ICSE Coaching Classes for Std. IX & X in Mumbai, Navi ...

GROUP (A)- CLASS WORK PROBLEMS

Q-1) If a ,b ,c are the position vectors of the

points A, B, C respectively Such that

3 5 8a b c+ =+ =+ =+ = , then find the ratio in which

i) C divides AB, ii) A divides BC

Ans. i) We are given that

3 +5 = 8a b c

∴∴∴∴3 +5 5 +3

= =8 5 + 3

a b b ac

∴∴∴∴ C divides AB internally in the ratio 5 : 3

ii) We are given that

3 +5 = 8a b c

∴∴∴∴ 3 = 8 – 5a c b

∴∴∴∴8 – 5 8 – 5

= =3 8 – 3

c b c ba

∴∴∴∴ A divides BC externally in the ratio 8 : 5.

Q-2) Show that the points whose position

vectors are 5 + 6 + 7a b c , 7 + 6 + 9a b c and

3 +6 +5a b c are collinear.

Ans. Let p = 5 + 6 +7a b c ,

q = 7 + 6 + 9a b c ,

r = 3 + 6 +5a b c ,

PQ = –q p

= ( ) ( )7 + 6 + 9 – 5 + 6 + 7a b c a b c

= 2 +2a c ... (i)

QR = –r q

= ( ) ( )3 + 6 +5 – 7 + 6 + 9a b c a b c

= 4 – 4a c

= ( )–2 2 + 2a c

= –2PQ [using i]

i.e. QR is scalar multiple of PQ

∴ PQ & PR 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

+ + + = 4AB AD CB CD MN .

Ans. Let , , , , ,a b c d m n be the position vectors of

vertices , , , , ,A B C D M N respectively.

Since M and N are midpoints of diagonal

AC and BD respectively.

m =+

2

a c,

n =+

2

b d

2m = +a c & 2n = +b d …(i)

Now + + +AB AD CB CD

= – + – + – + –b a d a b c d c

= 2 +2 – 2 – 2b d a c

= ( ) ( )2 + – 2 +b d a c (using i)

= ( ) ( )2 2 – 2 2n m

= 4 – 4n m

= ( )4 –n m

= 4MN

∴∴∴∴ + + +AB AD CB CD

= 4MN

Q-4) G and G′′′′are centroids of ∆∆∆∆ABC and ∆∆∆∆A′′′′B′′′′C′′′′.

Show + + = 3AA BB CC GG′ ′ ′ ′′ ′ ′ ′′ ′ ′ ′′ ′ ′ ′ .

Ans. Let , , , ', ', ',a b c a b c g and 'g are the position

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′

Vectors

Vectors

2 Mahesh Tutorials Science

∴∴∴∴ g =+ +

3

a b c

∴∴∴∴ 3g = + +a b c ...[i] And

∴∴∴∴ 'g ='+ '+ '

3

a b c

∴∴∴∴ 3 'g = '+ '+ 'a b c ...[ii]

L.H.S. = '+ '+ 'AA BB CC

= ( ) ( ) ( )'– + '– + '–a a b b c c

= ( ) ( )'+ '+ ' – + +a b c a b c

= 3 '– 3g g

= 3 'GG

Q-5) D, E, F, are mid - point of sides BC, CA, AB,

respectively of ∆∆∆∆ABC, show

i) + + =OE OF DO OA

ii) 2 1 1

+ + =3 3 2

AD BE CF AC

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.

∴∴∴∴ Using mid-point formula,

d =+

2

b c...(i);

e =+

2

c a...(ii)

f =+

2

a b...(iii)

(i) L.H.S. = + +OE OF DO

= + –OE OF OD

= + –e f d

AA

B C

EF

D

=+ + +

+ –2 2 2

c a a b b c

...[from (i), (ii) and (iii)]

=1

+ + + – –2

c a a b b c

= ( )12

2a = a

= OA = R.H.S.

Hence proved.

(ii) L.H.S.

=2 1

+ +3 3

AD BE CF

= ( ) ( ) ( )2 1– + – + –

3 3d a e b f c

= + 2 + 1 +

– + – + –2 3 2 3 2

b c c a a ba b c

=+ – 2 2 + – 2

+2 3 2

1 + – 2 +

3 2

b c a c a b

a b c

=( )2 + – 2+ – 2 + – 2

+ +2 6 6

c a bb c a a b c

=3 +3 – 6 + 2 – 4 + – 2

6

b c a a b b c

=1

3 – 36

c a

=1

–2

c a

=1

2AC

= R.H.S. Hence proved

Mahesh Tutorials Science 3

Vectors

Q-6) AB and CD are two chords of a circle

intersecting at right angles in P. Show that

+ + + = 2PA PB PC PD PO where O is centre

of the circle.

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

respect to centre O. Q and R are the mid-

points of AB and CD respectively.

∴∴∴∴ Using mid-point formula,

q =+

2

a b

∴∴∴∴ 2q = +a b ...(i)

r =+

2

c d

∴∴∴∴ 2r = +c d ...(ii)

L.H.S.

= + + +PA PB PC PD

= ( ) ( ) ( ) ( )– + – + – + –a p b p c p d p

= ( ) ( )+ + + – 4a b c d p

= 2 +2 – 4q r p [from (i) and (ii)

L.H.S. = ( )2 + – 2q r p ...(i)

Now, �� OQPR is a rectangle, hence it is also

parallelogram

∴∴∴∴ by parallelogram law

OP = +OQ OR

∴∴∴∴ p = +q r ...(ii)

Substituting (ii) in (i) we get,

A BQ

O

D

R

C

P

= ( )2 – 2P P

= – 2 = 2P PO

Hence proved.

Q-7) If , ,a b c and d are the position vectors of

the points A, B, C and D respectively such

that 2 + 7 = 5 + 4a b c d . Prove that AB and

CD intersect.

Ans. Given 2 + 7a b = 5 + 4c d

∴∴∴∴2 +7

2+7

a b

=5 + 4

5 + 4

c d

= e (say)

Let e is the position vector of E.E .

By section formula,

E divides AB internally in the ratio 7 : 2 and

CD internally in the ratio 4 : 5

∴∴∴∴ 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

i) + + =OA OB OC OH

ii) (((( ))))+ + = 2 + +HA HB HC AO BO CO

Ans. Let , , ,a b c o and h are p.v. of A,B,C,O and H

respectively.

We know that, if O,G and H are collinear then

G divides OH in the ratio 1:2 where G is

centroid and g is its position vector.

By section formula

g =+2

1+2

h o

∴∴∴∴ 3g = + 2h o … (i)

i) L.H.S.

= + +OA OB OC

= ( ) ( ) ( )– + – + –a b b o c o

Vectors


= + + – 3a b c o

=+ +

3 – 3 =3

a b cg o g

∴

= + 2 – 3h o o from equation (i)

= –h o

= OH

= R.H.S. Hence proved

ii) L.H.S.

= + +HA HB HC

= – + – + –a h b h c h

= + + – 3a b c h

= 3 – 3g h

= + 2 – 3h o h from eq. (i)

= 2 – 2o h

= ( )2 –o h

= 2HO

R.H.S. = ( )2 + +AO BO CO … (ii)

= ( ) ( ) ( )2 – + – + –o a o b o c

= ( )2 3 – + +o a b c

= 2 3 – 3o g

= ( )2 3 – + 2o h o

= 2 3 – – 2o h o

= 2 –o h

= 2HO … (iii)

From equations (ii) and (iii)

L.H.S. = R.H.S. Hence proved.

Q-9) ABCD is a quadrilateral M and N are mid -

points of the diagonal AC and BD

respectively. Show that + = 2AD BC MN .

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

∴∴∴∴ M and N are mid points of AB and CD

respectivrly.

m =+

2

a band

n =+

2

c d

2m = +a b and 2n = +c d …(i)

L.H.S. = +AD AC

= – + –d a c b

= ( ) ( )+ – +c d a b

= 2 – 2n m

= ( )2 –n m

= 2MN

Q-10)If +3 +ai j bk is unit vector, prove that

2 2+ + 8 = 0a b

Ans. Let

r = + 3 +ai j bk

⇒⇒⇒⇒ r = 2 2+ + 9a b

But r = 1

∴∴∴∴ 2 2+ + 9a b = 1

∴∴∴∴ 2 2+ + 9a b = 1

∴∴∴∴2 2+ + 8a b = 0

Q-11)If , ,a b c are position vectors of AB and C

where A ≡ ≡ ≡ ≡ (0, 2, –1), B ≡ ≡ ≡ ≡ (0, – 1,3),

C≡ ≡ ≡ ≡ (0, 1,2) Find x and y such that

= +c xa yb

Ans. a = 2 –j k ;

b = – + 3j k ;

c = + 2j k

∴∴∴∴ c = +xa yb

+ 2j k = ( ) ( )2 – + – + 3x j k y j k

+ 2j k = ( ) ( )2 – + – + 3x y j x y k


Vectors

∴∴∴∴ 2x – y = 1 ... (i) and

–x + 3y = 2 ... (ii)

Solving equations (i) and (ii) x = 1 and y = 1

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 = –b a

= – + +5i j k

AC = –c a

= –3 +3 +15i j k

AC = 3AB

–c a = ( )3 –b a ⇒⇒⇒⇒ –c a

= 3 – 3b a ⇒⇒⇒⇒ +2c a

= 3b

∴∴∴∴ b =+2

3

c a

=1. + 2

1+2

c a⇒⇒⇒⇒ B

divides seg AC internally in the ratio 1:2.

∴∴∴∴ 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.

Ans. Let C be ( )1 1 1, , .x y z

Then the postion vectors , , ,a b c g of the points

A, B, C, G are

a = 3 + –i j k ,

b = +5 +2i j k ,

c = 1 1 1+ +x i y j z k ,

g = 3 +3 +i j k ,

Since G is the centroid of the ∆∆∆∆ ABC, by the

centroid formula,

g =+ +

3

a b c

∴∴∴∴ 3g = + +a b c

∴∴∴∴ ( )3 3 + +i j k

= ( ) ( )

( )1 1 1

3 3 + – + +5 +2

+ + +

i j k i j k

x i y j z k

∴∴∴∴ 9 +9 +3i j k

= ( ) ( ) ( )1 1 13+1 – + 1+5 + + –1+2+x i y j z k

∴∴∴∴ 9 +9 +3i j k

= ( ) ( ) ( )1 1 1+ 4 + +6 + +1x i y j z k

By equality of vectors, we have,

9 = x1 + 4, 9 = y

1 + 6 and 3 = z

1 + 1

∴∴∴∴ x1 = 5,y

1 = 3 and z

1 = 2

∴∴∴∴ C(5,3,2)

∴∴∴∴ c = 5 +3 +j j k

Let P be the midpoint of BC.

Then by midpoint formula, we have,

∴∴∴∴ p =+

2

b c, where p is the position vector

of P.

∴∴∴∴ p =( ) ( )+5 +2 5 + 3 + 2

2

i j k i j k

= ( )1

6 + 8 + 4 3 + 4 + 22

i j k i j k

∴∴∴∴ midpoint of the side BC is (3,4,2).

GROUP (A)- HOME WORK PROBLEMS

Q-1) A and B are two points with position

vectors a and b . If point C and D have

position vectors 5 4−−−−a b and 3 5

,8

++++a b

prove A, B, C, D are collinear.

Ans. Let c and d respectively be position vectors

of points C and D

c = 5 – 4a b

d =3 +5

8

a b

c =5 – 4

5 – 4

a b

Vectors


d =3 +5

3+5

a b

⇒⇒⇒⇒ 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

(i, ii)

Q-2) Show that the point P, Q, R with position

vectors – 3 + ,a b c –2 + 3 + 4a b c and

– + 2b c are collinear.

Ans. Let , ,p q r respectively be position vectors of

points P, Q, R w.r.t a fixed point.

p = – 3 +a b c

q = –2 + 3 + 4a b c

r = – +2b c

PQ = –q p

= ( ) ( )–2 + 3 + 4 – – 3 +a b c a b c

PQ = –3 + 6 +3a b c

PQ = ( )–3 – 2 –a b c

–3

PQ= – 2 –a b c ...(i)

QR = –r q

= ( ) ( )– + 2 – –2 + 3 + 4b c a b c

= – + 2 +2 – 3 – 4b c a b c

= 2 – 4 – 2a b c

= ( )2 – 2 –a b c

= 23

−

PQ

QR =–2

3PQ

i.e QR is scalar multiple of PQ

∴∴∴∴ QR and PQ parallel or collinear

and point Q is common point

∴ 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 (((( ))))– = 2 – .p q s r Show

that the lines QS and PR trisect each other.

