 6 j + 4k
3i + j + 3k
Ans. i) To show that, vectors are co-planar we
have to prove that
(a b c )
(a b c )
3 –4
= 3 (–16 + 9) + 5 (4 + 3) + 2 (– 3 – 4)
= –3 (7) + (7) + 2 (–7)
= –21 + 35 – 14
=0
Vectors
(
a. b ×c
)
=
(i + j + k ) . ( –5i + 5k )
= (1) (–5) + (1) (0) + (1) (5)
=0
–3 5 2
–1 4 –3
1
–5i + 5k
=
= –5 + 0 + 5
= 0
=
= i ( –2 – 3 ) – j ( 2 – 2 ) + k ( 3 )
∴ a , b, c are caplaner.
Q-8) Find ‘p’ if the following vectors are coplanar
, b = 2i – 5 j + pk
, c = i – j – 6k
a = i – 3 j + 4k
Ans. a b c  = 0
Mahesh Tutorials Science
27
∴ 2k + 26 – 6k – 30 + ( –12 ) = 0
1 –3 4
2 –5 P
1
= 0
∴ –4k – 16 = 0
–1 6
1 (30 + P) + 3 (–12 – P) + 4 (–2 + 5) = 0
30 + P – 36 – 3P + 12 = 0
∴ k = – 4
, b = i + j + 2 and
Q-10) If a = 3i + 2 j + k
–2P + 6 = 0
are the co-terminus edges of
c = 3i – j + 2k
P=3
a tetrahedron. Find its volume.
Q-9) Find x is A (3, 2, -1), B (5, 4, 2), C (6, 3, 5),
D (1, 0, x) are coplanar.
Ans. Let a, b, c , d be position vectors of pts.
A, B, C, D respectively w.r.t a fixed point.
a = 3i + 2 j – k
a b c 


=
3
2
1
1
1
2
3 −1 2
= 3(4) – 2(– 4) + 1 (– 4)
b = 5i + 4 j + 2k
= 12 + 8 – 4
c = 6i + 3 j + 5k
= 16
d = i + kx
=
1
a b c 

6
volume of a Tetrahedron =
= b –a
AB
(5i + 4 j + 2k ) – (3i + 2j – k )
16
6
=
= 2i + 2 j + 3k
8
3
volume of a Tetrahedron =
= c –a
AC
=
(6i + 3 j + 5k ) – (3 j + 2j – k )
Q-11) If c = 3a + 2b , show that abc = 0 .
If angle between a and b is π 4 prove
that
= 3i + j + 6k
= d –a
AD
1
a b c 

6
Ans. volume of a Tetrahedron =
=
(
) (
=
–2 j – 2 j + k + 1 x
– 3i + 2 j – k
i + kx
)
c = 9a 2 + 6 2ab + 4b 2 a = a and b = b
Ans. We know that,
∴
(
)
points A, B, C, D are coplanar
AB , AC , AD will also be coplanar
 AB AC AD 


= 0
b ×b = 0
If in a scalar triple product, two vectors are
equal, then the secalar triple product is zero.
(
a b c  = a b × c


)
= a b × 3a + 2b 
(
2
2
3
3
1
6
–2 –2
= 0
(k +1)
∴ 2  k + 1 + 12 + 2 3k + 3 + 12 + 3 ( –6 + 2) = 0
(
)
)
(
= a 3b × a + 2b × b
(
= a 3b × a + 0
(
= 3a b × a
)
)
)
Vectors
Mahesh Tutorials Science
28
=
( 3a – 2b ) . ( 3a – 2b )
π
4
=
9a.a – 6a.b – 6b.a + 4b.b
c2 = C.C
=
9a.a – 12a .b + 4b.b + ∵ a .b = b.a
The measure of angle between a & b
is
=
(3a + 2b ) . (3a + 2b )
=
9a.a + 6a.b + 6b.a + 4b.b
=
9a .a +12a.b + 4b.b + ∵ a.b = b.a
(
(∵ a = a & b = b )
c =
(∵ a = a & b = b )
)
π
2
= 9a +12ab cos   + 4b
4
2
 1 
2
9a 2 – 12ab 
 + 4b
 2
=
9a 2 – 6 2ab + 4b 2
c2 =
a = a&b
 1 
2
9a +12ab 
 + 4b
2


and
= b
2
2
9a + 6 2ab + 4b
a = a&b
2
, v = 3i + k
and w = j – k
Q -13) If u = i – 2 j + k
and
= b
Find (u + w ).[(u × v ) × (v × w )]
Ans. u = i – 2 j + k
v = 3i + k
Q-12) If c = 3a – 2b , show that abc = 0 .
Also find c 2 when the measure of the angle
between a and b is π 4 .
w=
j – k
(u + w )
let a =
= i – j
Ans. We know that,
If in a scalar triple product, two vectors are
equal, then the scalar triple product is zero.
abc 


