3i − 4 j

TEXTUAL QUESTIONS aij =

4
3 − 4 −1 1
Exercise 7.1 a11 =

4
=
4
=
4
1. Construct an m × n matrix A = [aij], where aij is 3−8 −5 5

m
a12 =
= =
given by 4 4 4
(i − 2 j )2 3 − 12 −9 9
(i) aij = with m = 2, n = 3 a13 =
= =
2 4 4 4

co
3i − 4 j 3 − 16 −13 13
(ii) aij = with m = 3, n = 4 a14 =
= =
4 4 4 4
Solution : 3 ( 2) − 4 (1) 6−4 2
a21 =
= =
(i − 2 j )2 4 4 4

.
(i) Given aij = with m = 2, n = 3 3 ( 2) − 4 ( 2) 6−8 2

ks
2 a22 =
= =
we need to construct a 2 × 3 matrix. 4 4 4
3 ( 2) − 4 (3) 6 − 12
\ a11 =
(1 − 2 (1))2 =
( −1)2
=
1 a23 =
= =
6
2 4 4 4
2 2
3 ( 2) − 4 ( 4)

oo
6 − 16 10
a12 =
(1 − 2 (2))2 =
( −3)2
=
9 a24 =

4
=
4
=
4
2 2 2
3 (3) − 4 (1) 9−4 5
a13 =
(1 − 2 (3))2 =
( −5)2
=
25 a31 =

4
=
4
=
4
ab
2 2 2 3 (3) − 4 ( 2) 9−8 1
a32 =
= =
a21 =
(2 − 2 (1)) 2

=
0
=0
4 4 4
2 2 3 (3) − 4 (3) 9 − 12 3
a33 =
= =
(2 − 2 (2))2 ( −2)2 4 4 4 4
ur

a22 = = = 3 (3) − 4 ( 4)
2 2 2 9 − 16 7
a34 = = =
a23 =
(2 − 2 (3))2 =
( −4)2
=
16
4 4 4
2 2 2 1 5 9 13 
.s

 a11 a12 a13   4 4 4 4

\ A = \B = 2 2 6 10 
 a a23   4 4 4 4
21 a22
5 1 3 7 
w

1 9 25   4 4 4 4 
=  2 2 2
1 5 9 13
 0 4 16  1
2 2 = 2 2 6 10
4 
w

1 1 9 25 5 1 3 7 
=
2  0 4 16 
2. Find the values of p,
q, r, and s if
w

 
3i − 4 j  1 0 −4 
(ii) Given aij = with m = 3, n = 4. 2
 p −1 0 3
−31 − q  
4    
7 r + 1 9 3 
 =
  7 9
Let B be a 3×4 matrix with entries as 2 
 −2 8 s − 1   
 a11 a12 a13 a14   
 −2 8 − π 
B =  a21 a22 a23 a24   
 a31 a32 a33 a34 
Solution : 3 2 8
1 Also given A – 2B =   ... (2)
 p2 − 1 0 0 −4   −2 1 −7 
−31 − q 3   
   
Given  7 r +1 9  = 7 3 9  −12 12 0 
(1) × 2 ⇒4A – 2B = 
  2
 8 −4 −2
 −2 8 s − 1   −2 8 − π 
  (−)

m
  ( − ) (+ )
Since the matrices are equal, the corresponding entries on 3 2 8
(2) ⇒ A − 2B =  
both sides are equal.  −2 1 −7 
\ p2 – 1 = 1 ⇒ p2 = 2 ⇒ p = ± 2

co
[Equating a11]  −15 10 −8
Subtracting, 3A =  
– 31 – q3 = – 4 ⇒ – q3 = – 4 + 31  10 −5 5 
[Equating a13]
1  −15 10 −8
⇒ – q3 = 27 ⇒ A =
3  10 −5 5 

.
q3 = – 27 = (–3)3
Substituting the matrix A in (1) we get,

ks
⇒ q = –3
3− 2 1 1  −30 20 −16  −6 6 0 
Also r + 1 = 3 2 ⇒ r = 3 2 − 1 = = 2   –B =  
2 3  20 −10 +10  4 −2 −1
[Equating a22]

oo
1  −30 20 −16  −6 6 0 
s – 1 = – p ⇒ s = 1 – p
⇒   –   =B
3  20 −10 +10  4 −2 −1
[Equating a33]
p = ± 2 , q = – 3, r = 1/2, s = 1– p.

 20 −16 
 −10 + 6 −6 − 0
ab
2 x + y 4 x  ⇒ B =
3 3

3. Determine the value of x + y if   −10
 5 x − 7 4 x  20 − 4 +2
10
+1 
 7 7 y − 13  3 3 3 
= 
y x + 6 
 20 − 18 −16 
 −4 3 
ur

 2 x + y 4 x   7 7 y − 13 3
Solution : Given  = =  
 5x − 7 4 x   y x + 6   20 − 12 −10 + 6 10 + 3 
 3 3 3 
Equating the corresponding entries on both
.s

sides we get,  −4 2 −16 

 3 3 1  −12 2 −16
⇒B= \B= 
2x + y = 7 [Equating a11]  ... (1) 8 −4 13  3 8 −4 13 
4x = x + 6 [Equating a22]  ... (2)  3 3 3 
w

From (2), 4x – x = 6 ⇒ 3x = 6 ⇒ x = 6 ⇒ x = 2 1 a 4
3 5. If A =   , then compute A .
Substituting x = 2 in (1) we get,  0 1
1 a 
w

4+y = 7⇒ y=7–4 ⇒ y=3 Solution : Given A =  

\ x + y = 2 + 3 = 5
0 1 
4. Determine the matrices A and B if they satisfy 2A 1 a  1 a  1 + 0 a + a  1 2a 
⇒A2 =   =   = 
w

 
 6 −6 0  3 2 8 0 1  0 1  0 + 0 0 + 1  0 1 
–B+   = 0 and A – 2B =  
 −4 2 1  −2 1 −7  1 2a  1 2a 
A4 = A2.A2 =   
 6 −6 0 0 1  0 1 
Solution : Given 2A – B +   =0
 −4 2 1 1 + 0 2a + 2a  4 1 4a 
A4 =  ;A = 
0 + 1 
 −6 6 0  
⇒ 2A – B =  0 + 0 0 1 
  ... (1)
 4 −2 −1
 cosα −sinα   4 2
6. Consider the matrix Aα =   7. If A =   and such that (A – 2I)(A – 3I) = 0,
 sinα cosα   −1 x 
(i) Show that Aα Ab = A(α + b) find the value of x.
(ii) Find all possible real values of α satisfying  4 2
the condition Aα + AαT = I. Solution : Given A =  

m
 −1 x 
Also, (A – 2I) (A – 3I) = 0
cos α − sin α 
Solution : Given Aa =   4 2 1 0
 sin α cos α 

co
\ A – 2I =   – 2 
 −1 x  0 1
cos β − sin β 
Ab =    4 − 2 2 − 0
 sin β cos β  =  
 −1 − 0 x − 2
cos α − sin α  cos β − sin β 

.
(i) \ Aa Ab =    2 2 
 sin α cos α   sin β cos β  =  

ks
 −1 x − 2
cos α cos β − sin α sin β − cos α sin β − sin α cos β 
=    4 2 1 0
sin α cos β + cos α sin β − sin α sin β + cos α cos β  A – 3I =   – 3 
 −1 x   0 3
cos (α + β ) − sin (α + β )
=    4 − 3 2 − 0
 sin (α + β ) cos (α + β ) 

oo
=  
 −1 − 0 x − 9
since sin (α + β ) = sin α cos β + cos α sin β 
   1 2 
 cos (α + β ) = cos α cos β − sin α sin β  =  
 −1 x − 9
Aa.Ab = Aa+b
ab
2 2  1 2 
Hence proved. \ (A – 2I) (A – 3I) =   
 −1 x − 2  −1 x − 9
cos α − sin α   0 0
(ii)
Given Aa = 
 sin α cos α  =  
 0 0
ur

 cos α sin α   2−2 4 + 2 ( x − 9)   0 0

A aT =   ⇒  =  0 0
 − sin α cos α   −1 − 1( x − 2) −2 + ( x − 2) ( x − 9)  
Also, it is given that Aa + AaT = I  0 4 + 2 x − 18  0 0
.s

⇒  =
cos α − sin α   cos α sin α   1 0  −1 − x + 2 −2 + x − 11x + 18 0
2 0
⇒ + =
 sin α cos α   − sin α cos α  0 1  0 2 x − 14  0 0
⇒  = 
w

 − x + 1 x − 11x + 16 0 0
2
 2 cos α 0  1 0
⇒ =
 0 2 cos α  0 1 Equating the corresponding entries we get,
w

Equating the corresponding entries on both sides, we ⇒ 2x – 14 = 0 or – x + 1 = 0

get
1 ⇒ 2x = 14 or – x = – 1
π
2cos a = 1 ⇒cos a = = cos . ⇒ x = 7 or x = 1
w

2 3
π Since x = 1 alone satisfies the equation (A – 2I)
⇒ a = 2np ± , n∈Z 
3 (A – 3I) = 0, we get x = 1.
[ cos a = cos q ⇒ a = 2np ± q, nÎZ ]

π
\ a = 2np ± , n∈Z
3
  1 0 0  21 0 34
 0 1 0  
8. If A =  2
 , show that A is a unit matrix. = 12 8 23
a b −1  34 0 55 ... (3)
1 0 0  Substituting (2) and (3) in (1) we get,
 
Solution : Given A =  0 1 0 

m
 21 0 34 5 0 8   1 0 2 1 0 0
 a b −1 12 8 23 –6  2 4 5  +7  0 2 1  +k 0 1 0 = 0
       
1 0 0  1 0 0  34 0 55 8 0 13  2 0 3 0 0 1
  0 1 0 

co
A2=A.A =  0 1 0   
 a b −1  a b −1  21 − 30 + 7 + k 0 34 − 48 + 14 + 0 
 1 + 0 + 0 0 + 0 + 0 0 + 0 + 0  
⇒  12 − 12 + 0 + 0 8 − 24 + 14 + k 23 − 30 + 7 + 0 
 0 + 0 + 0 0 + 1 + 0 0 + 0 + 0
=    34 − 48 + 14 + 0 0 + 0 + 0 + 0 55 − 78 + 21 + k 
 
 a + 0 − a 0 + b − b 0 + 0 + 1

.
 0 0 0
 

ks
1 0 0 = 0 0 0
 
= 0 1 0 0 0 0
0 0 1
2
 −2 + k 0 0   0 0 0
\ A is a unit matrix.  0   0 0 0
−2 + k 0
 = 

oo
 
 1 0 2  0 0 −2 + k  0 0 0
 
9. If A =  0 2 1 and A3 – 6A2 + 7A + kI = 0, find Equating the corresponding entries both sides, we get
 2 0 3 –2+k = 0
k = 2
ab
the value of k. [Hy - 2018]
10. Give your own examples of matrices satisfying the
 1 0 2
  following conditions in each case:
Solution : Given A =  0 2 1 
(i) A and B such that AB ≠ BA.
 2 0 3 (ii) A and B such that AB = 0 = BA, A ≠ 0 and
ur

Also, A3 – 6A2 + 7A + kI = 0  ... (1) B ≠ 0.