Ans. –p q = 2 – 2s r

+2p r = +2q s

+2

3

p r=

+ 2

3

q s

+2

1+2

p r=

+ 2

1+2

q s= m

above is position vector of a point which

divies PR and QS internally in the ratio 2 : 1

i.e. point M (say) divides PR and QS internally

in the ratio 2 : 1

∴∴∴∴ PR and QS trisect each other.

Q-4) A, B, C are three points with position

vectors , ,a b c w.r.t origin O. If

7 = 4 + 3 ,c a b prove that A, B, C are

collinear.

Ans. 7c = 4 + 3a b

c =4 +3

7

a b

c =4 +3

4+3

a b

⇒⇒⇒⇒ point c divides seg AB internally in the ratio

3 : 4

∴∴∴∴ 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

2 = 5 – 3r p q then prove that P, Q, R are

collinear.

Ans. 2r = 5 – 3p q

r =5 – 3

2

p q

r =5 – 3

5 – 3

p q


Vectors

⇒⇒⇒⇒ point R divides seq. PQ externally in the

ratio 3 : 5.

⇒⇒⇒⇒ 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,

i) + + = 0AD BE CF

ii) + + = + +OA OB OC OD OE OF

iii) = 2BC FE

Ans. Let , , , , ,a b c d e f respectively be position

vectors of points A, B, C, D, E, F write a fixed

point .

point D. E. F are mid points of sides BC, AC

and AB respectvely

d =+

2

b c,

e =+

2

a c,

f =+

2

a b

i) + +AD BE CF

= – + – + –d a e b f c

= + + +

– + – + –2 2 2

b c a c a ba b c

= 2 – 2 +2 – 2 +2 – 2

2

a a b b c c

= 0

2

= R.H.S

ii) R.H.S= + +OD OE OF

= + +d e f

= + + +

+ +2 2 2

b c a c a b

= 2 + 2 +2

2

a b c

= ( )2 + +

2

a b c

= + +a b c

= + +OA OB OC

= L.H.S

iii) R.H.S = 2FE

= 2 ( )–e f

= 2+ +

–2 2

a c a b

= + – –

22

a c a b

= –c b

= BC

= L.H.S

A

B CD

EF

Q-7) The position vectors of four points A, B, C

and D are , , 2 3 , 2+ −+ −+ −+ −a b a b a b

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

c = 2 + 3a b ,

d = – 2a b

AC = –c a

= 2 +3 –a b a

= + 3a b

DB = –b d

= ( )– – 2b a b

= 3 –b a

BC = –c b

= 2 +3 –a b b

Vectors


= 2 + 2a b

Q-8) A divides seg PQ internally in the ratio

1 : 3 and B divides seg QR externally in

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.

∵ point A divides seg PQ internally in the ratio

1 : 3

∴∴∴∴ a =( )1 +3

1+3

q p

a =+3

4

q p...(i)

∵ point B divided seg QR externally in the

ratio 5 : 2

b = 5 – 2

5 – 2

r q

b =5 – 2

3

r q ...(ii)

∴∴∴∴ AB = –b a

=5 – 2 +3

–3 4

r q q p

=20 – 8 – 3 – 9

12

r q q p

= ( )1

20 –11 – 912

r q p

Q-9) The points C and D divides AB in ratio x : 1

inernally and externally respectively. Show

that A and B divides CD in ratio 1 –

1+

x

x

±±±±

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 =+

1+

a xb

x.... (i)

D ddivides AB externally in ratio x : 1

∴∴∴∴ By esternal division formula,

d =+

1 –

a xb

x.... (ii)

From (i) and (ii)

( )1+c x = +a xb .... (iii)

( )1 –d x = –a xb .... (iv)

Adding equation (iii) and (iv)

( ) ( )1+ + 1 –c x d x = 2a

∴∴∴∴ a =( ) ( )1+ + 1 –

2

x c x d

a =( ) ( )

( ) ( )

1+ + 1 –

1+ + 1 –

x c x

x x

∴∴∴∴ A divides OC in ratio

(1 + x) : (1 – x)

or A divides CD internally in ratio

(1 – x) : (1 + x)

i.e.1 –

1+

x

x

Now, equation (iii) - (iv)

∴∴∴∴ ( ) ( )1+ – 1 –c x d x = 2xb

∴∴∴∴ b =( ) ( )1+ – 1 –

2

x c x d

x

∴∴∴∴ b =( ) ( )

( ) ( )

1+ – 1 –

1+ – 1 –

x c x d

x x

∴∴∴∴ B divides CD externally in ratio

(1 – x) : (1 + x) i.e.1 –

1+

x

x

∴∴∴∴ A and B divides CD in ratio ±±±±1 –

1+

x

x

Q-10)If the vectors �� 2 – 3i q j k++++�� and

�� 4 – 5 6i j k++++�� are collinear, then find q.

Ans. Let a = � �2 – + 3i q j k� and

b = � �4 – 5 + 6i j k�

Since the vectors a and b are collinear, the

components of i� , �j and �k in a and b are

proportional.


Vectors

∴∴∴∴2

4=

–

–5

q=

3

6

1

2=

5

q=

1

2

∴∴∴∴ q =5

2

Q-11) If the points A(3, 0, p), B(–1, q, 3) and

C(–3, 3,0) are collinear, then find

i) the ratio in which the point C divides

the line segment AB

ii) the values of and q.

Ans. Let ,a b and c be the position vectors of A, B

and c respectively.

Then a = � �3 +0. +i j pk� ,

b = � �– + + 3i q j k� and

c = � �3 + 3 + 0.i j k�

i) As the points A, B, C, are collinear,

suppose the point C divides line segment AB

in the ratio λλλλ : 1.

∴∴∴∴ by the section formula,

c = � �λλλλ

λλλλ

. +1

.+1

b a

∴∴∴∴ � �–3 + 3 + 0.i j k�

=

� �( ) � �( )λλλλ

λλλλ

– + + 3 + 3 +0. +

+1

i q j k i j pk� �

∴∴∴∴ ( ) � �( )λλλλ +1 –3 + 3 +0.i j k�

=� �( ) � �( )λ λ λλ λ λλ λ λλ λ λ– + +3 + 3 +0. +i q j k i j pk� �

∴∴∴∴ ( ) ( ) �λ λλ λλ λλ λ–3 +1 + 3 +1 + 0.i i k� �

= ( ) ( ) �λ λλ λλ λλ λ– + 3 + 3 + 3 +i qi p k� �

By equality of vectors, we have,

–3(λ λ λ λ + 1) = – λ λ λ λ + 3 .... (i)

3(λ λ λ λ + 1) = λλλλ .... (ii)

0 = 3λλλλ + p .... (iii)

From equation (i), 3λ λ λ λ – 3 = –λλλλ + 3

∴∴∴∴ – 2λ λ λ λ = 6

∴∴∴∴ λ λ λ λ = – 3

∴∴∴∴ C divides segment AB externally in the ratio

3 : 1.

ii) Putting λ λ λ λ = –3 in equation (2), we get

3(– 3 + 1) = –3q

∴∴∴∴ –6 = –3q

∴∴∴∴ q = 2

Also, putting λ λ λ λ = –3 in equation (3), we get,

0 = –9 + p

∴∴∴∴ p = 9

Hence p = 9 and q = 2

Q-12) The position vectors of the points A and B

are �� 2 – +5i j k�� and ��–3 +2i j�� respectively.

Find the position vector of the point which

divides the line segment AB internally in

the ratio 1 : 4

Ans. The position vectors a and b of the points A

and B are

a = � �2 – +5i j k� and

b = �–3 +2i j�

Let C be the point, with position vector ,

divides the line segment AB internally in the

ratio 1 : 4

by the section formula for internal division,

∴∴∴∴ c =

� �1. + 4

1+ 4

b a

=

�( ) � �( )–3 +2 + 4 2 – +5

5

i j i j k� �

=� �( )1

5 – 2 + 205

i j k�

∴∴∴∴ c = � �2– + 45

i j k�

Q-13) If P is orthocentre, Q is the circmentre and

G is the centroid of a triangle ABC, then

prove that = 3QP QG

Ans. Let p and q be the position vectors of P and

G w.r.t the circumcentre Q.

i.e., QP = p and QG = g

we knoe that Q, G, P are collinear and G divides

segment QP internally in the ratio 1 : 2

Vectors


by section formula for internal division,

∴∴∴∴ g =1. + 2

1+2

p q=

3

p

∴∴∴∴ p = 3q

∴∴∴∴ QP = 3QG

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

a = ˆ ˆ ˆ+ 2 +2 ,ai j k

b = ˆ ˆ ˆ+ + ,ai bj k

c = ˆ ˆ ˆ+2 – 2 ,i j k

g = ˆ ˆ ˆ2 + –i j ck

∵∵∵∵ G is centroid of ∆∆∆∆ABC

∴∴∴∴ g =+ +

3

a b c

∴∴∴∴ 3g = + +a b c

( )3 2 + –i j ck

= ( ) ( )6 +3 – 3c = 2 +1 + 4 + +i j k a i b j k

Comparing the coefficeients of , ,i j k on both

sides

we get 2a + 1 = 6

4 + b = 3

–3c = 1

a = 5

2

b = –1

c = –1/3

Q-15) If A ≡ ≡ ≡ ≡ (1, 2, 3), B ≡≡≡≡ (, 4, 5) and C ≡≡≡≡ (p, q, 7) are

collinear points. Find p & q.

Ans. Let , , ,a b c g be position vectors pf points

A, B, C, G respectively

a = +2 +3i j k ;

b = 3 + 4 +5i j k ;

c = + +7pi q j k ;

∴∴∴∴ AB = –b a

= ( ) ( )3 + 4 +5 – + 2 + 3i j k i j k

= 2 +2 +2i j k ... (i)

AC = –c a

= ( ) ( )+ + 7 – + 2 + 3pi p j k i j k

= ( ) ( )–1 + – 2 + 4p i q j k ... (ii)

Given thaty the points are collinear

∴∴∴∴ AB and AC are collinear

AB = mAC [m : nonzero scalar]

2 +2 +2i j k = ( ) ( )–1 + – 2 +m p i q j uk

Comparing the of , ,i j k on both sides we get

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

( )1

– 22q = 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

i) The ratio in which the point C divides

the line AB

ii) The values of p and q.

Ans. Let , ,&a b c be the position vectors of A, B &

C respectively.

∴∴∴∴ a = ˆ ˆ ˆ4 +5 + . ,i j pk

b = ˆ ˆ ˆ+2 +4 ,qi j k

c = ˆ ˆ ˆ5 +8 +0 ,i j k

i) As the point A,B & C are collinear,

Suppose the point C divides line segment

AB in the ratio λλλλ:1

∴∴∴∴ By the section fromula.

c =λ

λ

. +1.

+1

b a

( )λλλλ ˆ ˆ ˆ+1 5 + 8 + 0i j k


Vectors

= ( ) ( )λλλλ ˆ ˆ ˆ ˆ ˆ ˆ+ 2 + 4 + 4 +5 +qi j k i j Pk

( ) ( )λ λλ λλ λλ λˆ ˆ ˆ5 +1 +8 +1 +0.i j k

= ( ) ( ) ( )ˆ ˆ ˆ+ 4 + 2+5 + 4+q i j P kλ λ λ

By equality of vectors, we have

5(λλλλ+1) = λλλλq + 4 ... (1)

8(λλλλ+1) = λλλλ2 + 5 ... (2)

0 = λλλλ4 + P ... (3)

From (1)

5λλλλ+ 5 = λλλλq + 4

5λ λ λ λ – λλλλq = – 1 ... (5)

From (2)

8λλλλ+ 8 = λλλλ2 + 5

6λ λ λ λ = – 3

x = –1

2... (6)

∴∴∴∴ C divides segment AB externally in the

ratio 1 : 2

ii) –1 –1

5 – = –12 2

q

Substitue λ–1

=2

in (5)

–5+

2 2

q= –1=

– 5 + q = – 2

q = 3

0 = λλλλ4 + p

4λλλλ + p = 0

–14

2

= p

p = – 2

∴∴∴∴ p = – 2 and q = 3

Q-17) The position vectors P and Q are ˆˆ ˆ– 2 +i j k

and ˆˆ ˆ+ 4 – 2i j k respectively. Find

coordinates & position vector of (((( ))))ˆR r

which divides the line segment PQ

internally in the ratio 2 : 1.

Ans. Let p and q be the position vectors of the

points P and Q respectively.

Then = – 2 +p i j k and = + 4 – 2q i j k

Since ( )ˆR r divides line segment PQ

internally in the ratio 2 :1, by section formula

for internal division,

r =2 +1.