(u × v )
b =
b ×b = 0
(
= a b ×c
= i + j + k
i
)
(
)
(
= a 3b × a – 2b × b
(
= a 3b × a – 0
(
= 3a b × a
)
)
)
π
is
4
k
= i (1) + k (1)
(v × w )
c =
i j
3 0
=
=3×0=0
The measure of angle between a & b
j
1 –2 1
0 1 –1
=
= a b × 3a – 2b 
k
1
0 1 –1
= i ( –1) – j ( –3) + k ( 3 )
–i + 3 j + 3k
=
2
c = c.c
Vectors
)
π
2
2
= 9a – 12ab cos   + 4b
4
 
2
=
(
To find u + w .  u × v × v × w 


(
) (
) (
)
Mahesh Tutorials Science
29
≡ a. b × c 
a b c 


=
1
=
(
–a . c × d
1 0
)
) (
) (
a d c 


=
c a d 


Also, b. c × d =
b c d 


(
)
)
(
–b. c × a
Q -14) If four points with position vectors a , b , c
and d are coplanar, prove that
(
)
(
)
(
) (
a. b × c = b. c × d + c. a × d a . b × d
a b d 


=
=–4
(
=
(
= 1(3 – 3) –1(3 + 1)
∴
b d a 


= a. d × c
1 1 1
–1 3 3
u + w . u ×v × v × w  = – 4


=
)
=
)
–a . b × c a b c 
(
)
∴
from (I)
∴
b c d  + c a d  + a b d  = a b c 

 
 
 

)
(
)
a . b × c = 1 , w here
Q-15 ) I f
Ans. i.e. To prove that
,
a = i + j + k
and c = i – j + 4k
,
b = 2i + q j + k
b c d  + c a d  + a b d  = a b c 

 
 
 

find q
Let, a,b,c and d be the position of the points
Ans. Given that
A, B, C and D respectively. Then,
AB
= b –a,
AC
= c –a ,
AD
= d –a
(
a. b × c
)
=1
∴
1 1 1 
2 q 1 

 =1
1 −1 4
∴
1(4q + 1) – 1 (8 – 1) + 1 ( – 2 – 1) = 1
∴
4q + 1 – 7 – 2 – q = 1
∴
3q – 8 = 1
∴
3q = 9
∴
q = 3.
The points A, B, C, D are coplanar.
∴
the vectors AB , AC , AD are coplanar.
∴
the  AB AC AD  = 0
∴
b – a c – a d – a  = 0


∴
∴
(b – a ) . (c – a ) × (d – a ) = 0
Q-16) Show that, if a , b , c are non-coplanar
(b – a ) . (c × d – c × a – a × d – a × a ) = 0
vectors, then the vectors 3a + b + 5c and
2a – 4b + 3c are non-coplanar.
where a × a = 0
∴
(b – a ) . (c × d – c × a – a × d ) = 0
∴
b. c × d – b. c × d – b. a × d –
Ans. Let
p = a – 2c , q = 3a + b + 5c ,
r = 2a – 4b + 3c
(
)
(
)
(
)
(
)
(
)
(
)
a. c × d + a. c × a + a. a × d = 0 ...(I)
(
)
(
∴
)
Now, a. c × d = 0, a. a × d = 0,
(
–b. a × d
)
(
= b. d × a
1 0 –2
 pqr  = 3 1 5


2 –4 3
= 1(3 + 20) – 0(9 – 10) – 2(–12 – 2)
= 23 + 28 ≠ 0
)
∴
p, q , r are non-coplanar.
Vectors
Mahesh Tutorials Science
30
Q-17) prove that
=
i) b + c  c + a × a + b  = 2a . b × c
(
) (
)
(
)
(
ii) a b + c a + b + c  = 0
iii) a + 2b – c  a – b × a – b – c 



= 3 a b c 
)
(
) (
)
=
a b c  + 2 a b c 




= 3 a b c 
=
(b c a ) + (c a b )
= RHS
=
(a b c ) + (a b c )
(
)
= 2a. b × c
)
= R.H.S.
ii) LHS =
a b + c a + b + c 


= a  b + c × a + b + c 
(
) (
)
b × a + b × b + b × c + c × a 

= a. 