 1 0 2  1 0 2 (iii) A and B such that AB = 0 and BA ≠ 0.
    3 −1
= 0 2 1 0 2 1
A2
 2 1 4
.s

2 2 
 2 0 3  2 0 3 Solution : (i) Let A =   and B =  
 4 1 5 1 3 
1 + 0 + 4 0 + 0 + 0 2 + 0 + 6
 3 −1
= 0 + 0 + 2 0 + 4 + 0 0 + 2 + 3  2 1 4 
w

  AB =   2 2
 2 + 0 + 6 0 + 0 + 0 4 + 0 + 9 
 4 1 5  
1 3 
5 0 8  6+2+4 − 2 + 2 + 12 12 12
w

  =  =  
=  2 4 5   ... (2) 12 + 2 + 5 −4 + 2 + 15  19 13
8 0 13  3 −1
   2 1 4
 5 0 8   1 0 2 BA =  2 2 
w

 4 1 5
   1 3   
A3 = A2.A =  2 4 5   0 2 1 
8 0 13  2 0 3 6−4 3 − 1 12 − 5  2 2 7 
 5 + 0 + 16 0 + 0 + 0 10 + 0 + 24  4+8 2 + 2 8 + 10  = 12 4 18
=    
=  2 + 0 + 10 0 + 8 + 0 4 + 4 + 15   2 + 12 1 + 3 4 + 15 14 4 19
 
8 + 0 + 26 0 + 0 + 0 16 + 0 + 39  \ AB ≠ BA
 0 −3  4 3 12. If A is a square matrix such that A2 = A, find the
(ii) Let A =   and B =   value of 7A - (I + A)3.
0 4   0 0 Solution : Given A is a square matrix and A2 = A
0 −3  4 3 0 + 0 0 + 0 0 0 Consider 7A – (I + A)3 = 7A – (I3 + 3I2A + 3IA2 + A3)
AB =  = =
0 4   0 0 0 + 0 0 + 0 0 0
[ (a + b)3 = a3 + 3a2b + 3ab2 + b3]
4 3 0 −3 0 + 0 −12 + 12

m

BA =  = = 7A – (I + 3A + 3A + A2.A)
2
0 0 0 4  0 + 0 0 + 0  [ I3 = I, I2 = I]
0 0
=  = 7A – (I + 3A + 3A + A.A)

0

co
0 [ A2 = A]
Hence AB = 0 = BA and A ¹ 0, B ¹ 0. = 7A – (I + 7A)
= 7A – I – 7A = – I
 0 1   1 0 

.
(iii) Let A =   and B =   13. Verify the property A(B + C) = AB + AC, when the
 0 0  0 0 matrices A, B, and C are given by

ks
 0 1  1 0  0 + 0 0 + 0  3 1 4 7
AB =    0 0 =  0 + 0 0 + 0  2 0 −3
 0 0         
A= 
1 4 5  , B =  −1 0 , and C =  2 1 .
 0 0    4 2  1 −1
=  

oo
 0 0   3 1
1 0  0 1   0 + 0 1 + 0   2 0 −3  −1 0
Solution : Given A =  ,B= 
BA = 
0 0   0 0 =  0 + 0 0 + 0  1 4 5  
      4 2
0 1 4 7 
= 
ab
  
 0 0 and C =  2 1 
\ AB = 0 and BA ¹ 0 1 −1

11. Show that f (x)f (y) = f (x + y), where  3 1 4 7 

 −1 0  2 1 
cosx −sinx 0 B + C =  +
ur


 sinx cosx 0  4 2 1 −1

f (x) =   .
 0 0 1  3 + 4 1 + 7   7 8
   
cos x − sin x 0 =  −1 + 2 0 + 1 = 1 1
.s

   4 + 1 2 − 1  5 1
Solution : Given f(x) =  sin x cos x 0
 0 0 1  7 8
 2 0 −3 
1 1
w

cos x − sin x 0 cos y − sin y 0 \ A(B + C) =    

 sin x cos x 0  sin y cos y 0 1 4 5   
f(x). f(y) =   5 1
 
 0 0 1  0 0 1 14 + 0 − 15 16 + 0 − 3
w

=  
cos x cos y − sin x cos y − cos x sin y − sin x cos y 0  7 + 4 + 25 8 + 4 + 5 
   −1 13
=  sin x cos y + cos x sin y − sin x sin y + cos x cos y 0
LHS = A(B + C) = 36 17   ... (1)
w

 0 0 1  
cos ( x + y ) − sin ( x + y ) 0  3 1
   2 0 −3 
=  sin ( x + y ) cos ( x + y ) 0 = f (x + y) AB =    −1 0

  1 4 5 
0 0 1  4 2
[ cos (x + y) = cos x cos y – sin x sin y = f (x + y)  6 + 0 − 12 2+ 0 − 6
=  
sin (x + y) = sin x cos y + cos x sin y] 3 − 4 + 20 1 + 0 + 10
  −6 −4 a–8 = –7⇒a=–7+8⇒ a=1
=   (3) × 2 ⇒ 2c + 8d = 4
19 11  (–) (–) (–)
4 7  (4) ⇒ 2c + 5d = 4
 2 0 −3 
AC =   2 1 3d = 0 ⇒ d=0
1 4 5   
1 −1 Substituting d = 0 in (3) we get,

m
8 + 0 − 3 14 + 0 + 3  5 17  c = 2
=  = 
 4 + 8 + 5 7 + 4 − 5  17 6  1 −2 ­­­­­­­­
\A =  
RHS = AB + AC 2 0 

co
 4 5
   2 −1 1
 −6 −4  5 17  15. If AT =  −1 0 and B =  , verify the
=  +   7 5 −2
19 11  17 6   2 3
following
 −6 + 5 −4 + 17   −1 13

.
=  = 36 17  (i) (A + B)T= AT + BT = BT + AT
19 + 17 11 + 6   

ks
(ii) (A – B)T = AT – BT (iii) (BT)T = B.
 ... (2)
 4 5
From (1) and (2), A(B + C) = AB + AC.    2 −1 1 
Solution : Given AT =  −1 0 and B =  
 2 3 7 5 −2
14. Find the matrix A which satisfies the matrix

Solution :
 1 2 3  −7 −8 −9
relation A  =
 4 5 6  2 4 6
.

1 2 3  −7 −8 −9
Given A  = 4 6 
oo
(i) Verify (A + B)T = AT + BT = BT + AT


4
(AT)T =  −1
 2
5
T

0 ⇒ A =

3
 4 −1 2
 5 0 3
 
ab
 4 5 6  2
4 −1 2  2 −1 1
 1 2 3 Now, A + B =  +
The order of the matrix   is 2 × 3 and the order of 5 0 3 7 5 −2
 4 5 6
 6 −2 3
 −7 −8 −9 = 
ur

the matrix  
 is also 2 × 3. 12 5 1
2 4 6
\ A must be of order 2 × 2.  6 12
 −2 5 
a b  \ (A + B)T =   ... (1)
.s

Let A =  
c d   3 1 
 a b   1 2 3  −7 −8 −9 2 7
\     =   
 c d   4 5 6 2 4 6  BT =  −1 5 
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 a + 4b 2a + 5b 3a + 6b   −7 −8 −9  1 −2
⇒ =    4 5  2 7   6 12
c + 4d 2c + 5d 3c + 6d   2 4 6   −1 0  −1 5   −2 5 
\ AT BT
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Equating the corresponding entries on both sides, we + =  +  =  

get  2 3  1 −2  3 1 
a + 4b = – 7 ... (1)  ... (2)
2a + 5b = – 8 ... (2)
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 2 7   4 5  6 12
c + 4d = 2 ... (3)      
2c + 5d = 4 ... (4) B + A =  −1 5  +  −1 0 =  −2 5 
T T

(1) × 2 ⇒ 2a + 8b = – 14  1 −2  2 3  3 1 

(–) (–) (+)  ... (3)
(2) ⇒ 2a + 5b = – 8 From (1), (2) and (3), (A + B)T = AT + BT = BT + AT
3b = – 6 ⇒ b=–2
Substituting b = – 2 in (1) we get,
 Verify (A – B)T = AT – BT
(ii)
 4 −1 2  2 −1 1 
Let P =
1
2
(A+A T )
A–B =  – 
 5 0 3 7 5 −2 1   4 −2  4 3   1 8 1 
=  +  =
2 0 1 2   3 −5  −2 −5  2 1 −10
= 
−5 5

m
 −2 1 8 1 
⇒ PT = =P
2 −2 2 1 −10
T 0 −5 
(A – B) =   ... (4) \ P is a symmetric matrix.

co
1 5  1
Let Q = A − AT 
 4 5  2 7   2 −2 2 
 −1 0  −1 5   0 −5 1   4 −2  4 3   1 0 −5
AT – BT =  =  −  =
 –   =   2   3 −5  −2 −5  2 5 0 
 2 3  1 −2 1 5 

.
1  0 5
QT = =–Q
 ... (5) 2  −5 0

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From (4) and (5), (A – B)T = AT – BT \ Q is a skew-symmetrix matrix.