2 +1

q p

=( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ2 + 4 – 2 + – 2 +

3

i j k i j k

=ˆ ˆ ˆ ˆ ˆ ˆ2 +8 – 4 + – 2 +

3

i j k i j k

= ( )1 ˆ ˆ ˆ3 + 6 – 33

i j k

∴∴∴∴ r = ˆ ˆ ˆ+ 2 –i j k

∴∴∴∴ coordinates of R are (1,2 – 1).

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.

a = +2 + 4i j k ,

b = 2 – +5i j k

AB = –b a

= 2 – +5 – – 2 – 4i j k i j k

= – 3 +i j k

a = ( ) ( ) ( )2 2 2

1 + 2 + 4

= 1+ 4+16

= 21

b = ( ) ( ) ( )2 2 2

2 + –1 + 5

= 4 +1+ 25

= 30

unit vectors along a =a

a

Vectors


=+ 2 + 4

21

i j k

unit vectors along b = 2 + +5

30

i j k

b

b

AB = ( ) ( ) ( )2 2 2

1 + –3 + 1

= 1+ 9 +1

= 11

unit vectors along AB =AB

AB

=– 3 +

11

i j k

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-

ordinates of B.

Ans. Let , ,a b c be the position vectors of points

A, B, C respectively w.r.t a fixed point.

A ≡≡≡≡ (3, 8, 6)

C ≡≡≡≡ (1, 5, 0)

a = 3 +8 +6i j k

c = +5i j

Now, C divides AB internally in the ratio

5 : 6

c =( ) ( )5 + 6

5 +6

b a

+5i j = ( ) ( )5 +6 3 +8 +6

11

b i j k

11 +55i j = 5 +18 + 48 +36b i j k

5b = –7 + 7 – 36i j k

b =7 7 36

– + –5 5 5i j k

co-ordinates of B ≡≡≡≡ –7 7 –36

, ,5 5 5

Q-20) If , ,a b c are position vectors of A, B and C

where A (1, 3, 0), B (2, 5, 0), C (4, 2, 0) such

that = +c xa yb then find x and y.

Ans. A ≡≡≡≡ (1, 3, 0)

B ≡≡≡≡ (2, 5, 0)

C ≡≡≡≡ (4, 2, 0)

a = +3i j

b = 2 +5i j

c = 4 +2i j

Now, c = +xa yb

4 +2i j = ( ) ( )+ 3 + 2 + 5x i j y i j�

4 +2i j = +3 +2 +5xi xj yi y j

4 +2i j = ( ) ( )+2 + 3 +5x y i x y j

comparing co-efficients of i and j on both

sides

x + 2y = 4 ...(i)

3x + 5y = 2 ...(ii)

multiplying equation, (i) by 3 and then

subtracting (ii) form (i).

3x + 6y = 12

3x + 5y = 2

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

ˆˆ ˆ= + +AB xi yj zk 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 = +3j k

b = 3 +2 –i j k∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧

AB = –b a


Vectors

+ +x i y j z k∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧

= 3 +2 – – – 3i j k j k∧ ∧ ∧ ∧ ∧∧ ∧ ∧ ∧ ∧∧ ∧ ∧ ∧ ∧∧ ∧ ∧ ∧ ∧

+ +x i y j z k∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧

= 3 + – 4i j k∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧

comparing co-efficients of , ,i j k∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧∧ ∧ ∧

on both side

x = 3, y = 1, z = – 4

L.H.S = x + y + 2

= 3 + 1 – 4

= 0

= R.H.S.

Q-22) If A (1, 4, 1), B (–2, 3, –5) and C (5, –7, 2) are

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

G respectively.

Let G be the centroid.

a = ˆ ˆ ˆ+ 4 +i j k

b = ˆ ˆ ˆ–2 +3 – 5i j k

c = ˆ ˆ ˆ5 – 7 +2i j k

∵ G is the centroid of ∆ ABC

g =+ +

3

a b c

3g = + +a b c

= ( ) ( )

( )

+ 4 + + –2 +3 +5

+ 5 – 7 + 2

i j k i j k

i j k

3g = 4 +0 – 2i j k

g =4

3i + 0 j +

2

3k

G 4 –2,0,

3 3

Centroid of ∆∆∆∆ABC

Q-23) If �� = + – 2 , = 2 – + 2a i j k b i j k� �� and

��= 3 – ,c i k�� then find the scalars m and n

such that = +c mb nb

Ans. c = +mb nb

∴∴∴∴ �3 –i k� =� �( ) � �( )+ – 2 + 2 – +2m i j k n i j k� �

∴∴∴∴ �3 –i k� = ( ) ( ) �

( ) �

2 + –

+ –2 +

m n i m n j

m n k

+ �

By equality of vectors,

m + 2n = 3 .... (i)

m – n = 0 .....(ii)

and –2m + n = – 1 .....(iii)

To solve equations (i) and (ii),

Subtracting (ii) from (i), we get,

3n = 3

∴∴∴∴ n = 1

Substituting n = 1 in equation (i), we get,

m + 2(1) = 3

∴∴∴∴ m = 1

Substituting m = 1, n = 1 in equation (iii), we

get,

LHS = –2(1) + (1)

= –2 + 1

= –1

= RHS.

Hence, m = 1 and n = 1.

Q-24) If G1 and G

2 are the centroids of the

triangles ABC and PQR respectively, then

prove that 1 2+ + = 3AP BQ CR G G .

Ans. Let , , , , , , ,a b c p q r g g1 2be the position vectors

of the points A, B, C, P, Q, R, G1 and G

2

respectively.

Since G1 and G

2 are the centroids of ∆∆∆∆ABC

and ∆∆∆∆PQR respectively,

∴∴∴∴ g1 =

+ +

3

a b c and g2

= + +

3

p q r

∴∴∴∴ a b c+ + = 3g1 and p q r+ + = 3g2

... (i)

Now, AP BQ CR+ +

= ( ) ( ) ( )– + – + –p a q b r c

= ( ) ( )+ + – + +p q r a b c

= 3 – 3g g2 1... [From (i)]

= ( )3 –g g2 1

= 3 –G G1 2

Vectors


Q-25) If , ,a b c are the position vectors of the

points A (1,3,0), B(2,5,0), C(4,2,0)

respectively and c ma nb= += += += + then find the

values of m and n.

Ans. The position vectors , ,a b c of the points A,

B,C are

ˆ ˆ ˆ ˆ ˆ ˆ3 0. , 2 5 0. ,

ˆ ˆ ˆ4 2 0. ,

= + + = + +

= + +

a i j k b i j k

c i j k

Now c ma nb= +

ˆ ˆ ˆ4 2 0.∴ + +i j k

( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ= + 3 + 0. + 2 + 5 + 0.m i j k n i j k

ˆ ˆ ˆ4 +2 + 0.∴ i j k

( ) ( )ˆ ˆ ˆ= + 2 + 3 +5 + 0.m n i m m j k


m + 2n = 4 .....(i)

and 3m + 5n = 2 ......(ii)

Multiplying equation (i) by 3, we get,

3m + 6n = 12

Subtracting equation (ii) from this equation,

we get, n = 10

Subtracting n = 10 in equation (i), we get,

m + 2(10) = 4 ∴∴∴∴ m = – 16

Hence m = – 16 and n = 10.

GROUP (B)- CLASS WORK PROBLEMS

Q-1) Show that the following points are collinear :

i) A(3, 2, –4), B(9, 8, –10), C(–2, –3, 1)

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 = � �3 + 2 – 4i j k� ,

b = � �9 +8 –10i j k� ,

c = � �2 – 3 +i j k� .

g = 3 +3 +i j k ,

∴∴∴∴ AB = –b a

=� �( ) � �( )9 +8 –10 – 3 + 2 – 4i j k i j k� �

= � �6 + 6 – 6i j k� ....(i)

AC = –c a

=� �( ) � �( )–2 – 3 + – 3 +2 – 4i j k i j k� �

= � �5 – 5 +5i j k�

=� �( )5

– 6 + 6 – 66

i j k�....(ii)

From (i) and (ii),

AC =5

–6

AB

i.e, AC is the scalar multiple of AB .

∴∴∴∴ they are parallel to each other. But they have

the point A in common.

∴∴∴∴ the vectors AB and AC lie on the same line.

the points a, B and C are collinear.

ii) Refer to the solution of Q.1(i)

PR = (–1)PQ .

Q-2) If the vectors �� 2 – + 3i q j k�� and �� 4 – 5 + 6i j k��

are collinear, then find the value of q.

Ans. Let a = � �2 – +3i q j k� and

b = � �4 – 5 +6i j k� .

Since the vectors a and b collinear, the

components of i� ,�j and �k a and b are

propertional.

∴∴∴∴2

4=

–

–5

q=

3

6

∴∴∴∴1

2=

5

q=

1

2

∴∴∴∴ q =5

2.

Alternative Method.

Let a = � �2 – + 3i q j k� and

b = � �4 – 5 + 6i j k� .

Since the vectors a and b are collinear,

× = 0a b


Vectors

∴∴∴∴

� �

2 – 3

4 –5 6

i j k

q

�

∴∴∴∴ ( ) � ( )� ( )

–6 +15 – 12 –12

+ –10 + 4 = 0

i q j

k q

�

∴∴∴∴ ( ) � � ( ) �–6 +15 – 0 –10 + 4q j j q k

= � �0 +0 +0i j k�

By equalify of vectors,

–6q + 15 = 0 and –10 + 4q = 0

∴∴∴∴ q =15

6=

5

2and

q =10

4=

5

2

Hence, q =5

2

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.

Ans. The position vectors , , , ,a b c d of the points A,

B, C, D are

a = � �– +i j k� ,

b = � �– + +i j k� ,

c = � �+ +i j k� ,

d = � �2 – 3 + 4i j k�

∴∴∴∴ AB = –b a

=� �( ) � �( )– + + – – +i j k i j k� �

= �–2 +2i j�

AC = –c a

=� �( ) � �( )+ + – – +i j k i j k� �

= �2 j and

AD = –d a

=� �( ) � �( )2 – 3 + 4 – – +i j k i j k� �

= � �– 2 +3i j k�

If A,B,C, D are coplanar, then there exist

scalars x,y such that

AB = . + .x AC y AD

∴∴∴∴ �–2 + 2i j� =�( ) � �( )2 + – 2 +3x j y i j k�

∴∴∴∴ �–2 + 2i j� = ( ) � �+ 2 – 2 – 3yi x y j yk�

By equality of vecotrs,

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

prove that the vectors

2 – 4 + 4 , – 2 + 4a b c a b c and – + 2 + 4a b c

are collinear

Ans. Let P,Q,R be the points whose position vectors

, ,p q r are given by

p = 2 – 4 + 4a b c ,

q = – 2 +0.a b c ,

r = – +2 + 4a b c .

∴∴∴∴ PQ = –q p

= ( ) ( )– 2 + 4 – 2 – 4 + 4a b c a b c

= – +2 + 0.a b c .... (i)

PR = –r q

= ( ) ( )+ 2 + 4 – 2 – 4 + 4a b c a b c

= ( )3 – + 2 + 0.a b c

∴∴∴∴ PQ = 3PQ .... [By (i)]

∴∴∴∴ ( )+ –3PQ PQ

= 0

Vectors


∴∴∴∴ PQ and PQ are collinear vectors,

Hence the points ( ) ( ),P p Q q and ( )R r are

collinear

Q-5) Express = 3 +2 + 4p i j k 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

, ,a b c

Let = + +p xa yb zc where , ,x y z

are scalars

∴∴∴∴ 3 +2 + 4i j k

= ( ) ( ) ( )+ + + + +x i j y j k z k i

= ( ) ( ) ( )+ + + + +x z i x y j y z k

+x z = 3;

+x y = 2;

+y z = 4

Solving we get

z =5

2;

y =3

2;

x =1

2

∴∴∴∴ p =1 3 5

+ +2 2 2a b c

Q-6) If ,a b and c are non-coplanar vectors, show

that the vectors 2 – + 4 , – + 4 – 3a b c a b c

and 5 – 6 +11a b c are coplanar.

Ans. p = 2 – + 4a b c ,

q = – + 4 – 3 ,a b c r

= 5 – 6 +11a b c

In order to show that vectors ,p q and r are

coplanar.

To find the scalars x and y such that

= +r x p yq

Now,

r = ( ) ( )2 – + 4 + – + 4 – 3x a b c y a b c

5 – 6 +11a b c

= ( ) ( ) ( )2 – + – + 4 + 4 – 3x y a x y b x y c

2x – y = 5 ...(i)

–x + 4y = –6 ...(ii)

4x – 3y = 11 ...(iii)

From (i) and (ii) x = 2 & y = – 1

For these values of x and y

L.H.S of (iii)

= 4(2) – 3(–1)

= 11

= R.H.S.