+c ×b +c ×c


(
)
(
)
(
= a. b × a + a. b × b + a . b × c
(
)
(
)
)
(
+ a. c × a + a . c × b + a. c × c
(
)
+0
(
)
+0
= 0 + 0 + a. b × c
– a. b × c
= 0
= RHS
iii) LHS
= a. a + 2b – c . a – b × a – b – c 
(
Vectors
a b c  + b c a 

 

b + c c + a a + a 


(
)
(
(
=
=
= 2 a bc
)
)
= 0 + a. b × c + 2b. c × a + 2 × 0 – 0 – 0
Ans. i) LHS = b + c .  c + a × a + b 
(
) ( ) ( )
+ 2b. (b × c ) – c . (c × a ) – c . (b × c )
a. c × a + a. b × c + 2b. c × a
=
(
(a + 2b – c ) . c × a + b × c 
)
a × a – a × b – a × c



=
(a + 2b – c ) .  – b × a + b × b + b × c 
=
(a + 2b – c ) . 
0 – a × b + c × a + a × b 

– o + b ×c


)
Mahesh Tutorials Science
31
GROUP (D)- CLASS WORK PROBLEMS
the medians of a triangle is called the
centroid of the triangle.
Q-1) Prove that medians of a triangle are
concurrent
Q-2) Prove that bisectors of angles of a triangle
are concurrent
Ans. Let a , b, c be the position vectors of the
Ans. Let a , b, c be the position vectors of the
vertices A, B, C and d, e, f be the position
vertices A, B, C of ∆ ABC and let the lengths
of the sides BC,CA and AB be x, y, z,
vectors of the midpoints D, E, F of the sides
BC, CA and AB respectively.
respectively. If segments AD, BE, CF are the
bisectors of the angles A, B, C, respectively,
A
then
A
F
G
00
E
F
B
D
E
P
C
x
x
B
Then by the midpoint formula,
D
C
d
=
b +c
,
2
D divides the side BC in the ratio AB : BC
i.e., z : y,
e
=
c +a
,
2
E divides the side AC in the ratio BA : BC
i.e., z : x,
f
a +b
=
2
and F divides the side AB in the ratio AC :
∴
2d
= b +c ;
2e
= c +a ;
2f
BC i.e., y : x,
Hence by section formula, the postion vectors
of the points D, E and F are
= a +b
∴
2d + a
= a +b +c
∴
2e + b
= a +b +c
∴
2f + c
= a +b +c
∴
2d + a
2 +1
=
2e + b
2 +1
= g
=
2f + c
2 +1
=
a +b +c
3
∴
... (Say)
This show that the point G whose position
vector is g , lies on the three medians AD,
BE and CF dividing each of them internally
∴
in the ratio 2 : 1.
Hence, the medians are concurrent in the
po in t G an d i ts p os it ion v e c to r is
(a + b + c ) / 3.
This point of concurrence of
∴
d =
zc + yb
;
z +y
e =
zc + xa
;
z +x
f =
yb + xa
;
y+x
(y + z ) d
= yb + zc ;
(z + x ) e
= zc + xa ;
(x + y ) f
=
( y + z ) d + xa
xa + yb.
=
=
(x + y ) f
=
xa + yb + zc
( z + x ) e + yb
+ zc
(y + z ) d + xa
(y + z ) + x
=
( z + x ) e + yb
(z + x ) + y
Vectors
Mahesh Tutorials Science
32
=
( x + y ) f + zc
(x + y ) + z
=
xa + yb + zc
x +y +z
∴
the mid points of the diagonal AC and
BD are the same. this shows that the
diagonals AC and BD bisect each other.
=
p
ii) conversely, suppose that the diagonals
AC and BD of ABCD bisect each other.
...(Say)
This show that the point P whose position
∴
vector is p , lies on the three bisectors AD,
i.e, they have the same midpoint.
the position vectors of these midpoints
are equal.
BE and CF dividing them in the ratios
(y + z) : x; (z + x) : y and (x + y) : z respectively.
∴
a +c
=
2
Hence, the three bisector segments are
concurrent in the point whose position vector
∴
a +c = b +d
∴
b –a = c –d
xa + yb + zc
.
is
x +y +z
∴
This point of concurrence of the bisectors is
∴
AB = DC
AB= DC and side AB || side DC
ABCD is parallelogram.
∴
called the incentre of the ∆ ABC.
Q-3) Using vector method, prove that the
diagonals of a parallelogram bisect each
other and conversely. OR prove that, A
quadrilateral is a parallelogram if and only
if its diagonals bisect each other.
Q-4) Prove that the median of a trapezium is
parallel to the parallel sides of the
trapezium and its length is half of the sum
of the lengths of the parallel sides.
Ans. Let a, b,c and d be respectively the position
vectors of the vertices A, B, C
Ans. i) Let a, b,c and d be respectively the
position vectors of the vertices A, B, C
and D of the parallelogram ABCD then
∴
Then the vectors AD and BC are parallel.
there exists a scalar k
such that AD = k . BC
D
∴
AD + BC = k .BC + BC
= (k + 1) BC
A
M
C
B
∴
AB
∴
b –a = c –d
∴
a +c = b +d
∴
a +c
b +d
=
2
2
M
B
D
N
(b + d ) by (i) they are equal
C
Let m and n be the position vectors of
the midpoints M and N of the nonparallel sides AB and DC respectively.
... (i)
the diagonal AC and BD are
Vectors
... (i)
= DC
then seg MN is the median of the trapezium.
by the mid point formula.
the position vectors of the mid points of
2
and D of the
trapezium ABCD with side AD || side BC.
AB = DC and side AB || side DC.
A
b +d
2
(a + c )
2
and
m=
∴
MN
a +b
d +c
and n =
2
2
= n –m
d +c  a +b 
= 
 – 