(iii) Verify (BT)T = B 1 8 1  1 0 −5
Now A = P + Q = +
 2 −1 1  2 1 −10 2 5 0 

oo
Given B =    ... (6)
7 5 −2 Thus A is expressed as the sum of symmetric
and skew-symmetric matrices.
2 7
   3 3 −1  3 −2 −4
\ B =  −1 5 
T
 −2 −2 1  T  3 −2 −5
(ii) Let B =  ⇒ B = 
ab
 1 −2 
 −4 −5 2   −1 1 2 
 2 −1 1 
Also, (BT)T =    ... (7)  3 3 −1  3 −2 −4 
7 5 −2 1 T  1  

Let P =  B + B  =   −2 −2 1  +  3 −2 −5 
From (6) and (7), (BT)T= B 2 2 
ur

  −4 −5 2   −1 1 2  
16. If A is a 3 × 4 matrix and B is a matrix such that  6 1 −5
both ATB and BAT are defined, what is the order 1
= 1 −4 −4
of the matrix B? 2 
.s

Solution : Given A is a 3 × 4 matrix.

 −5 −4 4 
AT is a 4 × 3 matrix.  6 1 −5
1
ATB and BAT are defined. ⇒ = 1 −4 −4 = P
PT
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To define ATB, B must be a 3 × 4 matrix. 2 

 −5 −4 4 
Also to define BAT, B must be a 3 × 4 matrix. \ P is a symmetric matrix.

Hence, the order of matrix B is (3 × 4)   3 3 −1  3 −2 −4 
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1 T 1    
17. Express the following matrices as the sum of a Let Q =  B − B  =   −2 −2 1  −  3 −2 −5 
2 2
symmetric matrix and a skew-symmetric matrix:   −4 −5 2   −1 1 2  
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 3 3 −1  0 5 3
 4 −2 1
(i)
 −2 −2 1
 3 −5 and (ii)  = −5 0 6
  . 2 
 −4 −5 2  −3 −6 0
 4 −2 4 3 0 −5 −3
(i) Let A =  ⇒ AT =  1
Solution : 
 3 −5

 −2 −5 ⇒ T
Q = 5 0 −6
2 
 3 6 0 
  0 5 3  1 2 2
1   2 1 −2
= −  −5 0 6 = – Q 19. If A =  T
 is a matrix such that AA = 9I ,
2
 −3 −6 0  x 2 y 
\ Q is a skew-symmetric matrix. find the values of x and y.
Now B = P + Q

m
 6 1 −5  0 5 3 1 2 2  1 2 x 
1  1   2 1 −2 T 2 1 2 
=  1 −4 −4 +  −5 0 6
2 2
Solution : Given A =   ⇒ A = 
 −5 −4 4   −3 −6 0  x 2 y   2 −2 y 

co
Thus, B is expressed as the sum of a symmetric Also, AAT = 9I
and a skew-symmetric matrix.
1 2 2  1 2 x  1 0 0
 2 −1  2 1 −2  2 1 2 
⇒ 9  0 1 0
 1 0   =  

.
18. Find the matrix A such that    x 2 y   2 −2 y  0 0 1

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 1 −8 −10
−  −3 4
   1+ 4 + 4 2+2−4 x + 4 + 2 y  9 0 0
AT =  1 2 −5 .
 
 9 22 15 ⇒ 2+2−4 4 +1+ 4 2 x + 2 − 2 y  =  0 9 0
 
 2 0 0 9
 2 −1  −1 −8 −10  x + 4 + 2 y 2 x + 2 − 2 y x 2
+ 4 + y 

oo
1 0 T  1 2 −5 
Solution : Given   A = 
 −3 4  3× 2  9 22 15  3×3  9 0 x + 2y + 4  9 0 0
   0 9 0
⇒ 0 9 2 x − 2 y + 2 =
AT is a matrix of order 2 × 3.  
 
a b c 
2 2
 x + 2 y + 4 2 x − 2 y + 2 x + y + 4  0 0 9
ab
Let AT =  
d e f  Equating the corresponding entries on both sides, we get
 2 −1  −1 −8 −10 x + 2y + 4 = 0
 1 0   a b c   1 2 −5 
\    d e f  =   2x – 2y + 2 = 0
ur

 −3 4   9 22 15  ⇒ x + 2y = – 4 ...(1)
 2a − d 2b − e 2c − f  ⇒ 2x – 2y = – 2 .. (2)
 a b c  3x = – 6 ⇒ x = – 2
⇒ 
 −3a + 4d −3b + 4e −3c + 4 f  Substituting x = – 2 in (1) we get,
.s

 −1 −8 −10 – 2 + 2y = – 4
 1 2 −5  ⇒ 2y = – 4 + 2 = – 2
=   ⇒ y = –1
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 9 22 15  Hence, x = – 2, y = – 1
Equating the corresponding entries on both
sides, we get 20. (i) what value of x, the matrix
For
 0 1 −2
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a = 1, b = 2, c = – 5 and 2a – d = – 1
 −1 0 
⇒2–d=–1⇒2+1=d⇒d=3 A=  x 3  is skew-symmetric.
2b – e = –8 ⇒ 4 – e = – 8 ⇒ 4 + 8 = e   2 −3 0 [Hy - 2018]
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⇒ e = 12
2c – f = – 10 ⇒ – 10 – f = – 10 ⇒ f = 0 0 p 3 
 
1 2 −5 2
(ii) If  2 q −1 is skew-symmetric, find the
\ AT = 3 12 0 
   
 r 1 0 
 1 3
  values of p, q, and r.
⇒ (AT)T = A =  2 12
 −5 0 
Solution : 0
 0 1 −2 ⇒ q2 = =0 ⇒ q=0
2
 3

(i) Given A =  −1 0 x  is a skew-symmetric Hence, p = – 2, q = 0 and r = – 3
matrix.  2 −3 0 
  21. Construct the matrix A =[ aij]3 × 3, where aij = i – j.

m
 0 −1 2  State whether A is symmetric or skew-symmetric.
 
⇒ AT =  1 0 −3 Solution : Given aij = i – j
 3  Let A = [aij]3×3
 −2 x 0 

co
Since A is a skew-symmetric matrix. \ a11 = 1 – 1 = 0 a21 = 2 – 1 = 1 a31 = 3 – 1 = 2
AT = – A a12 = 1 – 2 = – 1 a22 = 2 – 2 = 0 a32 = 3 – 2 = 1
 0 −1 2   0 1 −2 a13 = 1 – 3 = – 2 a23 = 2 – 3 = –1 a33 = 3 – 3 = 0
   
0 −1 −2

.
3
⇒  1 0 −3 = –  −1 0 x 
    1 0 −1

ks
3 ⇒ A =  
 −2 x 0   2 −3 0 
 2 1 0 
 0 −1 2   0 −1 2 
    0 1 2
⇒  1 0 −3 =  1 0 − x 
3  1
⇒ \ A =  −1
T 0

   

oo
3
 −2 x 0   −2 3 0   −2 −1 0
Equating the corresponding entries on both sides, we  0 −1 −2
get 1 0 −1
= –   =–A
x3 = 3 ⇒x= 33  2 1 0 
ab
0 p 3  Since AT = – A, A is a skew-symmetric matrix.
 2
 22. Let A and B be two symmetric matrices. Prove
(ii) Let B =  2 q −1
  that AB = BA if and only if AB is a symmetric
 r 1 0  matrix.
ur

0 2 r Solution : A and B are symmetric

  If (AB)T = AB
2
⇒ BT =  p q 1  ⇒ BT AT = AB [ (AB)T = BTAT]
  ⇒ BA = AB
 3 −1 0
.s

[ A and B are symmetric matrices ⇒ BT = B and AT = A]

Since B is a skew-symmetric matrix,
Hence proved.
BT = – B
23. If A and B are symmetric matrices of same order,
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0 2 r 0 p 3 
    prove that
2 2
 p q 1 = −  2 q −1 (i) AB + BA is a symmetric matrix.
    (ii) AB – BA is a skew-symmetric matrix.
 3 −1 0  r 1 0 
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Solution : Given A and B are symmetric matrices

0 2 r  0 − p −3
    (i) ⇒ AT = A and BT = B  ... (1)
2 2
⇒  p q 1 =  −2 − q 1 To prove that (AB + BA) is a symmetric matrix.
w

    Consider (AB + BA)T = (AB)T + (BA)T

 3 −1 0  − r −1 0 
Equating the corresponding entries on both sides, we = BTAT + ATBT
get [(AB)T = BTAT]
2 = –p⇒ p=–2 = BA + AB [using (1)]
r = –3 = AB + BA
q2 = – q2 ⇒ 2q2 = 0
 ⇒ (AB + BA)T = AB + BA  100 + 20 + 60  180 
\ (AB + BA) is a symmetric matrix.  
=  200 + 20 + 120 =  340 

(ii) Given A and B are symmetric matrices
 250 + 50 + 180   480
⇒ AT = A and BT = B ... (2)
To prove that (AB – BA) is a skew-symmetric matrix. \ Cost of I gift pack = ` 180
Cost of II gift pack = ` 340 and cost of III gift pack

m
Consider (AB – BA)T = (AB)T – (BA)T
= B TAT – ATB T = ` 480
[ (AB)T = BTAT]
= BA – AB
EXERCISE 7.2

co
 [using (2)] 1. Without expanding the determinant, prove that
= – (AB – BA) s a 2 b2 + c 2
⇒ (AB – BA)T = – (AB – BA) s b2 c 2 + a 2 = 0
\ (AB – BA) is a skew-symmetric matrix.