Thus, these values satisfy the (iii) equation

also.

∴∴∴∴ The given vectors are coplanar.

GROUP (B)- HOME WORK PROBLEMS

Q -1) a and b are non-collinear vectors. If

(((( ))))= – 2 +c x a b and (((( ))))= 2 +1 –d x 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

∴∴∴∴ ( )– 2 +x a b = ( )2 +1 –t x a b

∴∴∴∴ ( )– 2 +x a b = ( )2 +1 –t x a tb

∴∴∴∴ x – 2

= t(2x + 1) and

t = –1

∴∴∴∴ t = –1

∴∴∴∴ x – 2

= –(2x + 1)

= –2x – 1

∴∴∴∴ 3x = 1

∴∴∴∴ x =1

3


Vectors

Q-3) Express �� – – 3 + 4i j k�� as the linear


�� 2 + – 4 ,2 – +3i j k i j k� �� and �� 3 + – 2i j k�� .

Ans. Let a = � �2 + – 4i j k� ,

b = � �2 – + 3i j k� ,

c = � �3 + – 2i j k� and

p = � �– – 3 + 4i j k� .

Suppose p = + + .xa yb ze

Then, � �– – 3 + 4i j k�

=� �( ) � �( )

� �( )

2 + – 4 + 2 – + 3

+ 3 + – 2

x i j k y i j k

z i j k

� �

�

∴∴∴∴ � �– – 3 + 4i j k�

= ( ) ( ) �

( ) �

2 + + 3 + – +

+ –4 + – 2z

x 2y z i x y z j

x 3y k

�

Q-2) If , ,a b c are non-coplanar vectors then

prove that the vectors 2 – 4 +4 ,a b c

– 2 + 4a b c and – +2 +4a b c are collinear.

Ans. Let P, Q and R be the points whose position

vectors are 2 – 4 + 4 ,p a b c= – 2 + 4 ,q a b c=

– + 2 + 4r a b c=

Now,

( ) ( )– = – 2 + 4 – 2 – 4 + 4PQ q p a b c a b c=

∴∴∴∴ – + 2 +0.PQ a b c= ... (i)

and

( ) ( )= – = – + 2 + 4 – 2 – 4 + 4PR r p a b c a b c

( )= –3 + 6 + 0 = 3 – + 2 + 0.a b c a b c

∴∴∴∴ 3PR PQ= ...[From (i)]

∴∴∴∴ ( )+ –3 0PR PQ =

∴∴∴∴ PR and PQ are collinear vectors.

∴∴∴∴ P, Q and R are collinear.

∴∴∴∴ = 2 – 4 + 4 ,p a b c = – 2 + 4q a b c

= – + 2 + 4r a b c are collinear.


2x + 2y + 3z = –1

x – y + z = –3

–4x + 3y – 2z = 4

Solve these equations by using Cramer’s Rule

D =

2 2 3

1 –1 1

–4 3 –2

= 2(2 – 3) – 2(–2 + 4) + 3(3 – 4)

= – 2 – 4 – 3 = –9 ≠≠≠≠ 0

xD =

–1 2 3

–3 –1 1

4 3 –2

= –1(2– 3) – 2(6 – 4) + 3(–9 + 4)

= 1 – 4 – 15 = –18

yD =

2 –1 3

1 –3 1

–4 4 –2

= 2(6 – 4) + 1(–2 + 4) + 3(4 – 12)

= 4 + 2 – 24 = –18

zD =

2 2 –1

1 –1 –3

–4 3 4

= 2(–4 + 9) –2(4 – 12) –1(3 – 4)

= 10 + 16 + 1 = 27

∴∴∴∴ x = –18

= = 2,– 9

zD

Dy =

–18= = 2,

– 9

yD

D

z = 27

= = – 3– 9

zD

D

∴∴∴∴ = 2 + 2 – 3p a b c

Q-4) Express �� = 3 +2 + 4P i j k�� as the linear


�� = + , = + , = +a i j b j k c k i� ��

Ans. To express p as a linear combination of , ,a b c

Let p = + +xa yb zc

where x, y, z are scalars

∴∴∴∴ � �3 + 2 + 4i j k�

Vectors


Q-5) If a ,b ,c are non-coplanar vectors, then

show that the vectors

– + 3 – 5 , – + +a b c a b c and 2 – 3 +a b c are

collinear

Ans. Let p = – + 3 – 5a b c ,

q = – + +a b c and

r = 2 – 3 +a b c

Then in order to prove that these vectors are

coplanar, we should be also to find scalars x

and y such that

r = +x p yq

Now, +x p yq

= ( ) ( )– + 3 – 5 + – + + 5x a b c y a b c

= ( ) ( ) ( )– – + 3 + + –5 +x y a x y b x y c

and r = 2 – 3 +a b c

∴∴∴∴ 2 – 3 +a b c

= ( ) ( ) ( )– – + 3 + + –5 +x y a x y b x y c


–x – y = 2 .... (i)

3x + y = –3 .... (ii)

and –5x + y = 1 .... (iii)

Adding (i) and (ii), we get,

2x = –1

∴∴∴∴ x =1

–2

∴∴∴∴ –x – y = 2 gives 1

–2y = 2

∴∴∴∴ y =1

– 22

=3

–2

For these values of x and y,

–5x + y =1 3

–5 – + –2 2

=5 3

–2 2

= 1

Thus, these values satisfy the third equation

also.

∴∴∴∴ r =1 3

– + –2 2p q

∴∴∴∴ the given vectors are coplanar.

Q-6) Express �� – – 3 + 4i j k�� as the linear


�� 2 + – 4 ,2 – +3i j k i j k� �� and �� 3 + – 2i j k�� .

Ans. Let a = � �2 + – 4i j k� ,

b = � �2 – + 3i j k� ,

c = � �3 + – 2i j k� and

p = � �– – 3 + 4i j k� .

Suppose p = + + .xa yb zc

Then, � �– – 3 + 4i j k�

= � �( ) � �( )� �( )

2 + – 4 + 2 – + 3

+ 3 + – 2

x i j k y i j k

z i j k

� �

�

∴∴∴∴ � �– – 3 + 4i j k�

= ( ) ( ) �

( ) �

2 + + 3 + – +

+ –4 + – 2z

x 2y z i x y z j

x 3y k

�

=�( ) � �( ) �( )+ + + + +x i j y j k z k i� �

= ( ) ( ) � ( ) �+ + + + +x z i x y j y z k�

∴∴∴∴ x + z = 3;

x = y = 2;

y + z = 4

solving, we get

z =5

2,

y =3

2,

x =1

2

∴∴∴∴ p =1 3 5

+ +2 2 2a b c


Vectors


2x + 2y + 3z = –1

x – y + z = –3

–4x + 3y – 2z = 4

Solve these equations by using Cramer’s Rule

D =

2 2 3

1 –1 1

–4 3 –2

= 2(2 – 3) – 2(–2 + 4) + 3(3 – 4)

= – 2 – 4 – 3 = –9 ≠≠≠≠ 0

xD =

–1 2 3

–3 –1 1

4 3 –2

= –1(2– 3) – 2(6 – 4) + 3(–9 + 4)

= 1 – 4 – 15 = –18

yD =

2 –1 3

1 –3 1

–4 4 –2

= 2(6 – 4) + 1(–2 + 4) + 3(4 – 12)

= 4 + 2 – 24 = –18

zD =

2 2 –1

1 –1 –3

–4 3 4

= 2(–4 + 9) –2(4 – 12) –1(3 – 4)

= 10 + 16 + 1 = 27

∴∴∴∴ x = –18

= = 2,– 9

zD

Dy =

–18= = 2,

– 9

yD

D

z = 27

= = – 3– 9

zD

D

∴∴∴∴ = 2 + 2 – 3p a b c

Q-7) Express the vector �� + 4 – 4i j k�� as the linear


�� 2 – +3 , – 2 + 4i j k i j k� �� and �� – +3 – 5i j k�� .

Ans. Let a = � �2 – + 3i j k� ,

b = � �– 2 + 4i j k� ,

c = � �– – 3 – 5i j k� and

p = � �+ 4 – 4i j k� .

suppose p = + +xa yb zc .

Then,

� �+ 4 – 4i j k�

= � �( ) � �( )� �( )

2 – + 3 + – 2 + 4

+ – – 3 – 5

x i j k y i j k

z i j k

� �

�

� �+ 4 – 4i j k�

= ( ) ( ) �

( ) �

2 + – + – – 2 + 3

+ 3 + 4 – 5

x x y z i y x y z j

x y z k

�


2x + y – z = 1

–x – 2y + 32 = 4

3x + 4y – 5z = –1

We have to solve these equation by ussing

cramer’s Rule.

D =

2 1 –1

–1 –2 3

3 4 –5

= 2(10 – 12) – 1(5 – 9) – (–4 + 6)

= –4 + 4 – 2 = –2 ≠ ≠ ≠ ≠ 0

Dx

=

1 1 –1

4 –2 3

– 4 4 –5

= 1(10 – 12) –1 (20 + 12) – 1(16 – 8)

= –2 + 8 – 8

Dx

= –2

Dy

=

2 1 –1

–1 4 3

3 –4 –5

= 2(–20 + 12) – 1(5 – 9) – 1(4 – 12)

= –16 + 4 + 8

Dy

= –4

Dz

=

2 1 1

–1 – 2 4

3 4 –4

= 2(8 – 16) – 1(4 – 12) + 1(–4 + 6)

= –16 + 8 + 2

Vectors


Dz

= –6

x =xD

D=

–2

–2= 1

y =yD

D=

–4

–2= 2

z =zD

D=

–6

–2= 3

∴∴∴∴ p = + 2 + 3a b c

ii) ∴∴∴∴ abc =

3 2 1

1 1 2

3 1 2

−

= 1(1) –1(–1)

= 2

GROUP (C)- CLASS WORK PROBLEMS

Q-1) Find [ ]a bc if

= 2 + + 3 , = +2 + 4 ,

= + + 2

a i j k b i j k

c i j k

Ans. abc =

2 1 3

1 2 4

1 1 2

= 2(4 – 4) – 1(2 – 4) + 3(1 – 2)

∴∴∴∴ [ ]abc = –1

Q-2) = – + , = + – 4 , = 2 + +a i j pk b i j k c i j k

find p, if [ ] = 0abc

Ans. abc = 0

∴∴∴∴

1 1

1 1 4

2 1 1

p−

−= 0

∴∴∴∴ 1(1 + 4) + 1(1 +8) + p(1 – 2) = 0

14 – p = 0 ⇒⇒⇒⇒

∴∴∴∴ p = 14

Q-3) Find the volume of the parallelopiped if

= 2 , = 3a i b j and = 4c k are the

coterminus edges of the parallelepiped.

Ans. Volume of parallelepiped = [ ]abc

=

2 0 0

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 &

= 2 +3 +2c i j k are coplanar.

Ans. If [ ] = 0abc

then vectors ,a b and c are coplanar.

L.H.S. = a bc =

1 1 1

1 1 1

2 3 2

−

1( – 2 – 3) – 1(2 – 2) + 1(3 + 2)

= – 5 + 5 = 0

= R.H.S.

∴∴∴∴ , ,a b c

are coplanar vectors.

Q-5) Show that A, B, C, D are coplanar, if

A ≡≡≡≡ (2, 1, – 3), B ≡≡≡≡ (3, 3, 0), C ≡≡≡≡ (7, –1, 4) and

D ≡≡≡≡ (2, – 5, – 7).

Ans. We know that, four points A, B, C, and D are

coplanar if

[ ]AB AC AD = 0

AB = –b a

= ( ) ( ) ( )3 – 2 + 3 –1 + 0 + 3i j k

∴∴∴∴ AB = + 2 + 3i j k

AC = –c a

= ( ) ( ) ( )7 – 2 + –1 –1 + 4 + 3i j k

∴∴∴∴ AC = 5 – 2 + 7i j k

AD = –d a

= ( ) ( ) ( )2 – 2 + –5 –1 + –7 + 3i j k ⇒

= –6 – 4j k

∴∴∴∴ L.H.S. = AB AC AD


Vectors

=

1 2 3

5 –2 7

0 –6 –4

= 1(8 + 42) – 2(– 20) + 3 (– 30)

= 50 + 40 – 90 = 0 = R. H. S.

∴∴∴∴ , ,AB AC AD are coplanar vectors.

∴∴∴∴ The points A, B, C and D are coplanar.