 2   2 
Mahesh Tutorials Science
∴
33
=
1
d +c – a – b
2
=
1
d – a + c +d 

2
=
AD + BC
2
(
(
)
) (
∴
)
...[(by (i)]
2
∴
MN is a scalar multiple of BC
MN and BC are parallel vectors
∴
MN || BC where BC ||AD
∴
The median MN is parallel to the parallel
to sides AD and BC of the trapezium.
Now, AD and BC are collinear.
∴
∴
–a + r
=
(r + a ) . (r – a )
= r .r – r – a + a.r – a.a
= r
... (ii)
(k +1) BC
=
AP.BP
=
2
2
–a
=0
∴
AP ⊥ BP
∴
APB is a right angle.
Hence, the angle subtended on a semicircle
is the right angle.
Q-6) Using vectors, prove that altitudes of a
triangle are concurrent.
Ans. Let segment AD and CF be the altitudes of
∆ ABC, meeting each other in the point H.
A
AD + BC = AD + BC = AD + BC
F
[ ∵ AD and BC have same direction]
from (2) we have
E
H
∴
AD + BC
MN =
2
∴
1
AD + BC
MN =
2
(
B
)
D
C
Then it is enough to prove that HB is
perpendicular to AC .
Q-5) Using vectors, prove angle subtended in a
semi circle is right angle.
Choose H as the origin and let a , b, c be the
position vectors of the vertices A, B and C
respectively w.r.t. the origin H.
Ans. Let seg AB be a diameter of a circle with
centre C and P be any point on the circle
other than A and B.Then ∠ APB is an angle
Then HA = a , HB = b and HC = c ,
subtended on a semi-circle.
AB = b – a , BC = c – b and AC = c – a
Let AC = CB = a and CP = r
Then a = r
Now HA is perpendicular to BC
...(1)
∴
HA. BC = 0
∴
a. c – b = 0
∴
a.c – a.b = 0
P
(
)
r
A
a
C
–a
B
...(1)
Also HA is perpendicular to AB .
AP
=
AC + CP
= a +r
∴
HC . AB = 0
∴
c. b – a = 0
∴
c .b – c .a = 0
∴
c .b – a.c = 0
(
)
= r +a
BP
= BC + CP
=
(∵ c .a = a.c )
...(2)
–CB + CP
Vectors
Mahesh Tutorials Science
34
Adding (1) and (2), we get
z =
2
02 + ( –4 ) + 32
c .b – a.c = 0
∴
(
∴
b. c – a = 0
∴
HB.AC = 0
∴
HB is perpendicular to AC .
∴
)
c – a .b = 0
(
)
=
16 + 9
=
25 = 5
( )
If H h is the incentre of ∆ PQR, then
h =
the altitudes of ∆ ABC are concurrent.
xa + yb + zc
x +y +z
( ) ( ) (
4 4 j + 3 3k + 5 4 j + 3k
h =
Q-7) Using vector method. find the in centre of
a triangle is right angle. are A(0,4,0),
B(0,0,3) and C(0,4,3).
h =
)
4+3+5
16 j + 9k + 20 j +15k
12
Ans. The position vectors a , b, c of the vertices A,B
& C are
h =
36 j + 24k
12
a = 0i + 4 j + 0k
0 j + 36 j + 24k
12
b = 0i + 0 j + 3k
h =
c = 0i + 4 j + 3k
h = 0i + 3 j + 2k
AB
= b –a
=
H = (0,3,2)
(0i + 0 j + 3k ) – ( 0i + 4 j + 0k )
Q-8) Prove by vector method, sin(α + β ) =
sin α .cos β + cos α . sin β .
= 0i – 4 j = 3k
BC
Ans.
= c –b
=
Y
Q
B
(0i + 4 j + 3k ) – ( 0i + 0 j + 3k )
= 0i + 4 j + 0k
AC
O
= c –a
=
β
–2
X
(0i + 4 j + 3k ) – ( 0i + 4 j + 0k )
A
P
= 0i – 0 j + 3k
Let ∠XOP and ∠XOQ be in standard position
Let
x =
BC ,
and mXOP = –α, mXOQ = β .
y =
AC , and
Takes a pouint A on ray OP and a point B on
ray OQ such that OA = OB = 1.
z
Since cos (–α) = cos α
= AB
and sin (–α)
x =
=
y =
2
2
0 +4 +0
2
A is (cos (–α), sin (–α)),
16 = 4
i.e, (cos α, –sin α)
B is (cos β , sin β )
2
02 + ( –0 ) + 32
∴
=
Vectors
9
= 3
= –sin α,
OA
=
( cos α ) i – (sin α ) . j + 0.k
Mahesh Tutorials Science
35
These are equal.Hence the diagonals OC and
= OB ( cos β ) i + ( sin β ) . j + 0.k
OB
BA have the same midpoint.
∴
OA × OB
=
i
j
k
cos α
cos β
– sin α
sin β
0
0
∴
these diagonals bisect each other ...(1)
Now OC = a + b
and BA
=
= BC + CA
(cos α sin β + sin α cos β ) k
... (i)
= a –b
∴
=
OC .BA
(a – b ) . (a – b )
The angle between OA and OB is α + β .
= a .a – a.b + b.a – b.b
Also OA , OB lie in the XY-plane.
∴
= a2 – b2
the unit vector perpendicular to OA and OB