.
s c 2 a 2 + b2
24. A shopkeeper in a Nuts and Spices shop makes
s a 2 b2 + c2

ks
gift packs of cashew nuts, raisins and almonds.
Pack I contains 100 gm of cashew nuts, 100 gm of Solution : Let A = s b2 c2 + a2
raisins and 50 gm of almonds. s c2 a 2 + b2
Pack-II contains 200 gm of cashew nuts, 100 gm
Applying C2 → C2 + C3 we get,

oo
of raisins and 100 gm of almonds.
Pack-III contains 250 gm of cashew nuts, 250 gm s a 2 + b2 + c2 b2 + c2
of raisins and 150 gm of almonds. s a 2 + b2 + c2 c2 + a2
A =
The cost of 50 gm of cashew nuts is ` 50/-, 50 gm
of raisins is `10/-, and 50 gm of almonds is ` 60/-. s a 2 + b2 + c2 a 2 + b2
ab
What is the cost of each gift pack?
Taking ‘s’ common from C1 and (a2 + b2 + c2)
Solution : Gift pack matrix is as follows: common from C2 we get
 I II III  1 1 b2 + c2

Weight of Cashew nuts 100 200 250 2 2
ur

 A = s(a2 + b2 + c2) 1 1 c + a
Weight of Raisins 100 100 250
  1 1 a 2 + b2
Weight of Almonds  50 100 150 
= s(a2 + b2 + c2) (0) = 0[ C1 ≡ C2]
.s

Let us consider 50 gm of cashew nuts as one s a2 b2 + c2

packet, 50 gm of raisins as one packet and 2
50 gm of almonds as one packet, we get the Hence, s b c2 + a2 = 0
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matrix as s c2 a 2 + b2
 I II III 
b + c bc b 2c 2
No. of packets of cashewnuts  2 4 5 
 
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2. Show that c + a ca c 2a 2 = 0.
No. of packets of raisins 2 2 5  = A
  a + b ab a 2b 2
No. off packets of almonds 1 2 3 
Given cost matrix is [50 10 60] = B Solution : Applying R1→ aR1, R2→ bR2 and R3→ cR3
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we get,
\ Cost of gift pack
 2 4 5 ab + ac abc ab 2c 2
 
= AB = [50 10 60]  2 2 5 A = bc + ab abc a 2bc 2
1 2 3 ac + bc abc a 2b 2c
Taking out (abc) common from C2 and C3 we get,
 ab + ac 1 bc c −a 0
= (abc)2 bc + ab 1 ac LHS = 2abc 0 − a −b
ac + bc 1 ab c 0 −b
Applying C1 → C1 + C3 we get, Taking c, a, b common from C1, C2 and C3 respectively.
ab + bc + ca 1 bc

m
1 −1 0
2 ab + bc + ca 1 ac
= (abc)
= 2a2b2c2 0 −1 −1
ab + bc + ca 1 ab
1 0 −1

co
Taking out (ab + bc + ca) common from C1, we get Expanding along R1 we get,
1 1 bc  −1 −1 0 −1 
= 2a2b2c2 1 +1 
A = a2b2c2 (ab + bc + ca) 1 1 ac  0 −1 1 −1 
1 1 ab = 2a2b2c2 [(1 – 0) + (0 + 1)]

.
= a 2b 2c 2
(ab + bc + ca) (0) = 0

ks
= 2a2b2c2 [2]
[ C1 ≡ C2] = 4a2b2c2 = RHS
a2 bc ac + c 2 Hence proved.
3. Prove that a 2 + ab b2 ac = 4a2 b2 c2.

oo
4. Prove that
ab b 2 + bc c2 1+ a 1 1
 1 1 1
a2 bc ac + c 2 1 1+ b 1 = abc  1 + + +  .
 a b c
Solution : LHS = a 2 + ab b2 ac 1 1 1+ c
ab
ab 2
b + bc c2 1+ a 1 1
Taking out a, b, c common from C1, C2 and C3 respectively Solution : LHS = 1 1 + b 1
we get, 1 1 1+ c
Taking out a, b, c common from R1, R2 and R3
a c a+c
ur

respectively.
LHS = (abc) a + b b a
1 1 1
b b+c c +1
a a a
Applying C1 → C1 + C2 + C3 we get,
LHS = abc 1 1 1
.s

+1
2 (a + c ) c a+c b b b
1 1 1
= (abc) 2 ( a + b ) b a +1
c c c
w

2 (b + c ) b + c c
Applying R1 → R1 + R2 + R3 we get,
a+c c a+c
= 2abc a + b b a 1 1 1 1 1 1 1 1 1
w

1+ + + 1+ + + 1+ + +
b+c b+c c a b c a b c a b c
= abc 1 1 1
Applying C1 → C1 – C2 and C3 → C3 – C1 we get, +1
b b b
w

a + c −a 0 1 1 1
+1
= 2abc a + b − a −b c c c
b + c 0 −b
Applying C1 → C2 + C1 + C3 we get,
 2a 2b 2c
1 1 1
+ x y z [By Property 7]
1 1 1
= abc 1 + 1 + 1 + 1  +1 a b c
  b b b
a b c a b c
1 1 1 = 0 + 2 x y z [ R1 ≡ R2]

m
+1
c c c a b c
Applying C1 → C1 – C2 and C2 → C2 – C3 we get,
= 0 + 2(0) = 0 = RHS [ R1 ≡ R3]

co
0 0 1
7. Write the general form of a 3 × 3 skew-symmetric
1 matrix and prove that its determinant is 0.
 1 1 1
= abc 1 + + +  −1 1
b Solution : A square matrix A = [aij]3×3 is a skew-
 a b c
1 symmetric matrix if aij = – aji for all i, j and

.
0 −1 +1 the elements on the main diagonal of a skew-
c

ks
Expanding along R1 we get, symmetric matrix are zero.
 −1 1  0 a12 a13
= abc (1 + 1/a + 1/b + 1/c) 0 + 0 + 1  \ A = − a12 0 a23
 0 −1 
− a13 − a23 0
=
=
abc (1 + 1/a + 1/b + 1/c) [1]
abc (1 + 1/a + 1/b + 1/c) = RHS
Hence Proved.
oo Expanding along R1 we get,

|A| = 0 – a12
− a12
− a13
a23
− a12 0
0 + a13 − a − a
13 23
ab
sec 2θ tan 2θ 1 = – a12 (a13 a23) + a13(a12a23)
5. Prove that tan θ sec θ −1 = 0.
2 2
= – a12 a13 a23 + a12 a13 a23 = 0
38 36 2
Hence the determinant of a skew-symmetric
sec 2 θ tan 2 θ 1
ur

matrix is 0.
Solution : LHS = tan 2 θ sec 2 θ −1
a b aα + b
38 36 2
8. If b c bα + c = 0, prove that a, b, c
.s

Applying C2 → C2 + C3 we get,
aα + b bα + c 0
sec 2 θ 1 + tan 2 θ 1 sec 2 θ sec 2 θ 1 are in G.P. or a is a root of ax2 + 2bx + c = 0.
2 2 2 2
= tan θ −1 + sec θ −1 = tan θ tan θ −1 a b aα + b
w

38 38 2 38 38 2 Solution : Given b c bα + c = 0
[ 1 + tan2q = sec2q and sec2q –1= tan2q] aα + b bα + c 0
w

= 0  [ C1 ≡ C2] = RHS Expanding along R3 we get,

Hence Proved. b aα + b a aα + b
(aa + b) – (ba + c) +0=0
c bα + c b bα + c
w

x + 2a y + 2b z + 2c
6. Show that x y z = 0. ⇒ – (aa + b) (b2a + bc – aca – bc) – (ba + c) ( aba + ac
– aba – b2) =0
a b c
⇒ (aa + b) (b2a – aca) – (ba + c) (ac – b2) = 0
x + 2a y + 2b z + 2c x y z ⇒ a(aa + b) (b2 –ac ) + (ba + c) (b2 –ac) = 0
Solution : LHS = x y z = x y z ⇒ (b2 – ac) (aa2 + ba + ba + c) = 0
a b c a b c
⇒ (b2 – ac) (aa2 + 2ba + c) = 0 10. If a, b, c are pth, qth and rth terms of an A.P, find the
⇒ b2 –ac = 0 or aa2 + 2ba + c = 0 a b c
⇒ ac = b2 or aa2 + 2ba + c = 0 value of p q r .
⇒ a, b, c are in G.P. (or) a is a root of ax2 + 2bx + c = 0 1 1 1
Solution : Given a = tp, b = tq and c = tr.

m
1 a a 2 − bc
Let A be the first term and l be the last term of
9. Prove that 1 b b 2 − ca = 0. [Hy - 2018]
the A.P.
1 c c 2 − ab
p q r
\ a = (A + l ) , b = (A + l ) , c = (A + l ) 

co
... (1)
2 2 2
1 a a 2 − bc 1 a a2
b 2 − ca = 1 b b 2 p q r
Solution : LHS = 1 b (A + l ) (A + l ) (A + l )
a b c 2 2 2
1 c c 2 − ab 1 c c2

.
Consider p q r = p q r

ks
1 a bc 1 1 1
– 1 b ca  [By property 7] 1 1 1
1 c ab
 using (1)
Multiplying and dividing R1, R2 and R3 of

oo
 A + l
second determinant by a, b, c respectively. Taking  common from R1 we get,
 2 
1 a a2 a a2 abc p q r
1 A+l A+l
LHS = 1 b b
2
–
abc
b b2 abc = p q r = (0) = 0  [ R1 ≡ R2]
2 2
ab
2
1 c c c c2 abc 1 1 1
Taking abc common from C3 of second a b c
determinant \ p q r =0
1 1 1
1 a a2 a a2 1
ur

2 abc a2 + x2 ab ac
= 1 b b – b b2 1 2 2
abc 11. Show that ab b +x bc is divisible
1 c c2 c c2 1
ac bc c + x2
2
.s

Applying C2 ↔ C3 in the second determinant, by x4.