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

D ≡ (2,4,3)

Given that the points A, B, C, D are

coplanar

AB = –b a

= ( ) ( ) ( )4 – 2 + 1+1 + 3 –1i j k

AB = 2 + + 2i j k

AC = –c a

= ( ) ( ) ( )– 2 + 1+1 + 1 –1m i j k

AC = ( )– 2 + 2m i j

AD = –d a

= ( ) ( ) ( )2 – 2 + 4 +1 + 3 –1i j k

= 5 + 2j k

∴∴∴∴ AB AC AD = 0

2 1 2

2 2 0

0 5 2

m −= 0

2(4) – 1(2m – 4) + 2(5m –10) = 0 ⇒⇒⇒⇒

∴∴∴∴ 8 – 2m + 4 + 10m – 20 = 0

∴∴∴∴ 8m = 8 ⇒⇒⇒⇒

∴∴∴∴ m = 1

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.

Ans. The points O, A, B and P are coplanar. (given)

∴∴∴∴ [ ]OA OB OP = 0

∴∴∴∴ OA = a = + 2 + 3i j k

OB = b = 2 + 3 + 4i j k

OP = p = + +xi y j zk

∴∴∴∴ [ ]OA OB OC = 0

∴∴∴∴

1 2 3

2 3 4

x y z= 0

∴∴∴∴ 1(3z –4y) – 2(2z – 4x) + 3(2y – 3x) = 0

∴∴∴∴ 3z –4y – 4z + 8x + 6y – 9x = 0

∴∴∴∴ –x + 2y – z = 0

∴∴∴∴ x – 2y – z = 0

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

AD as concurrent edges.

Ans. We know that, the volume of parallelopiped

with concurrent edges AB, AC and

AD = , ,AB AC AD

AB = –b a

= ( ) ( ) ( )+ ++ ++ ++ +2 –1 –1 –1 3 –1i j k

∴∴∴∴ AB = – 2 + 2i j k

AC = –c a

= ( ) ( ) ( )3 –1 + –2 –1 + –2 –1i j k

∴∴∴∴ AC = 2 – 3 – 3i j k

AD = ( ) ( ) ( )– 3 –1 + 3 –1 + 4 –1d a i j k=

∴∴∴∴ AD = 2 + 2 + 3i j k

∴∴∴∴ volume of parallelopiped

=

1 2 2

2 3 3

2 2 3

−

− −

= 1 (– 9 + 6) + 2 (6 + 6) + 2 (4 + 6)

= – 3 + 24 + 20 = 41 cubic units.

Q-9) Prove that [ + + + ] = 2[ ]a b b c c a abc

Ans. Let = [ + + + ]a b b c c a

= ( ) ( ) ( )××××+ .[ + + ]a b b c c a

= ( ) × × × ×× × × ×× × × ×× × × ×+ .[ + + + ]a b b c b a c c c a

= . [ + + ]

+ . [ + + ]

a b c b a c a

b b c b a c a

× × ×× × ×× × ×× × ×

× × ×× × ×× × ×× × ×

Vectors


××××[ = 0]c c∵

= ( ) ( ) ( )

( ) ( ) ( )

× × ×× × ×× × ×× × ×

× × ×× × ×× × ×× × ×

. + . + .

+ . + . + .

a b c a b a a c a

b b c b b a b c a

= ( ) ( )× ×× ×× ×× ×. + .a b c b c a + 0 + 0 + 0 + 0

(in box product if two vectors are equal , its

value is zero)

= [ ] + [ ]a b c b c a

= [ ] + [ ]a b c a b c

( )∵∵∵∵ =abc b c a

= 2[ ] = . . .a b c R H S

Q-10) If = + 3c a b , show that , = 0abc Also if

2 2 2= + 3 3 +c a ab ab , Prove that the

angle between a and b is 6ππππ

Ans. c = + 3a b

i.e c is linear combination of and a b

∴∴∴∴ , ,a b c are coplanar

∴∴∴∴ abc = 0

c = + 3a b

∴∴∴∴ c = + 3a b

Squaring both sides.

2c =

2+ 3a b

2c = ( ) ( )+ 3 . + 3a b a b

2c = . + 3 . + 3 . + .a a a b b a ab b

2c =

2 2+ 6 . + 9a a b b

2c = ( )

22 π+ 6 6 cos + 9

6a a b

2c =

2 23+ 6 . + 9

2a a b b

c2 =2 23

6 . + 92

a ab b+

c2 = 2 2+ 3 3 + 9a ab b

Q-11) Show that the statement

× ×× ×× ×× ×( – ).[( – ) ( – )] = 2 .a b b c c a a b c

is true only when ,a b and c are coplanar.

Ans. If , ,a b c are coplanar then

[ ]a b c = 0

i.e., ( )××××.a b c = 0

L.H.S.

= ( ) ( ) ( )××××– . – –a b b c c a

= ( ) × × × ×× × × ×× × × ×× × × ×– .[ – – + ]a b b c b a c c c a

= ( ) × × ×× × ×× × ×× × ×– .[ + + ]a b b c a b c a

= × × ×× × ×× × ×× × ×

× × ×× × ×× × ×× × ×

.[ + + ]

– .[ + + ]

a b c a b c a

b b c a b c a

= ( ) ( ) ( )

( ) ( ) ( )

× × ×× × ×× × ×× × ×

× × ×× × ×× × ×× × ×

. + . + .

. + . .

a b c a a b a c a

b b c b a b b c a− +

= ( ) ( )× ×× ×× ×× ×. – .a b c b c a

= [ ] – [ ]a b c b c a

= [ ] – [ ]a b c a b c = 0

R.H.S

= 2 . ×a b c

= 2[ ]a b c

if , ,a b c are planar than [ ]a b c = 0

∴∴∴∴ R.H.S. = (0)

= 0

∴∴∴∴ L.H.S = R.H.S

When , ,a b c are co planar

Q-12) , ,a b c are three non-coplanar vectors. If

= , = , =b c c a a b

p q ra bc a bc a b c

× × ×× × ×× × ×× × ×

then prove that

i) . . . = 3a p b q c r+ ++ ++ ++ +

ii) (((( )))) (((( )))) (((( ))))+ . + + . + + . = 3a b p b c q c a r


Vectors

Ans. i) L.H.S.

= . + . + .a p b q c r

=( ) ( ) ( )× × ×× × ×× × ×× × ×. . .

+ +[ ] [ ] [ ]

a b c b c a c a b

abc abc abc

=[ ] + [ ]+ [ ]

[ ]

abc bc a c ab

abc

=[ ] + [ ]+ [ ]

[ ]

abc abc abc

abc

=3[ ]

[ ]

abc

abc

= 3

ii) L.H.S. ( ) ( ) ( )+ . + + . + + .a b p b c q c a r

= ( )( ) ( ) ( )

( )( )

× ×× ×× ×× ×

××××

+ .. +[ ] [ ]

+ + .[ ]

b c b c c aa b

abc abc

a bc a

abc

+

Since a bc is a scalar.

Let a bc = k

( ).a b p+ = ( )( )+

+ .b c

a bk

=1

+a bc abck

= { }1

+ 0a b ck

=1a b c

k

= ××××1

0 kk

= 1

Similarly ( )+ .b c q = 1

and ( )+ .c a r = 1

( ) ( ) ( )+ . + + . + + .a b p b c q c a r

= 1 + 1 + 1 = 3

iii) . .a p =( )

( )

××××⇒⇒⇒⇒.

b caabc

.c q =( )

( )

××××.c q

cabc

=( )

( )⇒⇒⇒⇒

abc

abc=

( )

( )

c c a

abc

= 1

.b p =( )

( )

××××⇒⇒⇒⇒

.b b c

abc= ( )

0

abc

=( )

( )⇒⇒⇒⇒

bbc

abc

.c v =( )

( )

××××.c a b

abc

=( )

⇒⇒⇒⇒0

abc=

( )

( )

c a b

abc

= 0

⇒⇒⇒⇒ =( )

( )

abc

abc

⇒⇒⇒⇒ .b p =( )

( )

××××.b c a

abc

=( )

( )

bc a

abc ⇒⇒⇒⇒ =

( )

( )

abc

abc

.a r =( )

( )

××××.a c a

abc

=( )

( )

abc

abc

= ( )0

abc = 0

Vectors


Q-13) If the vectors

+ + ; + + ; + 1ai j k i b j k i cka b c≠ ≠ ≠≠ ≠ ≠≠ ≠ ≠≠ ≠ ≠

are coplanar then show that

1 1 1+ + =1.

1 – 1 – 1 –a b c

Ans. Let

p = + +ai j k ,

q = + +i bj k ,

r = + +i j ck

and , ,p q r are coplanar.

∴∴∴∴ [ ]pqr = 0

∴∴∴∴ [ ]pqr =

1 1

1 1

1 1

a

b

c

= 0

Applying C1 – C

2 and C

2 – C

3

1 0 1

0 1 1

1 1

a

b

c c c

−

−

− −

= 0

∴∴∴∴ by taking (a – 1), (b – 1), (1 – c) common

∴∴∴∴ (a – 1)(b – 1)(1 – c ) =

11 0

–1

10 1

–1

1 11 –

a

b

c

c

= 0 ⇒⇒⇒⇒

∴∴∴∴ a b ≠≠≠≠ c ≠1

∴∴∴∴ (a –1) (b – 1) (c – 1) ≠ 0 ⇒⇒⇒⇒

11 0

–1

10 1

–1

1 11 –

a

b

c

c

= 0

1 1– –

1 – – 1 – 1

c

c b a= 0 ⇒⇒⇒⇒

1 1+ +

1 – –1 1 –

c

c b a= 0

1 11+ +

1 – 1 – 1 –

c

c b a

+

= 1 ⇒⇒⇒⇒

1 – + 1 1+ +

1 – 1 – 1 –

c c

c b a= 1

1 1 1+ +

1 – 1 – 1 –c b a= 1 ⇒⇒⇒⇒

1 1 1+ +

1 – 1 – 1 –a b c= 1

Q-14) Show that if four points A, B, C D with

position vectors , , ,a b c d are coplanar then

+ + =bcd cad abd abc

Ans. ∵ four points A, B, C, D are coplanar

∴∴∴∴ AB AC AD are also coplanar

∴∴∴∴ AB AC AD = 0

∴∴∴∴ ( )××××.AB AC AD = 0

( ) ( ) ( )××××– . – –b a c a d a = 0

( ) × × × ×× × × ×× × × ×× × × ×– . – – +b a c d c a a d a a = 0

( ) ( ) ( )

( ) ( ) ( )

× × ×× × ×× × ×× × ×

× × ×× × ×× × ×× × ×

. – . – .

– . + . – .

b c d b c a b a d

a c d a c a a a d= 0

×××× = 0a a

– – –

+ +

bc d bc a bad a c d

a c a aad

= 0

– – – + 0 + 0bc d abc abd c a d

= 0

+ + =bc d abd c a d abc

GROUP (C)- HOME WORK PROBLEMS

Q-1) Find a b c

i) ��

��

= 2 – 2 +3 ; =10 +3 ;

= – +2

a i j k b i k

c i j k

� ��

��

ii) = + , = –a i j b j k and = +c k i


Vectors

Ans. i) a = � ��2 – 2 + 3i j k

b = ��10 + 3i k

c = � �� + + 2i j k

abc =

2 2 3

10 0 3

1 1 2

−

−

= 2 (0 + 3) + 2 (20 – 3) + 3 (–10 – 0)

= 6 + 34 – 30

= 10

ii) ∴∴∴∴ abc =

3 2 1

1 1 2

3 1 2

−

= 1(1) –1(–1)

= 2

Q-2) Find ‘p’ if a b c = 0 where,

�� = + + , = – +a i j k b i j k� �� and

�� = 2 +3 +c i j pk��

Ans. i) a = � �� + +i j k

b = � �� – +i j k

c = � ��2 + 3 +i j pk

abc = 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

P = 2

Q-3) Find the volume of parallelopiped whose

co-terminus edges are

i) ��

��

= – 2 – , = 3 +2 + ,

= + +5

a i j k b i j k

c i j k

� ��

��

ii) ��

��

= +2 , = 2 + + ,

= 2 +2 +3

a i k b i j k

c i j k

� ��

��

Ans. i) a = � � �– 2 +j j k

b = � � �3 – 2 +j j k

c = � � �+ + 5j j k

abc =

1 –2 –1

3 2 1

1 1 5

= 1 (10 – 1) + 2 (15 – 1) + 1 (3 – 2)

= 9 + 28 + 1

= 38

ii) abc =

1 0 2

2 1 1

2 3 3

= 1 (3 – 2) – 0 (6 – 2) + 2 (4 – 2)

= 1 – 0 + 4

= 5

Q-4) If A ≡≡≡≡ (1, 1, 1), B ≡≡≡≡ (2, 1, 3), C ≡≡≡≡ (3, 2, 2) and

D ≡≡≡≡ (3, 2, 4), find the volume of the

parallelopiped with AB, AC and AD as

concurrent edges.