is k .
∴
(∵ a = b )
= 0
OA × OB
= [OA.OB sin (α + β )] k
= sin (α + β ). k
∴
OC ⊥ BA i.e., the diagonals OC and BA are
at right angles.
...(2)
∴
from (1) and (2), the diagonals of a rhombus
...(ii)
from (i) and (ii),
(∵ a.b = b.a )
bisect each other at right angles.
sin (α + β ) = sin α cos β + cos α sin β .
Conversely : Let the diagonals OC and BA of
the quadrilateral.OACB bisect each other at
GROUP (D) – HOME WORK PROBLEMS
right angles.
Since the diagonals bisect each other, the
Q-1) By vectors method. prove that th e
quadrilateral OACB is a parallelogram.
diagonals of rhombus are perpendicular to
each other.
Ans. Let
OACB be a rhombus and OA = a ,
OB = b . Then OA = OB
a
B
C
a+
b
Now,
∴
OC ⊥ BA
∴
OC . BA = 0
∴
(a + b ) . (a – b ) = 0
∴
a.a – a.b + b.a – b.b = 0
∴
a
∴
a
∴
a
b
a
O
a
b
–b
A
2
2
2
2
– a.b + b.a – b = 0
–b
2
=0
2
=b =0
∴
a =b
∴
a = b.
∴
a =b
Now OC = a + b .
∴
l (OA ) = l (OB )
∴
the position vectors of the vertices O, A, B
and C w.r.t. the origin O are 0, a , b and a + b
respectively.
By the midpoint formula, the position vectors
of the midpoints of the diagonals. BA and OC
i.e ., adjacent side s OA and OB of the
parallelogram OABC are equal.
∴
OABC is a rhombus.
Hence, a quadrilateral is rhombus if and only
if diagonals bisect each other at right angle.
(a + b ) / 2 are and (a + b + 0 ) i.e., (a + b ) / 2 .
Vectors
Mahesh Tutorials Science
36
Q-2) Using vectors prove that, if the diagonals
a.a + a.b + b.a + b.b
of a parallelogram are at right angles then
it is rhombus.
= b.b – b.a – a.b + a.a
a 2 + b 2 + 2a .b
Ans. Let ABCD be a parallelogram
Let
and
=
AC
BD
= b 2 – 2a.b + a 2
AB + BC
= a +b
2a.b + 2a.b
= BA + AD
= 0 ⇒ 4 a.b = 0 ⇒ a.b = 0
=
( )
–a + b
∴ the diagonals AC and BD are perpendicular
∴ AC .BD = 0
⇒
(a + b )( –a + b ) =
a ⊥ b ⇒ AB ⊥ AD ⇒ AB ⊥ AD
∴
The parallelogram ABCD is a rectangle.
Q-4) Show by vector method that the sum of the
square of the diagonals of a parallelogram
0
is equal to the sum of the sequare of its
sides.
–a.a + a.b – b.a + b.b = 0
b2 = a2 ⇒ b = a
Ans. Let ABCD be a parallelogram.
⇒
l(AB) = l(BC) = l(CD) = l(AD)
∴
the parallelogram ABCD is a rhombus .
p
C
Q-3) Show by vector method. If the diagonal of
q
a parallelogram are congruent then it is a
rectangle.
Ans. Let ABCD be parallelogram
q
A
Let AB = p and AD = q
From ∆ ABC, by triangle law,
AC
=
Since ABCD is a parallelogram,
AB + BC
DC = AB = p and BC = AD = q
= a +b
AC = AB + BC = p + q
From ∆ ABC, by triangle law,
BD
B
p
Let AB = a , AD = b
= BC + CD
∴
(
)(
AC 2 = AC .AC = p + q . p + q
= b –a
= p.p + p.q + q. p + q .q
a
D
= p2 + q2 + 2p.q
C
b
∴
= BC + CD
= BC – DC
BD2
= BD.BD
=
= p2 + q2 + 2p.q
We have to prove that ABCD is a rectangle
i.e. to prove that a ⊥ b
∴
Since the diagonals AC and BD are congruent
AC = BD ⇒ AC
2
= BD
2
AC 2 + BD
2
=
Vectors
( q – p ) . (q – p )
(∵ p.q = q.p )
( p2 + q 2 + 2 p.q )
+ ( p 2 + q 2 – 2 p.q )
= p2 + q2 + p2 + q2
(a + b ) . (a + b ) = (b – a ) . (b – a )
= q–p
= q.q – q. p – p.q + p. p
B
a
(∵ p.q = q.p )
BD
b
A
)
Mahesh Tutorials Science
37
= AB2 + BC 2 + CD 2+ DA 2
(i + j – 3k ) – (i – 2 j + k )
=
This show that the sum of the squares of the
diagoanals of a parallelogram is equal to the
sum of the squares of its sides.
= 0i + 3 j – 4k
= r –q
QR
Q-5) If A (1, 2, 3); B (3, -1, 5); C (4, 0, -3) are
=
three non-collinear points by using vector
method then show A is point on circle
= 0i – 3 j + Ok
having BC as diameter.
Ans. A ≡ (1, 2, 3), B ≡ (3, −1, 5), C ≡ (4, 0, −3)
(i – 2 j – 3k ) – (i + j – 3k )