1 a a2 a 1 a2 a2 + x2 ab ac
b2 + x2
w

2
= 1 b b + b 1 b
2 Solution : LHS = ab bc

1 c c2 c 1 c2 ac bc c + x2
2

Applying C1 ↔ C2 in the second determinant Multiplying R1,R2, R3 by a, b, c respectively,

w

and dividing the determinant by abc we get,

1 a a2 1 a a2
2 2
a 3 + ax 2 a 2b a 2c
= 1 b b – 1 b b = 0 = RHS 1
w

= ab 2 b3 + bx 2 b 2c
1 c c 2
1 c c 2 abc
ac 2 bc 2 c3 + cx 2
Hence proved.
Taking a, b, c common from C1, C2 and C3 we
get,
 \ a = ARp–1 ⇒ log a = log A + (p – 1) log R
a2 + x2 a2 a2
abc b = ARq–1 ⇒ log b = log A + (q – 1) log R
= b2 b2 + x2 b2
abc c = ARr–1 ⇒ log c = log A + (r – 1) log R
c2 c2 c2 + x2
Applying R1 → R1 + R2 + R3 we get, log a p 1

m
\ Let A = log b q 1
a 2 + b2 + c2 + x2 a 2 + b2 + c2 + x2 a 2 + b2 + c2 + x2
log c r 1
= b2 b2 + x2 b2
log A + ( p − 1) log R p 1

co
c2 c2 c2 + x2
\ A = log A + ( q − 1) log R q 1
Taking (a2 + b2 + c2 + x2) common from R1, log A + ( r − 1) log R r 1
we get
Applying C2 → C2 – C3 we get,
1 1 1

.
log A + ( p − 1) log R p −1 1

ks
2
LHS = (a2 + b2 + c2 + x2) b b2 + x2 b2
A = log A + ( q − 1) log R q −1 1
c2 c2 c2 + x2
log A + ( r − 1) log R r −1 1
Applying C1 → C1 – C2 and C2 → C2 – C3 we
get, Applying C1 → C1 – (log A) C3 – (log R)C2 we get,

= (a2 + b2 + c2 + x2) −x
0

0
2
0
x
−x
2

2
oo
1
b
c + x2
2
2
0 p −1 1
A = 0 q −1 1 = 0
0 r −1 1
ab
Taking out x2 common from C1 and C2 we get, \A = 0 Hence proved.
0 0 1 1 log x y log x z
2
= x4(a2 + b2 + c2 + x2) −1 1 b 13. Find the value of log y x 1 log y z
ur

2 2
0 −1 c + x log z x log z y 1
Expanding along R1 we get, if x, y, z ≠ 1.
 −1 1  1 log x y log x z
.s

LHS = x4(a2 + b2 + c2 + x2) 1 

 0 −1  Solution : Let A = log y x 1 log y z
= x4(a2 + b2 + c2 + x2) (1) log z x log z y 1
w

= x4(a2 + b2 + c2 + x2) which is Given x, y, z ≠ 1

divisible by x4. Expanding along R1 we get,
w

12. If a, b, c are all positive, and are pth, qth and rth 1 log y z log y x log y z log y x 1
A= 1 – logxy + logxz
log a p 1 log z y 1 log z x 1 log z x log z y

terms of a G.P., show that log b q 1 = 0.

w

= 1 – logy z . logz y – logx y = 1 – logy z . logz y

log c r 1 (logy x – logy z. logz x)+ logx z (logy x logz y logz x)
Solution : Given a, b, c are pth, qth and rth terms of a
= 1 – 1 – logxy ( log x – log x ) + log x (logzx log x )
G.P. y y z z

Let A be the first term and R be the common [ logzx.logxz = 1 and logyz.logzx = logyx]
ratio of the G.P. = 0 – logxy (0) + logxz (0) = 0
 1 log x y log x z 1
1− 2
\ log y x 1 log y z = 0 4n  1 
=  
3  2
log z x log z y 1 4

1

∵ Sn =
( )
a 1− rn 


m
 
 2 α n
1 1 

1− r 

 , prove that ∑ det ( A ) =  1 − n  .
k
14. If A = 
 0 1 k =1 3 4   1 
2 1−
  n
 1   4n 
2
( )

co
1  \ ∑ det A k
=    3   [From (1)]
 2 α k =1
2  
Solution : Given A =  1
  4 

 0 2  1 4 1 1 1
1 1  1 2 = × 1 − n  = 1 − 
3  4n 


.
⇒ |A| = × =   4 3  4 
2 2  2

ks
 1  1  Hence proved.
 2 α  2 α
15. Without expanding, evaluate the following
Also A2 =    
0 1 0 1 determinants :
 2   2  2 3 4

= 
 1  2
 

 0

2
1
  
 
α

2 


2 oo (i) 5 6 8
6 x 9 x 12 x
x+ y y+z z+ x
ab
(ii) z x y
4
 1 1 1 1
\ |A2| =  
2 2 3 4
n
\ ∑ det (A k ) = det (A) + det (A2) + det Solution : (i) Let A = 5 6 8
ur

k =1
 (A3) + ........ + det (An) 6 x 9 x 12 x
2 4 2n Taking (3x) common from R3 we get,
 1  1  1
=   +   + ............ +   2 3 4
 2  2  2
.s

2 2 2( n −1) 
 1  1  1 A = 3x 5 6 8 = 3x (0) = 0
=   1 +   + .......... +    ... (1)
 2   2  2  2 3 4
[ R1 º R3]
2 2( n −1)
 1  1
w

1 +   + ............... +   x+ y y+z z+x

 2  2
2
 1 (ii) Let B = z x y
is a G.P. with a = 1 and r =   .
 2 1 1 1
w

( )
 n
2 1− 1 2 Applying C1 → C1 – C2 and C2 → C2 – C3,
 1  2 
\ Sn =   1  we get
2  1− 1 2 
( )
w

 2  x−z y−x z+x

1 B = z−x x− y y
n 2 1− 0 0 1
=
4  1
×  
1 2 − ( z − x) − ( x − y ) z + x
1− z−x x− y y
4 =
0 0 1
 Taking (z – x) and (x – y) common from C1 We know that, determinant of a skew-
and C2 we get, symmetric matrix is zero.
−1 −1 1 \ |A| = 0
= (z – x) (x – y) 1 1 y 1 4 20
0 0 1 19. Determine the roots of the equation 1 −2 5 = 0.

m
Expanding along R3 we get,
1 2 x 5 x2
 −1 −1 
B = (z – x) (x – y) 1 
 1 1  1 4 20

co
= (z – x) (x – y) (– 1 + 1) = 0 Solution : Let A = 1 −2 5
\B = 0 1 2 x 5x2
16. If A is a square matrix and | A | = 2, find the value Given A = 0
of |AAT|.

.
1 4 20
Solution : Given A is a square matrix and

ks
|A| = 2 1 −2 5 = 0
2
\ |AAT| = |A| |AT| = |A| . |A| [ |A|T = |A|] 1 2x 5x
= 2 (2) = 4 By property 1
Applying R1 → R1 – R2 and R2 → R2 – R3 we get,
17. If A and B are square matrices of order 3 such
that | A | = –1 and |B| = 3, find the value of |3AB|.
Solution : Given A and B are square matrices of order 3.
Also, |A| = – 1 and |B| = 3
oo 0

1
6
0 −2 − 2 x 5 − 5 x 2
2x
15

5x 2
= 0
–2
−2
–1
+1
ab
Consider |3AB| = 33|A|.|B| Expanding along C1 we get,
= 27 (–1) (3) = – 81
6 15
[ A is a square matrix of order 3] 0+0+1 = 0
\ |3AB| = – 81 −2 − 2 x 5 − 5 x 2
ur

18. If l = –2, determine the value of 6 15

⇒ = 0
0 2λ 1 −2 − 2 x 2 5 − 5x2
λ 2 0 3λ + 1 .
2
⇒ 6(5 – 5x2) – 15(– 2 – 2x) = 0
.s

−1 6λ − 1 0 ⇒ 30 – 30x2 + 30 + 30x = 0
⇒ – 30x2 + 30x + 60 = 0
Solution : Given l = – 2
Dividing by – 30 we get,
w

0 2λ 1 ⇒ x2 – x – 2 = 0
2 2
Let A = λ 0 3λ + 1 ⇒ (x – 2) (x + 1) = 0
−1 6λ − 1 0 ⇒ x = 2 or – 1
w

Hence the roots are –1, 2.