Ans. A ≡≡≡≡ (1, 1, 1), B = (2, 1, 3), C ≡≡≡≡ (3, 2, 2),

D ≡≡≡≡ (3, 2, 4)

let , , ,a b c d be position vectors of points A,

B, C, D respectively with respect to a fixed

point.

∴∴∴∴ a = � �� + +i j k

b = � ��2 + + 3i j k

c = � ��3 + 2 + 2i k k

d = � ��3 + 2 – 4i j k

AB = –b a

= � �( ) � �( )� �2 + + 3 – + +i j k i j k

= �� + 2i k

AC = –c a

=� �( ) � �( )� �3 + 2 – 2 – + +i j k i j k

= � ��2 + +i j k

vol. of parallelopiped = AB AC AD

Vectors


=

1 0 2

2 1 1

2 1 3

= 1 (3 – 1) – 0 (6 – 2) + 2 (2 - 2)

= 2 – 0 + 0

= 2

Q-5) If the volume of the parallelopiped whose

coterminus edges are

�� –3 +2 + , 2 + – , – + 3 +2i j nk i j k i j k� � ��

is 7 find n.

Ans. Let a = � ��–3 + 2 +i j nk

b = � ��2 + –i j k

c = � ��– + 3 – 2i k k

( )abc = 7

–3 2

2 1 –1

–1 3 2

n

= 7

–3 (2 + 3) – 2 (4 – 1) + n (6 + 1) = 7

–5 – 6 + 6n + n = 7

7n = 7 + 21

7n = 28

n = 4

Q -6) Show that the following vectors are

coplanar.

i) �� 3 – 5 +2 , – + 4 – 3 , +3 – 4i j k i j k i j k� � ��

ii) �� 3 + +3 , +3 +2 , 2 +6 + 4i j k i j k i j k� � ��

Ans. i) To show that, vectors are co-planar we

have to prove that

( )abc = 0

( )abc =

–3 5 2

–1 4 –3

1 3 –4

= 3 (–16 + 9) + 5 (4 + 3) + 2 (– 3 – 4)

= –3 (7) + (7) + 2 (–7)

= –21 + 35 – 14

= 0

( )abc = 0

so,

vectors are co-planar

ii) To show that, vectors are co-planar we

have to prove that

( )abc = 0

( )abc =

3 1 3

1 3 2

2 6 4

= 3 (12 – 12) – 1 (4 – 4) + (6 – 6)

= 0

( )abc = 0

∴∴∴∴ vectors are co-planar

Q-7) If �� = + + , = – +a i j k b i j k� ��

and �� = 2 +3 +2c i j k�� find (((( )))).a b c××××

Ans. a = � �� + +i j k

b = � �� – +i j k

c = � ��2 + 3 + 2i j k

××××b c =

� ��

1 –1 1

2 3 2

i j k

= ( ) � ( ) � ( )� –2 – 3 – 2 – 2 + 3i j k

= ��–5 + 5i k

( )××××.a b c =� �( ) �( )� �+ + . –5 + 5i j k i k

= (1) (–5) + (1) (0) + (1) (5)

= –5 + 0 + 5

= 0

∴∴∴∴ , ,a b c are caplaner.

Q-8) Find ‘p’ if the following vectors are coplanar

�� = – 3 + 4 , = 2 – 5 + , = – – 6a i j k b i j pk c i j k� � ��

Ans. abc = 0


Vectors

1 –3 4

2 –5

1 –1 6

P= 0

1 (30 + P) + 3 (–12 – P) + 4 (–2 + 5) = 0

30 + P – 36 – 3P + 12 = 0

–2P + 6 = 0

P = 3

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 = � ��3 + 2 –i j k

b = � ��5 + 4 + 2i j k

c = � ��6 +3 +5i j k

d = �� +i kx

AB = –b a

=� �( ) � �( )� �5 + 4 +2 – 3 + 2 –i j k i j k

= � ��2 +2 + 3i j k

AC = –c a

=� �( ) � � �( )�6 +3 +5 – 3 + 2 –i j k j j k

= � ��3 + + 6i j k

AD = –d a

=�( ) � �( )� �+ – 3 +2 –i kx i j k

=� � �( )+ +–2 – 2 1j j k x

∴∴∴∴ points A, B, C, D are coplanar

, ,AB AC AD will also be coplanar

AB AC AD = 0

�( )

2 2 3

3 1 6

–2 –2 +1k

= 0

∴∴∴∴�( ) � ( ) + + + + + + + =

2 1 12 2 3 3 12 3 –6 2 0k k

∴∴∴∴ � � ( )+ + =2 26 – 6 – 30 –12 0k k

∴∴∴∴ � =–4 –16 0k

∴∴∴∴ �k = – 4

Q-10) If �� = 3 + 2 + , = + + 2a i j k b i j� �� and

�� = 3 – + 2c i j k�� are the co-terminus edges of

a tetrahedron. Find its volume.

Ans. volume of a Tetrahedron =1

6abc

a bc =

3 2 1

1 1 2

3 1 2−

= 3(4) – 2(– 4) + 1 (– 4)

= 12 + 8 – 4

= 16

volume of a Tetrahedron =1

6abc

=16

6

volume of a Tetrahedron =8

3

Q-11) If = 3 + 2 ,c a b show that = 0abc .

If angle between a and b is 4ππππ prove

that

2 2= 9 + 6 2 + 4 =c a ab b a a and =b b

Ans. We know that,

××××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

= ( )×××× 3 +2a b a b

= ( )× ×× ×× ×× ×3 + 2a b a b b

= ( )××××3 + 0a b a

= ( )××××3a b a

Vectors


The measure of angle between a & b

is ππππ

4

c2 = C.C

= ( ) ( )3 + 2 . 3 + 2a b a b

= 9 . + 6 . + 6 . + 4 .a a a b b a b b

= ( )∵∵∵∵9 . +12 . + 4 . + . = .a a a b b b a b b a

=

( )

2 2ππππ

∵∵∵∵

9 +12 cos + 44

= & =

a ab b

a a b b

=2 21

9 +12 + 42

a ab b

2 2 29 + 6 2 + 4a ab b

a = &a b = b

Q-12) If = 3 – 2 ,c a b show that = 0abc .

Also find c2 when the measure of the angle

between a and b is 4ππππ .

Ans. We know that,

××××b b = 0

If in a scalar triple product, two vectors are

equal, then the scalar triple product is zero.

abc = ( )××××a b c

= ( )×××× 3 – 2a b a b

= ( )× ×× ×× ×× ×3 – 2a b a b b

= ( )××××3 – 0a b a

= ( )××××3a b a

= 3 × × × × 0 = 0

The measure of angle between a & b

is ππππ

4

c2 = c.c

= ( ) ( )3 – 2 . 3 – 2a b a b

= 9 . – 6 . – 6 . + 4 .a a a b b a b b

= ( )∵∵∵∵9 . – 12 . + 4 . + . = .a a a b b b a b b a

=

( )

2 2ππππ

∵∵∵∵

9 –12 cos + 44

= & =

a ab b

a a b b

=2 21

9 –12 + 42

a ab b

c2 = 2 29 – 6 2 + 4a ab b and

a = &a b = b

Q -13) If �� = – 2 +u i j k�� , ��= 3 +v i k�� and �� = –w j k

Find ( + ).[( ) ( )]× × ×× × ×× × ×× × ×u w u v v w

Ans. u = � �� – 2 +i j k

v = ��3 +i k

w = � �–j k

let a = ( )+u w

= �� –i j

b = ( )××××u v

= � �� + +i j k

=

� ��

1 –2 1

0 1 –1

i j k

= ( ) � ( )+� 1 1i k

c = ( )××××v w

=

� ��

3 0 1

0 1 –1

i j k

= ( ) � ( ) � ( )� –1 – –3 + 3i j k

= � ��– + 3 + 3i j k

To find ( ) ( ) ( )× × ×× × ×× × ×× × ×+ .u w u v v w

c = and


Vectors

≡≡≡≡ ××××.a b c

= abc

=

1 1 0

1 1 1

–1 3 3

= 1(3 – 3) –1(3 + 1)

= – 4

∴∴∴∴ ( ) ( ) ( )× × ×× × ×× × ×× × ×+ .u w u v v w = – 4

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

Ans. i.e. To prove that

+ + =bc d c a d abd abc

Let, , ,a b cand d be the position of the points

A, B, C and D respectively. Then,

AB = –b a ,

AC = –c a ,

AD = –d a

The points A, B, C, D are coplanar.

∴∴∴∴ the vectors , ,AB AC AD are coplanar.

∴∴∴∴ the = 0AB AC AD

∴∴∴∴ – – – = 0b a c a d a

∴∴∴∴ ( ) ( ) ( )××××– . – – = 0b a c a d a

∴∴∴∴ ( ) ( )× × × ×× × × ×× × × ×× × × ×– . – – – = 0b a c d c a a d a a

where ×××× = 0a a

∴∴∴∴ ( ) ( )× × ×× × ×× × ×× × ×– . – – = 0b a c d c a a d

∴∴∴∴ ( ) ( ) ( )× × ×× × ×× × ×× × ×. – . – . –b c d b c d b a d

( ) ( ) ( )× × ×× × ×× × ×× × ×. + . + . = 0a c d a c a a a d ...(I)

Now, ( ) ( )× ×× ×× ×× ×. = 0, . = 0,a c d a a d

( )××××– .b a d = ( )××××.b d a

= bda

= abd

( )××××– .a c d = ( )××××.a d c

= a dc

= c a d

Also, ( )××××.b c d = bc d

( )××××– .b c a = ( )××××– .a b c abc

∴∴∴∴ from (I)

∴∴∴∴ + + =bc d c a d abd abc

Q-15) If (((( )))). = 1××××a b c , where �� = ,+ ++ ++ ++ +��a i j k

�� = 2 + ++ ++ ++ +��b i q j k and �� = – 4++++��c i j k ,

find q

Ans. Given that

( ).a b c×××× = 1

∴∴∴∴

1 1 1

2 1

1 1 4

q

−

= 1

∴∴∴∴ 1(4q + 1) – 1 (8 – 1) + 1 ( – 2 – 1) = 1

∴∴∴∴ 4q + 1 – 7 – 2 – q = 1

∴∴∴∴ 3q – 8 = 1

∴∴∴∴ 3q = 9

∴∴∴∴ q = 3.

Q-16) Show that, if , ,a b c are non-coplanar

vectors, then the vectors 3 + + 5a b c and

2 – 4 +3a b c are non-coplanar.

Ans. Let = – 2 ,p a c = 3 + + 5 ,q a b c

= 2 – 4 + 3r a b c

∴∴∴∴

1 0 –2

= 3 1 5

2 –4 3

pqr

= 1(3 + 20) – 0(9 – 10) – 2(–12 – 2)

= 23 + 28 ≠≠≠≠ 0

∴∴∴∴ , ,p q r are non-coplanar.

Vectors


Q-17) prove that

i) (((( )))) (((( )))) (((( ))))+ + + = 2 . × ×× ×× ×× ×

b c c a a b a b c

ii) + + + = 0 a b ca b c

iii) (((( ))))+2 – – – –

= 3

××××

a b c a b a b c

a b c

Ans. i) LHS = ( ) ( )××××+ . + +b c c a a b

= + + +b c c a a a

= ( ) ( )+b c a c a b

= ( ) ( )+a b c a b c

= ( )2 a b c

= ( )××××2 .a 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 aa

c b c c

= ( ) ( ) ( )

( ) ( ) ( )

× × ×× × ×× × ×× × ×

× × ×× × ×× × ×× × ×

. + . + .

+ . + . + .

a b a a b b a b c

a c a a c b a c c

= 0 + 0 + ( )××××.a b c + 0

– ( )××××.a b c + 0

= 0

= RHS

iii) LHS

= ( ) ( )××××. + 2 – . – – –a a b c a b a b c

= ( )× × ×× × ×× × ×× × ×

× × ×× × ×× × ×× × ×

– –+ 2 – .

– + +

a a a b a ca b c

b a b b b c

= ( )× × ×× × ×× × ×× × ×

××××

0 – + ++ 2 – .

– +

a b c a a ba b c

o b c

= ( ) × ×× ×× ×× ×+ 2 – . +a b c c a b c

=

( ) ( ) ( )

( ) ( ) ( )

× × ×× × ×× × ×× × ×

× × ×× × ×× × ×× × ×

. + . + 2 .