= r –p
PR
let a , b, c be position vectors of points A, B
=
(i – 2 j – 3k ) – (i – 2 j + k )
& C respectively w.r.t a fixed point.
= 0i + 0 j – 4k
a = i + 2 j + 3k
Let x =
b = 3i – j + 5k
y =
c = 4i – 3k
AB
= b –a
=
(3i – j + 5k ) – (i + 2 j + 3k )
PQ
x =
02 + ( –3 ) + 02
=
= c –a
=
y =
( 4i – 3k ) – (i + 2 j + 3k )
=
= 3i – 2 j – 6k
z =
AB . AC =
(2i – 3 j + 2k ) . (3i – 2 j – 6k )
( )
(
)
( )
= 2 ( 3 ) i . i + ( –3 )( –2 ) j – j + 2 ( –6 ) k .k
= 6 (1) + 6 (1) − 12 (1)
= 0
PR
z =
= 2i – 3 j + 2k
AC
QR ,
2
9
= 3
2
02 + 02 + ( –4 )
16 = 4
2
02 + 02 + ( –4 )
=
9 +16
=
25 = 5
( )
If H h is the incentre of the triangle PQR,
then
seg AB ⊥ seg AC
i.e. seg BC subtends right angle at A
h =
i.e. seg BC is diameter of a semicircle.
Q-6) Using vector method. find incentre of a
triangle whose vertices are P(1, – 2,1) ;
Q(1, 1, – 3) and R(1, – 2, – 3).
Ans. The position vectors p, q , r of the vertices
P,Q,R are
x p + yq + zr
x +y + z
h
=
3i – 6 j + 3k + 4i + 4 j – 12k + 5i – 10 j – 15k
3+4+5
=
12i – 12 j – 24k
12
p
= i – 2j +k
h = i – j – 2k
q
= i – j – 3k
h = (1, –1, –2)
r
= i – 2 j – 3k
PQ
= q–p
Vectors
Mahesh Tutorials Science
38
Q -7) Using vectors, prove that the perpendicular
bisectors of the sides
concurrent.
of a triangle are
Ans.
A
∴
c +a
. c –a =0
2
∴
as above c2 – b2 = 0
∴
from (1) and (2), we get
(
)
...(2)
b2 = a2
F
B
∴
b2 = a2 = 0
∴
(b + a ) . (b – a ) = 0
∴
(b + a ) . (b – a ) = 0
E
C
D
Let D, E, F be the midpoints of the sides BC,
CA and and AB of ∆ ABC.
Let the perpendicular bisector of the sides
∴
f = OF
is perpendicular bisector of the sides of ∆ ABC
are concurrent.
Choose O as the origin and let a , b, c , d, e and
This point of concurrrent of the perpendicular
f be the position vectors of the points A, B,
bisector of the sides of a triangle is called the
circumcentre of the triangle.
C, D, E, F respectively.
Q-8) Prove by vector meth od, th at, a
perpendicular to AB = b – a .
quadrilateral is a square if and only if
diagonals are congruent and bisect each
By the midpoint formula,
other at right angles.
b +c
,
d =
2
Ans. Let ABCD be a square.
Let a , b, c , d be the position vectors A, B, C,
c +a
,
e =
2
f =
)
f. b –a =0
BC and AC meet each other in the point O.
He re we have to prove that OF = f i s
(
∴
D respectively.
a +b
,
2
D d
C c
A a
B b
Now OD = d is perpendicular to BC = c – b .
(
)
∴
d. c – b = 0
∴
b +c
. c –b =0
2
∴
(c + b ) . (c – b ) = 0
Since ABCD is a sequare, we have,
∴
c .c + b.c – c .b – b.b = 0
AB = DC
∴
c2 – b2 = 0
∴
b –a =c –d
(∵ c.c = c 2 ,b.b = b2and c.b = b.c )
∴
b +d =a +c
c2 – b2
∴
b +d a +c
=
= p ... (Say)
2
2
∴
(
)
...(1)
Also, OE = e is pcpendicular to AC = c – a
∴
(
)
e. c – a = 0
This shows that the point P whose position
vector is p is the midpoint of BD as well as
of AC.
Vectors
Mahesh Tutorials Science
∴
∴
39
the diagonals BD and AC bisect each other
Since the diagonals bisect each other, the
at P.
quadrilateral ABCD is a parallelogram.
the diagonals of a square bisect each other
Now, AC ⊥r BD = 0
...(i)
Now, AC = BC = BC + AB
and BD = BC + CD = BC + BA ... ∵ CD = BA 
∴
AC .BD = 0
∴
( BC + AB ) . ( BC – AB ) = 0
∴
BC .BC – BC .AB + AB .BC – AB .AB = 0
∴
BC
∴
BC
∴
BC
∴
BC = AB
= BC – AB
∴
(BC + AB ) . ( BC – AB )
AC .BD =
= BC .BC – BC .AB + AB .BC – AB .AB
2
=
BC
– AB .BC + AB .BC – AB
=
l ( BC )  – l ( AB ) 
2
2
2
... ∵ l ( AB ) = l ( BC ) 
=0
∴
AC is perpendicular to BD
∴
diagonals AC and BD are at right angles
2
=
( BC )
( )
2
( )
+ AB
and BD.BD
=
2
... ∵ AB ⊥r BC 