0 −4 1
= 4 0 13 20. Verify that det(AB) = (det A) (det B) for
 4 3 −2  1 3 3
w

−1 −13 0  1 0 7  −2 4 0
A=   and B =  .
 [ Put l = – 2 ]
 2 3 −5  9 7 5
Since a12 = – a21 , a13 = – a31 , a23 = – a32 and
the elements in the main diagonal are zero, A 4 3 −2
is a skew-symmetric matrix. Solution : Given A = 1 0 7
2 3 −5
  1 3 3 21. Using cofactors of elements of second row,
 −2 4 0 5 3 8
and B =   evaluate | A |, where A = 2 0 1 .
 9 7 5
1 2 3
 4 3 −2  1 3 3
 5 3 8
AB = 1 0 7   −2 4 0

m
 
   Solution : Given A =  2 0 1
 2 3 −5  9 7 5 1 2 3
 4 − 6 − 18 12 + 12 − 14 12 + 0 − 10 3 8
Co-factor of 2 = A21 = (–1)1+2

co
  2 3
=  1 + 0 + 63 3 + 0 + 49 3 + 0 + 35 
 2 − 6 − 45 6 + 12 − 35 6 + 0 − 25  = – (9 – 16) = 7
−20 10 2 5 8
Co-factor of 0 = A22 = (–1)2+2
det (AB) = 64 52 38 1 3

.
−49 −17 −19 = 15 – 8 = 7

ks
Expanding along R1 we get, 5 3
Co-factor of 1 = A23 = (–1)2+3
52 38 1 2
det (AB) = – 20 – 10 ×
−17 −19 = – (10 – 3) = – 7
64 38 64 52 \ |A| = a21 A21 + a22 A22 + a23 A23

oo
 +2
−49 −19 −49 −17 = 2(7) + 0(7) + 1(–7)

= – 20(–988 + 646) – 10(–1216 + 1862) + 2(–1088 + 2548) = 14 – 7 = 7
= – 20(–342) – 10(646) + 2(1460)
Exercise 7.3
ab
= – 6840 – 6460 – 292
= 3300 ... (1) Solve the following problems by using Factor Theorem :
4 3 −2
|A| = 1 0 7 x a a
1. Show that a x a = (x – a)2 (x + 2a).
ur

2 3 −5
a a x
0 7 1 7
= 4 –3 –2 x a a
3 −5 2 −5
Solution : Let A = a x a  ... (1)
.s

1 0
 a a x
2 3
Putting x = a in (1) we get,
= 4(0 – 21) – 3(–5 –14) – 2(3 + 0) = –84 + 57 – 6 = –33
a a a
w

1 3 3
A = a a a =0
|B| = −2 4 0
a a a
9 7 5
w

[ R1 º R2 º R3]
4 0 −2 0 −2 4
=1 –3 +3
7 5 9 5 9 7 \ (x – a)2 is a factor of A.
w

Putting x = – 2a in (1) we get,

= 1(20 + 0) –3 (–10 + 0) + 3 (–14 – 36) = 20 + 30 –150
= – 100 −2a a a
\ |A| |B| = –33 (–100) = 3300 ... (2) A = a −2a a
From (1) and (2) det (AB)= det (A) det (B) a a −2a
Applying C1 → C1 + C2 + C3 we get,
 0 a a c a−c a
A = 0 −2a a =0 A = −c c + a − a = 0
0 a −2a c c−a a
\ (x + 2a) is also a factor of A.
[ C1 ∝ C3]
Since the leading diagonal of A is of degree 3, (b – 0) = b is a factor of A.

m
only 3 factors are available and their may be a Putting c = 0 in (1) we get,
constant k.
b a a−b
x a a
A = b a b−a =0

co
\ A = a x a = k(x – a)2 (x + 2a) −b − a a + b
a a x [ C1 ∝ C2]
Putting x = – a in the above equation, we get \ (c – 0) = c is a factor of A.
−a a a

.
Since the leading diagonal A is of degree 3, only 3 factors

ks
a −a a
= k (– a – a)2 (– a + 2a) are available and there may exist a constant k.
a a −a
b+c a−c a−b
0 a a
\ b − c c + a b − a = k(abc)
⇒ 0 − a a = k(4a2) (a)
c−b c−a a+b

⇒ 2a
a a
−a a
2a a − a

= 4ka3 [Expanding along C1]

oo
[Applying C1 → C1 + C2]
Putting a = 1, b = 1 and c = 1 in the above equation, we get
2 0 0
0 2 0 = k(1)(1)(1)
ab
0 0 2
⇒ 2a(a2 + a2) = 4ka3
⇒ 2a(2a2) = 4ka3 ⇒ 8 = k
3 b+c a−c a−b
⇒ 4 a = 4 k a3 ⇒ k=1
ur

x a a \ b − c c + a b − a = 8abc
c−b c−a a+b
\ A = a x a = (x – a)2 (x + 2a)
a a x x+a b c
.s

b+c a−c a−b 3. Solve a x+b c = 0.

2. Show that b − c c + a b − a = 8 abc. a b x+c
c−b c−a a+b
w

x+a b c
b+c a−c a−b Solution : Let A = a x+b c  ... (1)
Solution : Let A = b − c c + a b − a a b x+c
w

c−b c−a a+b Putting x = 0 in (1) we get,

Putting a = 0 in (1) we get, a b c
b + c − c −b
A = a b c = 0[ R1 º R2 º R3]
w

A = b−c c b =0 a b c
c−b c b
\ (x – 0)2 = x2 is a factor of A.
[ C2 ∝ C3]
(a – 0) = a is a factor of A. Putting x = – (a + b + c) in (1) we get,
Putting b = 0 in (1) we get,
 −b − c b c ⇒ (a – b) is a factor of D.
Putting b = c in (1) we get,
A = a −a − c c
a b −a − b 2c a a2

0 b c D = c+a c c2 = 0
c2

m
= 0 − a − c c a+c c
0 b −a − b  [ R2 º R3]
 [Applying C1 → C1 + C2 +C3] ⇒ (b – c) is a factor of D.

co
= 0 Putting c = a in (1) we get,
\ x + (a + b + c) is a factor of A. b + a a a2
Since the leading diagonal of A is of degree D = 2a b b2 = 0
3, only 3 factors are available and there may a + b a a2

.
exist a constant k.  [ R1 º R3]

ks
x+a b c
⇒ (c – a) is a factor of D.
\ a x+b c = k(x2) (x + a + b + c)
Putting a = – (b + c) in (1) we get,
a b x+c
b+c −b − c ( − b − c )2

oo
Putting x = – a we get,
0 b c D = c −b− c b b2
a −a + b c = k(a2) (– a + a + b + c) −b −c+ b c c2
a b −a + c b + c − (b + c ) (b + c )2
ab
Expanding along R1 we get, −b b b2
=
– b[(–a2) + (ac ac)] + c( ab + a2 – ab ) = k(a2) (b + c)
ab– ab
−c c c2
⇒ a2b + a2c = k(a2) (b + c)
= 0 [ R2 ∝ R3]
⇒ a 2 ( b + c ) = k( a 2 ) ( b + c )
ur

\ (a + b + c) is a factor of D.
⇒ k = 1
\ 1 (x2) (x + a + b + c) = 0 Since the leading diagonal of D is of degree 4, only 4
factors and a constant k are available.
⇒ x = 0 or x = – (a + b + c)
.s

Hence, the values of x are 0, 0, – (a + b + c). b+c a a2

b + c a a2 \ c+a b b
2
= k (a + b + c) (a – b) (b – c) (c – a)
4. Show that c + a b b 2 = (a + b + c) (a – b) (b – c)×
w

2
a+b c c
a + b c c2
(c – a). Putting a = 2, b = 1, c = 0 we get,
b + c a a2 1 2 4
w

Solution :
2
Let D = c + a b b  ... (1) 2 1 1 = k(3) (1) (1) (–2)
3 0 0
a+b c c2
w

Expanding along R3 we get,

Putting a = b we get,
2 4
b + c b b2 ⇒ 3 = – 6k
2
1 1
D = c+b b b =0
2 ⇒ 3(2 – 4) = – 6k
2b c c
[ R1 º R2] ⇒ −6 = −6 k
⇒ k = 1 ⇒ 3(–16) – 5(–10) + 5(10) = 13k
⇒ – 48 + 50 + 50 = 13k
b+c a a2
2
⇒ 52 = 13k
\ c+a b b = (a + b + c) (a – b) (b – c) (c – a) ⇒ k = 4
2
a+b c c 4− x 4+ x 4+ x

m
Hence proved. \ 4 + x 4 − x 4 + x = 4 (x2) (x + 12)
4− x 4+ x 4+ x 4+ x 4+ x 4− x
5. Solve 4 + x 4 − x 4 + x = 0. ⇒ 4(x2) (x + 12) = 0

co
4+ x 4+ x 4− x ⇒ x = 0 or x = – 12
4− x 4+ x 4+ x 1 1 1
Solution : Let A = 4 + x 4 − x 4 + x  ... (1) x y z = (x – y) (y – z) (z – x)
6. Show that

.
4+ x 4+ x 4− x
x2 y2 z2

ks
Putting x = 0 in (1) we get,
1 1 1
4 4 4
A = 4 4 4 = 0  [ R1 º R2 º R3] Solution : Let D = x y z  ... (1)
4 4 4 x2 y 2 z 2

oo Putting x = y in (1) we get,

2
⇒ x is a factor of (1)
Putting x = – 12 we get, 1 1 1
4 + 12 4 − 12 4 − 12 D = y y z =0
A = 4 − 12 4 + 12 4 − 12
ab
2 2
y y z2
4 − 12 4 − 12 4 + 12 [  C1 º C2]
16 −8 −8 \ (x – y) is a factor of (1).
= −8 16 −8 Putting y = z in (1) we get,
ur

−8 −8 16 1 1 1
0 −8 −8
1 z z = 0 [  C2 º C3]
= 0 16 −8 = 0 2 2
1 z z
0 −8 16
.s

[Applying C1 → C1 + C2 + C3] \ (y – z) is a factor of (1)

\ (x + 12) is also a factor of (1). Putting z = x in (1) we get,
Since the leading diagonal of A is of degree 3, only 3
w

1 1 1
factors and a constant k are available
x y x = 0 [  C1 º C3]
4− x 4+ x 4+ x
2 2 2
x y x
w

\A = 4+ x 4− x 4+ x
4+ x 4+ x 4− x \ (z – x) is a factor of (1)
= k (x2) (x + 12) Since the leading diagonal of D is of degree 3,
w

there are 3 factors and a constant k.