+ 2 . – . – .

a c a a b c b c a

b b c c c a c b c

= ( ) ( )× × ×× × ×× × ×× × ×0 + . + 2 . + 2 0 – 0 – 0a b c b c a

= +a b c b c a

= + 2a b c a b c

= 3 a b c

= RHS


Vectors

GROUP (D)- CLASS WORK PROBLEMS

Q-1) Prove that medians of a triangle are

concurrent

Ans. Let , ,a b c be the position vectors of the

vertices A, B, C and , ,d e f be the position

vectors of the midpoints D, E, F of the sides

BC, CA and AB respectively.

Then by the midpoint formula,

d =+

2

b c,

e =+

2

c a,

f =+

2

a b

∴∴∴∴ 2d = +b c ;

2e = +c a ;

2f = +a b

∴∴∴∴ 2 +d a = + +a b c

∴∴∴∴ 2 +e b = + +a b c

∴∴∴∴ 2 +f c = + +a b c

∴∴∴∴2 +

2+1

d a=

2 +

2+1

e b=

2 +

2+1

f c=

+ +

3

a b c

= g ... (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

point G and its position vector is

( )+ + /3.a b c This point of concurrence of

A

F E

CB

G

D

the medians of a triangle is called the

centroid of the triangle.

Q-2) Prove that bisectors of angles of a triangle

are concurrent

Ans. Let , ,a b c be the position vectors of the

vertices A, B, C of ∆∆∆∆ABC and let the lengths

of the sides BC,CA and AB be x, y, z,

respectively. If segments AD, BE, CF are the

bisectors of the angles A, B, C, respectively,

then

D divides the side BC in the ratio AB : BC

i.e., z : y,

E divides the side AC in the ratio BA : BC

i.e., z : x,

and F divides the side AB in the ratio AC :

BC i.e., y : x,

Hence by section formula, the postion vectors

of the points D, E and F are

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 = + .xa yb

∴∴∴∴ ( )+ +y z d xa = ( )+ +z x e yb

= ( )+ +x y f zc

= + +xa yb zc

∴∴∴∴( )

( )

+ +

+ +

y z d xa

y z x=

( )( )

+ +

+ +

z x e yb

z x y

A

F E

CB D

P

00

xx

Vectors


=( )

( )

+ +

+ +

x y f zc

x y z

=+ +

+ +

xa yb zc

x y z= p ...(Say)

This show that the point P whose position

vector is p , lies on the three bisectors AD,

BE and CF dividing them in the ratios

(y + z) : x; (z + x) : y and (x + y) : z respectively.

Hence, the three bisector segments are

concurrent in the point whose position vector

is + +

.+ +

xa yb zc

x y z

This point of concurrence of the bisectors is

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.

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

AB = DC and side AB |||||||| side DC.

∴∴∴∴ AB = DC

∴∴∴∴ –b a = –c d

∴∴∴∴ +a c = +b d

∴∴∴∴+

2

a c=

+

2

b d ... (i)

the position vectors of the mid points of

the diagonal AC and BD are ( )+

2

a c and

( )+

2

b d by (i) they are equal

∴∴∴∴ the mid points of the diagonal AC and

BD are the same. this shows that the

diagonals AC and BD bisect each other.

ii) conversely, suppose that the diagonals

AC and BD of �� ABCD bisect each other.

i.e, they have the same midpoint.

∴∴∴∴ the position vectors of these midpoints

are equal.

∴∴∴∴+

2

a c=

+

2

b d

∴∴∴∴ +a c = +b d

∴∴∴∴ –b a = –c d

∴∴∴∴ AB = DC

∴ AB= DC and side AB |||||||| side DC

∴∴∴∴ �� ABCD is parallelogram.

CB

AD

M

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 and D of the

trapezium ABCD with side AD |||||||| side BC.

Then the vectors AD and BC are parallel.

∴∴∴∴ there exists a scalar k

such that AD = k . BC

∴∴∴∴ AD+BC= .BC +BCk

= (k + 1) BC ... (i)

Let m and n be the position vectors of

the midpoints M and N of the non-

parallel sides AB and DC respectively.

then seg MN is the median of the trapezium.

by the mid point formula.

+=

2

a bm and

+=

2

d cn

∴∴∴∴ MN = –n m

=+ +

–2 2

d c a b

CB

A D

M N


Vectors

∴∴∴∴ = ( )1+ – –

2d c a b

= ( ) ( )1– + +

2d a c d

=AD+BC

2... (ii)

=( )+1 BC

2

k...[(by (i)]

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.

∴∴∴∴ + = + = +AD BC AD BC AD BC

[∵ AD and BC have same direction]

∴∴∴∴ from (2) we have

∴∴∴∴ MN = +

2

AD BC

∴∴∴∴ MN = ( )1+

2AD BC

Q-5) Using vectors, prove angle subtended in a

semi circle is right angle.

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

subtended on a semi-circle.

Let = =AC CB a and =CP r

Then =a r ...(1)

AP = AC+CP

= + ra

= +ar

BP = BC+CP

= –CB+CP

= – +a r

∴∴∴∴ AP.BP = ( ) ( )+ . –r a r a

= . – – + . – .r r r a a r a a

= 2 2– = 0r a

∴∴∴∴ ⊥⊥⊥⊥AP BP

∴∴∴∴ APB is a right angle.

Hence, the angle subtended on a semicircle

is the right angle.

a –a

r

CA B

P

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.

Then it is enough to prove that HB is

perpendicular to AC .

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.

Then ,HA a HB b= = and HC c= ,

– , –AB b a BC c b= = and –AC c a=

Now HA is perpendicular to BC

∴∴∴∴ . 0HA BC =

∴∴∴∴ ( ). – 0a c b =

∴∴∴∴ . – . 0a c a b = ...(1)

Also HA is perpendicular to AB .

∴∴∴∴ . 0HC AB =

∴∴∴∴ ( ). – 0c b a =

∴∴∴∴ . – . 0c b c a =

∴∴∴∴ . – . 0c b a c = ( ). .c a a c=∵ ...(2)

A

B C

EF

D

H

Vectors


Adding (1) and (2), we get

. – . 0c b a c =

∴∴∴∴ ( )– . = 0c a b

∴∴∴∴ ( ). – = 0b c a

∴∴∴∴ HB.AC = 0

∴∴∴∴ HB is perpendicular to AC .

∴∴∴∴ the altitudes of ∆∆∆∆ABC are concurrent.

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).

Ans. The position vectors , ,a b c of the vertices A,B

& C are

a = 0 + 4 + 0i j k

b = 0 + 0 + 3i j k

c = 0 + 4 + 3i j k

AB = –b a

= ( ) ( )0 + 0 + 3 – 0 + 4 + 0i j k i j k

= 0 – 4 =3i j k

BC = –c b

= ( ) ( )0 + 4 + 3 – 0 + 0 + 3i j k i j k

= 0 + 4 + 0i j k

AC = –c a

= ( ) ( )0 + 4 + 3 – 0 + 4 + 0i j k i j k

= 0 – 0 +3i j k

Let x = BC ,

y = AC , and

z = AB

x = 2 2 20 + 4 +0

= 16 = 4

y = ( )22 20 + –0 +3

= 9 = 3

z = ( )22 20 + –4 +3

= 16 + 9

= 25 = 5

If H ( )h is the incentre of ∆∆∆∆PQR, then

h =+ +

+ +

xa yb zc

x y z

h =( ) ( ) ( )4 4 +3 3 +5 4 3

4+3 +5

j k j k+

h =16 + 9 + 20 +15

12

j k j k

h = 36 + 24

12

j k

h =0 +36 +24

12

j j k

h = 0 +3 +2i j k

H = (0,3,2)

Q-8) Prove by vector method, sin(α α α α + ββββ) =

sin αααα .cos ββββ + cos αααα . sin ββββ .

Ans.

Let ∠∠∠∠XOP and ∠∠∠∠XOQ be in standard position

and mXOP = –αααα, mXOQ = ββββ.

Takes a pouint A on ray OP and a point B on

ray OQ such that OA = OB = 1.

Since cos (–αααα) = cos αααα

and sin (–αααα) = –sin αααα,

A is (cos (–αααα), sin (–αααα)),

i.e, (cos αααα, –sin αααα)

B is (cos ββββ, sin ββββ)

∴∴∴∴ OA = ( ) ( )α αα αα αα αcos – sin . + 0.i j k

Y

B

Q

X

PA

O –2–2

β


Vectors

OB = ( ) ( )β ββ ββ ββ βcos + sin . + 0.OB i j k

∴∴∴∴ ××××OA OB =

� �

α αα αα αα α

β ββ ββ ββ β

cos – sin 0

cos sin 0

i j k�

= (cos αααα sin β β β β + sin αααα cos ββββ)k

... (i)

The angle between OA and OB is α α α α + ββββ.

Also OA ,OB lie in the XY-plane.

∴∴∴∴ the unit vector perpendicular to OA and OB

is k .

∴∴∴∴ ××××OA OB = [OA.OB sin (α α α α + ββββ)]k

= sin (α α α α + ββββ).k ...(ii)

from (i) and (ii),

sin (α α α α + ββββ) = sin αααα cos ββββ + cos αααα sin ββββ.

GROUP (D) – HOME WORK PROBLEMS

Q-1) By vectors method. prove that the

diagonals of rhombus are perpendicular to

each other.

Ans. Let �OACB be a rhombus and =OA a ,

=OB b . Then OA = OB

∴∴∴∴ =a b

∴∴∴∴ a = b.

Now = +OC a b .

∴∴∴∴ 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

( )+ /2a b are and ( )+ + 0a b i.e., ( )+ /2a b .

These are equal.Hence the diagonals OC and

BA have the same midpoint.

∴∴∴∴ these diagonals bisect each other ...(1)

Now OC = +a b

and BA = +BC CA

= –a b

∴∴∴∴ .OC BA = ( ) ( )– . –a b a b

= . – . + . – .a a a b b a b b

= a2 – b2 ( )∵∵∵∵ . = .a b b a

= 0 ( )∵∵∵∵ =a b

∴∴∴∴ ⊥⊥⊥⊥OC BA i.e., the diagonals OC and BA are

at right angles. ...(2)

∴∴∴∴ from (1) and (2), the diagonals of a rhombus

bisect each other at right angles.

Conversely : Let the diagonals OC and BA of

the quadrilateral.OACB bisect each other at

right angles.

Since the diagonals bisect each other, the

quadrilateral OACB is a parallelogram.

Now,

∴∴∴∴ ⊥⊥⊥⊥OC BA

∴∴∴∴ .OC BA = 0

∴∴∴∴ ( ) ( )+ . –a b a b = 0

∴∴∴∴ . – . + . – .a a a b b a b b = 0

∴∴∴∴2 2– . + . –a a b b a b = 0

∴∴∴∴2 2–a b = 0

∴∴∴∴2 2=a b = 0

∴∴∴∴ =a b

∴∴∴∴ ( ) ( )=l OA l OB

i.e., adjacent sides 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.

O A

CB

a

b

a

b

ab+

a

b–

Vectors


Q-2) Using vectors prove that, if the diagonals

of a parallelogram are at right angles then

it is rhombus.

Ans. Let ABCD be a parallelogram

Let AC = +AB BC

= +a b

and BD = +BA AD

= – +a b

∴∴∴∴ the diagonals AC and BD are perpendicular

∴∴∴∴ .AC BD = 0

⇒⇒⇒⇒ ( ) ( )+ – +a b a b = 0

– . + . – . + .a a a b b a b b = 0

b2 = a2 ⇒⇒⇒⇒ b = a

⇒⇒⇒⇒ l(AB) = l(BC) = l(CD) = l(AD)

∴∴∴∴ the parallelogram ABCD is a rhombus .

Q-3) Show by vector method. If the diagonal of

a parallelogram are congruent then it is a

rectangle.

Ans. Let ABCD be parallelogram

Let = , =AB a AD b

From ∆ ∆ ∆ ∆ ABC, by triangle law,

AC = +AB BC

= +a b

From ∆ ∆ ∆ ∆ ABC, by triangle law,

BD = +BC CD

= –b a

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 ⇒⇒⇒⇒ 2 2=AC BD

( ) ( )+ . +a b a b = ( ) ( )– . –b a b a

. + . + . + .a a a b b a b b

= . – . – . + .b b b a a b a a

2 2+ + 2 .a b a b

= 22 – 2 . +b a b a

2 . + 2 .a b a b

= 0 ⇒⇒⇒⇒ ( )4 .a b = 0 ⇒⇒⇒⇒ .a b = 0

⊥⊥⊥⊥ ⇒⇒⇒⇒ ⊥⊥⊥⊥ ⇒⇒⇒⇒ ⊥⊥⊥⊥a b AB AD AB AD

∴∴∴∴ The parallelogram ABCD is a rectangle.

b

BA

DC

a

b

a

Q-4) Show by vector method that the sum of the

square of the diagonals of a parallelogram

is equal to the sum of the sequare of its

sides.