2
∴
∴
∴
=
BC
=
( BC )
( )
– 0 – 0 + AB
2
( )
+ AB
2
ABCD is a rhombus.
AC = BD
∴
AC
∴
AC .AC = BD.BD
∴
( BC .AB ) . ( BC + AB )
AC
= BD
=
∴
2
= BD
2
( BC – AB ) . (BC – AB )
BC .BC + BC .AB + AB .BC + AB .AB
= BC .BC – BC .AB – AB .BC + AB .AB
... ∵ AB ⊥ BC 


r
∴
AB .BC + AB .BC
=
2
AC .AC = BD.BD
2
2
∴
(BC – AB ) . ( BC – AB )
= BC.BC – BC .AB – AB .BC + AB .AB
2
=0
Also, l(AC) = l(BD)
(BC + AB ) . ( BC + AB )
+ 0 + 0 + AB
= AB
2
=0
l(BC) = l(AB)
= BC.BC + BC .AB + AB .BC + AB .AB
BC
2
– AB
2
parallelogram ABCD are equal.
... (ii)
=
2
– .AB .BC + AB .BC – AB
i.e., the adjacent sides AB and BC of the
∴
Also, AC .AC =
2
2
∴
l(AC) = l(BD)
∴
from (i), (ii) and (iii), the diagonals of a square
are congruent and bisect each other at right
...(iii)
– AB .BC – AB .BC
∴
4AB .BC = 0
∴
AB .BC = 0
∴
AB ⊥r BC
i.e., the adjacent sides of a rhombus ABCD
are perpendicular to each other.
Hence, ABCD is a square.
angles.
Conversely : Let the diagonals AC and BD of
the quadrilaeral ABCD are congruent and
bisect each other at right angles.
Vectors
Mahesh Tutorials Science
40
Q-9) Using vector method, find the incentre of
a triangle whose vertices are P(0,2,1),
P,Q,R are
∴
p
= 2 j + k , q = 2k and
r
=
PQ
= q–p
∴
 3 1 
H =  – , ,1
 2 2 
Q-10)Prove that the segment joining the midpoints of the diagonals of a trapezium is
parallel to the parallel sides and equals half
its difference.
–2i + 2k
(
=
–2i – 2 j + 2k
=
–2i – 2 j – k
Ans. Let M and N be the midpoints of the diagonals
AC and BD respectively of the trapezium
ABCD in which side AD|| side BC.
)
A
= 2k
B
∴
( –2i + 2k ) – (2i + k )
=
–2i – 2 j + k
∴
PR
z =
PQ
and
2
m =
∴
x =
0+0+2
y =
( –2)2 + ( –2)2 +12
= 3
and y =
( –2)2 + ( –2)2 +12
= 3
= 2
( )
If H h is the incentre of the triangle PQR, then
x p + yq + zr
x +y +z
=
h
=
(
) ( ) (
2 2 j + k + 3 –2i + 3 –2i + 2k
... (i)
)
=
1 4 j + 2k – 6i – 6i + 6k
8
=
1
–12i + 4 j + 8k
8
(
)
b +d
a +c
and n =
2
2
MN = n – m =
b +d a +c
–
2
2
=
1
d –a – c –b 