Putting x = 1, we get
1 1 1
3 5 5
5 3 5 = k (1)2 (1 + 12) \ x y z = k(x – y) (y – z) (z – x)
2 2 2
5 5 3 x y z
⇒ 3(9 – 25) – 5(15 – 25) + 5 (25 – 15) = 13k
[ Expanding along R1]
 Putting x = 0, y = 1, z = – 1 in the above k 2 1
equation we get, 1
⇒ 4 = absolute value of 2 4 1
1 1 1 2
3 2 1
0 1 −1 = k(–1) (2) (–1)
1
0 1 1 ⇒ 4 = absolute value of [k(4 – 2) –
2

m
1 −1
⇒ 1 = 2k [Expanded along C1]  2(2 – 3) + 1(4 – 12)] [Expanded along R1]
1 1
1
1(1 + 1) = 2k ⇒ 2 = 2 k ⇒ k = 1 ⇒ 4 = absolute value of [2k + 2 – 8]

co
2
1 1 1 1
⇒ 4 = absolute value of [2k – 6]
\ x y z = (x – y) (y – z) (z – x) 2
1
x2 y2 z2 ⇒ 4 = ± ( 2k − 4)
2

.
Case (i)
Exercise 7.4

ks
1
when 4 = ( 2k − 6)
2
1. Find the area of the triangle whose vertices are ⇒ 8 = 2k – 6
(0, 0), (1, 2) and (4, 3). ⇒ 14 = 2k
Solution : Let the vertices of the triangle be A(0, 0) ⇒ k = 7
B (1, 2) C(4, 3)

Area of the DABC = absolute value of x2

1
2
x1

x3
oo
y1 1
y2 1
y3 1

⇒
Case (ii)
1
when 4 = – ( 2k − 6)
2
8 = – 2k + 6
ab
⇒ 8 – 6 = – 2k
0 0 1 ⇒ 2 = – 2k
1
= absolute value of 1 2 1 ⇒ k = –1
2 \ The values of k are –1 or 7.
4 3 1
3. Identify the singular and non-singular matrices:
ur

1 1 2
= absolute value of 0 + 0 + 1 4 3   1 2 3  2 −3 5
2 
 4 5 6 6 0 4
(i)   (ii)  
 [Expanded along R1]
7 8 9  1 5 −7 
.s

1
= absolute value of [3 − 8] 0 a − b k
2 
1 b − a
= absolute value of [ −5] (iii)  0 5
2 
w

 − k −5 0
= absolute value of (–2.5)
= 2.5 Sq.units  1 2 3
 
Solution : (i) Let A =  4 5 6
2. If (k, 2), (2, 4) and (3, 2) are vertices of the triangle
w

of area 4 square units then determine the value 7 8 9

of k. 5 6 4 6 4 5
Solution : Let the vertices of the triangle be A(k, 2) |A| = 1 –2 +3
8 9 7 9 7 8
w

B(2, 4) and C(3, 2)

 [Expanded along R1]
Also area of DABC = 4 sq. units.
We know that, area of DABC = 1(45 – 48) – 2(36 – 42) + 3(32 – 35)

x1 y1 1 = – 3 – 2(–6) + 3(–3)
1 = – 3 + 12 – 9 = – 12 + 12 = 0
= absolute value of x2 y2 1
2 Since |A| = 0, the given matrix is singular.
x3 y3 1
  2 −3 5  b −1 2 3
 
(ii) Let B =  6 0 4  |B| = 3 1 2 =0
1 5 −7  1 −2 4
0 4 6 4 6 0 1 2 3 2 3 1
|B| = 2 +3 +5 = (b – 1) –2 +3 = 0
5 −7 1 −7 1 5 −2 4 1 4 1 −2

m
 [Expanded along R1]
 [Expanded along R1]
⇒ (b – 1) (4 + 4) – 2(12 – 2) + 3(– 6 – 1) = 0
= 2(0 – 20) + 3(–42 – 4) + 5(30 – 0)
(b – 1) (8) – 2(10) + 3(–7) = 0

co
= – 40 + 3(–46) + 5(30) 8b – 8 – 20 – 21 = 0
= – 40 – 138 + 150 = – 28 ≠ 0 8b – 49 = 0
Since |B| ≠ 0, the given matrix is non-singular. ⇒ 8b = 49
 0 a − b k 49
⇒ b =

.
b − a 0 5 8
(iii) Let C = 

ks
 b − 1 2 3 
 − k −5 0  49 
If b = then  3 1 2 is singular.
b−a 5 8
|C| = 0 – (a – b) +  1 −2 4 2
−k 0 0 cos θ sin θ

oo
k × b − a 0 = – (a – b) (5k) + k [–5 (b – a)] 5. If cos 2θ = 0 , determine cos θ sin θ 0
− k −5 = – (a – b) (5k) + 5k (a – b) = 0 sin θ 0 cos θ
Since |C| = 0, the given matrix is singular. Solution : Given cos 2q = 0
2
4. Determine the values of a and b so that the 0 cos θ sin θ
ab
following matrices are singular: Let A = cos θ sin θ 0
 b − 1 2 3 sin θ 0 cos θ
 7 3  3
(i) A=   (ii) B =  1 2 0 cos θ sin θ 0 cos θ sin θ
 −2 a  
 1 −2 4 = cos θ sin θ 0 cos θ sin θ 0
ur

Solution :
sin θ 0 cos θ sin θ 0 cos θ
 7 3
(i) Given A =   0 + cos 2 θ + sin 2 θ 0 + sin θ + 0 0 + 0 + sin θ cos θ
 −2 a  = 0 + sin θ cos θ + 0 cos 2 θ + sin 2 θ + 0 sin θ cos θ + 0 + 0
Since A is a singular matrix
.s

0 + 0 + sin θ cos θ sin θ cos θ + 0 + 0 sin 2 θ + 0 + cos 2 θ

|A| = 0
∵sin 2θ 1 − cos 2θ = 1 = 1
2
7 3 
= 0
−2 a
w

1 1
⇒ 7a + 6 = 0 1
2 2
 1 
⇒ 7a = – 6 1 1  1 1  1  1 1
= 1 = 1 1 −  − 1 +  − 
w

6 2 2  4  2  2 −  2  4 2 
⇒ a = – 4
7 1 1
b − 1 2 3  1
2 2
 1 2
w

(ii) B =  3 3 1 1 6 −1−1 4 1
= − − = = =
 1 −2 4 4 8 8 8 8 2
Since B is a singular matrix, |B| = 0 6. Find the value of the product:
6  7 3 log 3 64 log 4 3 log 2 3 log 8 3
∴ If a = – then   is singular. ×
7  −2 a  log 3 8 log 4 9 log 3 4 log 3 4
 log 3 64 log 4 3 log 2 3 log8 3 1 2 3 8
Solution : × Hint : 2X +   =  
log 3 8 log 4 9 log 3 4 log 3 4 3 4   7 2
log 3 64 ⋅ log 2 3 + log 4 3 ⋅ log 3 4 log 3 64 ⋅ log8 3 + log 4 3 ⋅ log 3 4  3 8  1 2  2 6 
= ⇒ 2X =  – = 
log 3 8 ⋅ log 2 3 + log 4 9 ⋅ log 3 4 log 3 8 ⋅ log8 3 + log 4 9 ⋅ log 3 4 7 2 3 4  4 −2

m
1 3 
= 6 log3 2 log 2 3 + 1 2 log3 8 log8 3 + 1 X =  
3 log 3 2 log 2 3 + 1 1 + 2 log 4 3 log 3 4  2 −1
1 3 

co
6 +1 2 +1 7 3
= = = 21 – 15 = 6  [Ans: (1)  ]
3 + 2 1+ 2 5 3  2 −1
3. Which one of the following is true about the
EXERCISE 7.5  1 0 0
 
matrix 0 0 0 ?

.
CHOOSE THE CORRECT OR THE 0 0 5

ks
MOST SUITABLE ANSWER FROM (1) a scalar matrix
THE GIVEN FOUR ALTERNATIVES. (2) a diagonal matrix
1 (3) an upper triangular matrix
1. If aij = (3i – 2j) and A = [aij]2 × 2 is

oo
2 (4) a lower triangular matrix
 1  1 1
[Ans: (2) a diagonal matrix]
 2 2 2 − 2
(1)   (2)   4. If A and B are two matrices such that A + B and AB are
 − 1 1 2 1 both defined, then
 2   
ab
(1) A and B are two matrices not necessarily of same
   1 1 order
2 2 − 2 2 
(3)   (4)   (2) A and B are square matrices of same order
1 − 1  1 2 (3) Number of columns of A is equal to the number
 2 2   
ur

of rows of B
1
Hint : aij = (3i − 2 j ) (4) A = B.
2 Hint : For addition both A and B must be of same order
1 1 1 1 to get AB, number of columns of A should be
.s

a11
(3 − 2) = , a12 = (3 – 4) = –
= equal to number of rows of B.
2 2 2 2
1 1 2 If A and B ae square matrices of same order

a21=
(6 – 2) = 2, a22 = (6 – 4) = 2 = 1
both condition are satisfied.
w

2 2
1 1 1 1 [Ans: (2) A and B are square matrices of same order]
 2 − 2  2 − 2
\A = 
 [Ans: (2)  ]  λ 1
w

2 2
    5. If A =   , then for what value of λ , A = 0?
2 1 2 1  −1 − λ 
 1 2  3 8 (1) 0 (2) ±1 (3) –1 (4) 1
2. What must be the matrix X, if 2X +  = ?
w