Ans. Let � ABCD be a parallelogram.

Let =AB p and =AD q

Since ABCD is a parallelogram,

= =DC AB p and = =BC AD q

= + = +AC AB BC p q

∴∴∴∴ ( ) ( )2 = . = + . +AC AC AC p q p q

= . + . + . + .p p p q q p q q

= p2 + q2 + 2p.q ( ). = .p q q p∵

BD = +BC CD = –BC DC = –q p

∴∴∴∴ BD2 = .BD BD = ( ) ( )– . –q p q p

= . – . – . + .q q q p p q p p

= p2 + q2 + 2p.q ( ). = .p q q p∵

∴∴∴∴ AC 2 + BD 2 = ( )( )

2 2

2 2

+ +2 .

+ + – 2 .

p q p q

p q p q

= p2 + q2 + p2 + q2

A B

C

q

p

q

p


Vectors

= AB2 + BC 2 + CD 2+ DA 2

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.

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

having BC as diameter.

Ans. A ≡≡≡≡ (1, 2, 3), B ≡≡≡≡ (3, −−−−1, 5), C ≡≡≡≡ (4, 0, −−−−3)

let , ,a b c be position vectors of points A, B

& C respectively w.r.t a fixed point.

a = +2 +3i j k

b = 3 – +5i j k

c = 4 – 3i k

AB = –b a

= ( ) ( )3 – +5 – + 2 + 3i j k i j k

= 2 – 3 + 2i j k

AC = –c a

= ( ) ( )4 – 3 – + 2 + 3i k i j k

= 3 – 2 – 6i j k

.AB AC = ( ) ( )2 – 3 + 2 . 3 – 2 – 6i j k i j k

= ( ) ( ) ( ) ( ) ( ) ( ) ( )2 3 . + –3 –2 – + 2 –6 .i i j j k k

= 6 (1) + 6 (1) − − − − 12 (1)

= 0

seg AB ⊥⊥⊥⊥ seg AC

i.e. seg BC subtends right angle at A

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

p = – 2 +i j k

q = – – 3i j k

r = – 2 – 3i j k

PQ = –q p

= ( ) ( )+ – 3 – – 2 +i j k i j k

= 0 +3 – 4i j k

QR = –r q

= ( ) ( )– 2 – 3 – + – 3i j k i j k

= 0 – 3 +i j Ok

PR = –r p

= ( ) ( )– 2 – 3 – – 2 +i j k i j k

= 0 +0 – 4i j k

Let x = QR ,

y = PR

z = PQ

x = ( )0 + –3 + 022 2

= 9 = 3

y = ( )0 +0 + –422 2

= 16 = 4

z = ( )0 +0 + –422 2

= 9 +16

= 25 = 5

If H ( )h is the incentre of the triangle PQR,

then

h =+ +

+ +

x p yq zr

x y z

h

=3 – 6 +3 + 4 + 4 –12 +5 –10 –15

3 + 4 +5

i j k i j k i j k

=12 –12 – 24

12

i j k

h = – – 2i j k

h = (1, –1, –2)

Vectors


Q -7) Using vectors, prove that the perpendicular

bisectors of the sides of a triangle are

concurrent.

Ans.

Let D, E, F be the midpoints of the sides BC,

CA and and AB of ∆ ∆ ∆ ∆ ABC.

Let the perpendicular bisector of the sides

BC and AC meet each other in the point O.

Choose O as the origin and let , , , ,a b c d e and

f be the position vectors of the points A, B,

C, D, E, F respectively.

Here we have to prove that =OF f is

perpendicular to = –AB b a .

By the midpoint formula,

d =+

2

b c,

e =+

2

c a,

f =+

2

a b,

Now =OD d is perpendicular to = –BC c b .

∴∴∴∴ ( ). – = 0d c b

∴∴∴∴ ( )+. – = 0

2

b cc b

∴∴∴∴ ( ) ( )+ . – = 0c b c b

∴∴∴∴ . + . – . – . = 0c c b c c b b b

∴∴∴∴ c2 – b2 = 0

( )2 2∵∵∵∵ . = , . = . = .c c c b b b andc b b c

∴∴∴∴ c2 – b2 ...(1)

Also, =OE e is pcpendicular to = –AC c a

∴∴∴∴ ( ). – = 0e c a

∴∴∴∴ ( )+. – = 0

2

c ac a

∴∴∴∴ as above c2 – b2 = 0 ...(2)

∴∴∴∴ from (1) and (2), we get

b2 = a2

∴∴∴∴ b2 = a2 = 0

∴∴∴∴ ( ) ( )+ . – = 0b a b a

∴∴∴∴ ( ) ( )+ . – = 0b a b a

∴∴∴∴ ( ). – = 0f b a

∴∴∴∴ =f OF

is perpendicular bisector of the sides of ∆∆∆∆ ABC

are concurrent.

This point of concurrrent of the perpendicular

bisector of the sides of a triangle is called the

circumcentre of the triangle.

B CD

EF

A

Q-8) Prove by vector method, that, a

quadrilateral is a square if and only if

diagonals are congruent and bisect each

other at right angles.

Ans. Let ABCD be a square.

Let , , ,a b c d be the position vectors A, B, C,

D respectively.

Since ABCD is a sequare, we have,

=AB DC

∴∴∴∴ – = –b a c d

∴∴∴∴ + = +b d a c

∴∴∴∴+ +

= =2 2

b d a cp ... (Say)

This shows that the point P whose position

vector is p is the midpoint of BD as well as

of AC.

dD cC

bBaA


Vectors

∴∴∴∴ the diagonals BD and AC bisect each other

at P.

∴∴∴∴ the diagonals of a square bisect each other

...(i)

Now, = = +AC BC BC AB

and ∵∵∵∵= + = + ... =BD BC CD BC BA CD BA

= –BC AB

∴∴∴∴ .AC BD = ( ) ( )+ . –BC AB BC AB

= . – . + . – .BC BC BC AB AB BC AB AB

=2 2– . + . –BC AB BC AB BC AB

= ( ) ( )2 2–l BC l AB

= 0 ... ( ) ( )=l AB l BC ∵

∴∴∴∴ AC is perpendicular to BD

∴∴∴∴ diagonals AC and BD are at right angles

... (ii)

Also, .AC AC = ( ) ( )+ . +BC AB BC AB

= . + . + . + .BC BC BC AB AB BC AB AB

= ( )22

⊥⊥⊥⊥∵∵∵∵+ 0 +0 + ... rBC AB AB BC

= ( ) ( )2 2+BC AB

and .BD BD = ( ) ( )– . –BC AB BC AB

= . – . – . + .BC BC BC AB AB BC AB AB

= ( )22

⊥⊥⊥⊥∵∵∵∵– 0 – 0 + ... rBC AB AB BC

= ( ) ( )2 2+BC AB

∴∴∴∴ .AC AC = .BD BD

∴∴∴∴2 2=AC BD

∴∴∴∴ l(AC) = l(BD) ...(iii)

∴∴∴∴ from (i), (ii) and (iii), the diagonals of a square

are congruent and bisect each other at right

angles.

Conversely : Let the diagonals AC and BD of

the quadrilaeral ABCD are congruent and

bisect each other at right angles.

Since the diagonals bisect each other, the

quadrilateral ABCD is a parallelogram.

Now, r⊥⊥⊥⊥ = 0AC BD

∴∴∴∴ . = 0AC BD

∴∴∴∴ ( ) ( )+ . – = 0BC AB BC AB

∴∴∴∴ . – . + . – . = 0BC BC BC AB AB BC AB AB

∴∴∴∴2 2– . . + . – = 0BC AB BC AB BC AB

∴∴∴∴2 2– = 0BC AB

∴∴∴∴2 2=BC AB

∴∴∴∴ =BC AB

∴∴∴∴ l(BC) = l(AB)

i.e., the adjacent sides AB and BC of the

parallelogram ABCD are equal.

∴ ABCD is a rhombus.

Also, l(AC) = l(BD)

∴ =AC BD

∴2 2=AC BD

∴ . = .AC AC BD BD

∴ ( ) ( ). .BC AB BC AB+

= ( ) ( )– . –BC AB BC AB

∴ . + . + . + .BC BC BC AB AB BC AB AB

= . – . – . + .BC BC BC AB AB BC AB AB

∴ . + .AB BC AB BC

= – . – .AB BC AB BC

∴ 4 . = 0AB BC

∴ . = 0AB BC

∴∴∴∴ r⊥⊥⊥⊥AB BC

i.e., the adjacent sides of a rhombus ABCD

are perpendicular to each other.

Hence, ABCD is a square.

Vectors


Q-9) Using vector method, find the incentre of

a triangle whose vertices are P(0,2,1),

Q(– 2,0,0) and R(–2,0,2).

Ans. The position vectors , ,p q r of the vertices

P,Q,R are

p = � � �2 + , = 2j k q k and

r = �–2 + 2i k�

∴∴∴∴ PQ = –q p

=� �( )–2 – 2 +2i j k�

= � �–2 – 2 –i j k�

QR = –r p

=�( ) ( )–2 + 2 – –2i k i� �

= �2k and

PR = –r p

=�( ) �( )–2 +2 – 2 +i k i k� �

= � �–2 – 2 +i j k�

Let x = QR ,

y = PR and

z = PQ

∴∴∴∴ x = 20 + 0 + 2 = 2

y = ( ) ( )2 2 2–2 + –2 +1 = 3

and y = ( ) ( )2 2 2–2 + –2 +1 = 3

If H ( )h is the incentre of the triangle PQR, then

h =+ +

+ +

x p yq zr

x y z

=

� �( ) ( ) �( )2 2 + + 3 –2 + 3 –2 +2

2+3+3

j k i i k� �

=� � �( )1

4 + 2 – 6 – 6 + 68

j k i i k� �

=� �( )1

–12 + 4 + 88

i j k�

∴∴∴∴ h =� �3 1

– + +2 2i j k�

∴∴∴∴ H =3 1

– , ,12 2

Q-10)Prove that the segment joining the mid-

points of the diagonals of a trapezium is

parallel to the parallel sides and equals half

its difference.

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.

Then the vectors AD and BC are collinear.

∴∴∴∴ there exists a non-zero scalar k, such that

AD = .k BC

∴∴∴∴ –AD BC = . –k BC BC = ( )–1k BC ... (i)

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,

m = +

2

a c and n =

+

2

b d

∴∴∴∴ MN = –n m = + +

–2 2

b d a c

= ( ) ( )1– – –

2d a c b

= ( )1–

2AD BC =

–1

2

kBC ... [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.

Now, AD and BC are collinear vectors.

∴∴∴∴ –AD BC = –AD BC = –AD BC

∴∴∴∴ from (ii) we have,

A D

B C

MN


Vectors

MN = ( )1–

2AD BC

∴∴∴∴ MN = 1

–2AD BC =

1– .

2AD BC

Q-11) If the lengths of two non-parallel sides of

a trapezium are equal then, prove by vector

method that lengths of their diagonals are

equal.

Ans.

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

∴∴∴∴ ∠∠∠∠D =∠∠∠∠C ...(ii)

Now, = +AC AD DC and = +BD BC CD

∴∴∴∴ ( ) ( ). = + . +AC AC AD DC AD DC

= . + . + . + .AD AD AD DC DC AD DC DC

= 2 2

+2 . +AD AD DC DC

( )∵∵∵∵ . = .AD DC DC AD

= ( )2 2

ππππ+2 . cos – +AD AD DC D DC

= 2 2

– 2 . cos +AD AD CD D DC ...(iii)

and ( ) ( ). = + . +BD BD BC CD BC CD

= 22

+2 . +BC BC CD CD

( )∵∵∵∵ . = .BC CD CD BC

= ( )22

ππππ+2 cos – +BC BC CD C CD

= 22

– 2 cos +BC BC CD C CD

=22

– 2 . cos +AD AD CD D CD ...(iv)

[By (i) and (ii)]

DM N

C

BA

π –D π –C

From (iii) and (iv), we get,

. = .AC AC BD BD

∴∴∴∴2 2

=AC BD

∴∴∴∴ =AC BD

∴∴∴∴ l (AC) = l (BD)

Hence, the diagonal of the tranpezium are

euqal.

ICSE Coaching Classes for Std. IX & X in Mumbai, Navi ...

Documents

Transcript of ICSE Coaching Classes for Std. IX & X in Mumbai, Navi ...