2
=
1
AD – BC
2
(
(
) (
)
)
=
k –1
BC
2
... [By (1)]... (2)
Thus MN is a scalar multiple of BC .
∴
MN and BC are collinear vectors, i.e.,
parallel vectors.
∴
MN BC , where BC AD .
∴
seg MN is parallel to the parallel sides AD
and BC of the trapezium.
2+3+3
(
AD – BC = k .BC – BC = (k – 1) BC
Let a , b, c , d ,m and n be the position vectors
of the points A, B, C, D, M and N respectively.
Since M and N are the midpoints of the
digonals AC and BD, by the midpoint formula,
QR ,
y =
there exists a non-zero scalar k, such that
AD = k .BC
=
Let x =
C
Then the vectors AD and BC are collinear.
and
= r –p
PR
M
N
( –2i + 2k ) – ( –2i )
=
Vectors
D
= r –p
QR
∴
3 1 i + j +k
2
2
h =
Q(– 2,0,0) and R(–2,0,2).
Ans. The position vectors p, q , r of the vertices
–
∴
Now, AD and BC are collinear vectors.
)
∴
∴
AD – BC = AD – BC = AD – BC
from (ii) we have,
Mahesh Tutorials Science
1
MN = 2 AD – BC
(
41
From (iii) and (iv), we get,
)
AC . AC = BD . BD
1
1
AD – BC =
AD – BC .
2
2
2
∴
AC
Q-11) If the lengths of two non-parallel sides of
∴
AC = BD
a trapezium are equal then, prove by vector
method that lengths of their diagonals are
∴
∴
MN =
equal.
Ans.
2
l (AC) = l (BD)
Hence, the diagonal of the tranpezium are
euqal.
A
B
M
N
D
π –D
= BD
C
π –C
Let ABCD be a trapenzium such that AB is
parallel to CD and l (AD) = l (BC)
i.e., AD = BC ...(i)
Let AM and BN be the perpendiculars from
A and B to DC respectively.
∴
∴
∆ AMD ≅ ∆ BNC
∠C ...(ii)
∠D =∠
Now, AC = AD + DC and BD = BC + CD
∴
(
)(
AC . AC = AD + DC . AD + DC
)
= AD . AD + AD . DC + DC . AD + DC . DC
2
= AD + 2AD . DC + DC
2
(∵ AD . DC = DC . AD )
2
= AD + 2 AD . DC cos ( π – D ) + DC
2
2
2
= AD – 2 AD . CD cos D + DC ...(iii)
(
)(
and BD . BD = BC + CD . BC + CD
2
= BC + 2BC .CD + CD
)
2
(∵ BC .CD = CD . BC )
2
= BC + 2 BC CD cos ( π – C ) + CD
2
= BC – 2 BC CD cos C + CD
2
2
2
2
= AD – 2 AD . CD cos D + CD ...(iv)
[By (i) and (ii)]
Vectors