 3 4  7 2 λ 1
Hint : A =  
 1 3  1 −3  −1 − λ 
(1)   (2)  
 2 −1  2 −1 A2 = 0

 2 6  2 −6 λ 1 λ 1 λ 2 − 1 0 
(3)   (4)   ⇒  −1 − λ   −1 − λ  =   =0
 4 −2  4 −2      0 −1 + λ 2 
⇒ l2 – 1 = 0 1 2 2
⇒ l2 = 1
\ 2 1 −2 = 27

l = +1 [Ans: (2) +1]
a 2 b
 1 −1 a 1
6. If A =   and B =   and (A + B)
2 ⇒ 1(b + 4) – 2(2b + 2a) + 2(4 – a) = 27
 2 −1  b −1 ⇒ b + 4 – 4b – 4a + 8 – 2a = 27

m
= A2 + B2 , then the values of a and b are ⇒ – 6a – 3b + 12 = 27
Only (–2, –1) satisfies this equation.
(1) a = 4, b = 1 (2) a = 1, b = 4 [Ans: (4) (–2, –1)]

co
(3) a = 0, b = 4 (4) a = 2, b = 4 8. If A is a square matrix, then which of the following
is not symmetric?
1 −1 a 1  (1) A + AT (2) AAT (3) ATA (4) A – AT
Hint : A =   ,B=  
 2 −1  b −1 Hint : (A–AT)T = AT – (AT)T

a +1 0  = AT – A = – (A – AT)

.
A+B =  
b + 2 −2 [Ans: (4) A – AT]

ks
a +1 0  a +1 0  9. If A and B are symmetric matrices of order n,
(A + B)2 =   
b + 2 −2 b + 2 −2 where (A ≠ B), then
 (a + 1)2 0 (1) A + B is skew-symmetric
=  

oo
(2) A + B is symmetric
( a + 1) (b + 2) − 2 (b + 2) 4 (3) A + B is a diagonal matrix
1 −1 1 −1  −1 0  (4) A + B is a zero matrix
A2 =    =  
 2 −1  2 −1  0 −1 Hint : (A+ B)T = AT+ BT = A + B
 a 2 + b a − 1 [Ans: (2) A + B is symmetric]
ab
a 1  a 1     a x
B2
=   = T
 b −1  b −1  ab − b b + 1 10. If A = 
y  and if xy = 1, then det (AA ) is
 a
Since (A + B)2 = A2 + B2 equal to
 (a + 1)2 0  a 2 + b a − 1 (1) (a – 1)2 (2) (a2 + 1)2
ur

 = 
( a + 1) (b + 2) − 2 (b + 2) 4  ab − b b + 1 (3) a2 –1 (4) (a2 –1)2
 −1 0   a 2 + b − 1 a − 1 a x a y 
+  Hint : A =  ⇒ AT =  
 =  
 y a  x a
.s

 0 −1  ab − b b 
a–1 = 0 det (AAT) = det (A) . det (AT) = det (A). det (A)
⇒ a = 1, b = 4 [Ans: (2) a = 1, b = 4] a x
det (A) = = a2 – xy = a2 – 1[ xy = 1]
w

1 2 2  y a
  det (AAT) = (a2 – 1) (a2 – 1) = (a2 – 1)2
7. If A =  2 1 −2 is a matrix satisfying the
a [Ans: (4) (a2 –1)2]
2 b 
w

11. The value of x, for which the matrix

equation AAT = 9I , where I is 3 × 3 identity matrix,
then the ordered pair (a, b) is equal to e x − 2 e 7+ x 
A =  2+ x  is singular is [March - 2019]
e 2 x +3 
w

(1) (2, –1) (2) (–2, 1) e

(3) (2, 1) (4) (–2, –1) (1) 9 (2) 8 (3) 7 (4) 6
Hint : Given AA = 9I
T Hint : Since A is singular, |A| = 0
⇒ |AAT| = |9I| [ |A| = |AT|] e x−2 e7 + x
3
|A|.|A| = 9 |I| \ = 0
e 2+ x e2 x+3
|A|2 = 93 = 9 × 9 × 9 = 92 × 32 = 272
ex–2.e2x + 3 – e2 + x.e7 + x = 0
|A| = 27
 e3x + 1 – e9 + 2x = 0 R1× a, R2× b, R3× c and divide by abc
⇒ e3x + 1 = e9 + x
x1 y1 a
3x + 1 = 9 + 2x 1 1 abc 1
⇒ x = 8[Ans: (2) 8] = x2 y2 b= × =
2abc 2 abc 4 8
x3 y3 c
12. If the points (x, –2), (5, 2), (8,8) are collinear, then

m
x is equal to [Hy- 2018]  2a x1 y1 a x1 y1 
1  abc abc 
(1) –3 (2)
3
(3) 1 (4) 3  ∵ 2b x2 y2 =
2
⇒b x2 y2 =
4 
 2c x3 y3 c x3 y3 
Hint : Since the points are collinear, area of the triangle  

co
is 0. 1
x −2 1  [Ans: (3) ]
1 8
Absolute value of 5 2 1=0
2 α β
8 8 1 14. If the square of the matrix   is the unit
 γ −α 

.
1  2 1 5 1 5 2
matrix of order 2, then α, β and γ should satisfy
Absolute value of x +2 +1 =0

ks
2  8 1 8 1 8 8 the relation.
1
Absolute value of [x (2 – 8) + 2(5 – 8) + 1(40 – 16)] = 0 (1) 1+ α 2 + βγ = 0 (2) 1– α2 – βγ = 0
2
(3) 1 – α2 + βγ = 0 (4) 1 + α2 – βγ = 0
1
Absolute value of [–6x – 6 + 24] = 0 α
β  α β  1 0

oo
2 Hint : Given  = 
1    1
Absolute value of [–6x + 18] = 0 γ
−α   γ −α  0
2 2
 α + βγ αβ − αβ  1 0
⇒   = 
Absolute value of –6x + 18 = 0 ⇒ 6x = 18 ⇒ x = 3  αγ − αγ
2
βγ + α  0 1
 
ab
 [Ans: (4) 3] α 2 + βγ 0  1 0
x1 y1 ⇒   =  
2a  0
2
βγ + α  0 1
abc
13. If 2b x2 y2 = ≠ 0, then the
2 ⇒ a2 + bg = 1
2c x3 y3 ⇒ 1- a2- bg = 0
ur

area of the triangle whose vertices are  [Ans: (2) 1– α2 – βγ = 0]

 x1 y1   x2 y2   x3 y3 
 ,  ,  ,  ,  ,  is a b c ka kb kc
a a b b c c 15. If Δ = x y z , then kx ky kz is
.s

1 1
(1) (2) abc p q r kp kq kr
4 4
1 1 (1) D (2) kD (3) 3kD (4) k3D
(3) (4) abc
w

8 8 Hint :Taking k common form R1, R2 and R3 we get,

Hint : Area of the triangle
ka kb kc a b c
x1 y1
1 kx ky kz = k3 x y z = k3D [Ans: (4) k3D]
w

a a
= Absolute value of 1 x2 y2 kp kq kr p q r
1
2 b b
3− x −6 3
x3 y3
w

1 16. A root of the equation −6 3 − x 3 = 0 is

c c
x1 y1 3 3 −6 − x
1
a a (1) 6 (2) 3 (3) 0 (4) –6
1 x2 y2
Consider 1 3 − x −6 3
2 b b
x3 y3 Hint : −6 3 − x 3 = 0
1
c c 3 3 −6 − x
 −x −x −x 19. If ⋅ denotes the greatest integer less than or
−6 3 − x 3 = 0 equal to the real number under consideration and
–1 ≤ x < 0, 0 ≤ y < 1, 1 ≤ z < 2 , then the value of the
3 3 −6 − x
 x + 1  y  z
[ Applying R1 → R1 + R2 + R3]
determinant  x   y  + 1  z  is

m
1 1 1
 x  y  z + 1
– x −6 3 − x 3 = 0
(1)  z  (2)  y  (3)  x (4)  x  +1
3 3 −6 − x
−1 + 1 0 1

co
⇒ x = 0  x  + 1  y   z 
is a root of the equation. [Ans: (3) 0] Hint :  x   y  + 1  z  = −1 0 + 1 1
 0 a −b  x   y   z  + 1 −1 0 1+1
  ∵ − 1 ≤ x < 0 ⇒  x  = −1
17. The value of the determinant of A =  − a 0 c

.
is  
 b −c 0   0 ≤ y < 1 ⇒  y  = 0

ks
(1) –2abc (2) abc  1 ≤ z < 2 ⇒  z  = 1
(3) 0 (4) a2 + b2 + c2 0 0 1
−1 1
−1 1 1 = 1  [Expanded along R1]
0 a −b −1 0
−a c −a 0 −1 0 2
Hint : |A|= − a 0 c = 0 – a
b −c 0
b 0
–b
b −c

= – a (– bc) – b (ac) = abc – abc = 0 oo

[Ans: (3) 0]
18. If x1, x2, x3 as well as y1, y2, y3 are in geometric
= 1[0 + 1] = 1 =  z  [Ans: (1)  z  ]

a 2b 2c
20. If a ¹ b, b, c satisfy 3 b c = 0, then abc =
ab
progression with the same common ratio, then the 4 a b
points (x1, y1), (x2 , y2 ), (x3 , y3 ) are (1) a + b + c (2) 0
(1) vertices of an equilateral triangle (3) b 3
(4) ab + bc
(2) vertices of a right angled triangle a 2b 2c
ur

(3) vertices of a right angled isosceles triangle Hint : 3 b c = 0 and a ¹ b

(4) collinear 4 a b
Hint : x1, x2, x3 are in G.P. ⇒ a(b2 – ac) – 2b(3b – 4c) + 2c(3a – 4b) = 0
.s

2
Let it be represented as a, ar, ar y1, y2, y3 are in G.P ⇒ ab2 – a2c – 6b2 + 8b c + 6ac – 8b c = 0
Let it be represented by b, br, br2
⇒ ab2 – 6b2 – a2c + 6ac = 0⇒b2 (a – 6) – ac(a – 6) = 0
 (They have same common ratio)
w

⇒ (a – 6) (b2– ac) = 0⇒ a = 6 or b2 = ac
x1 y1 1 a b 1
1 1 ⇒ b2 = ac ⇒ b3 = abc [Ans: (3) b3]
Area of = 2 2 x y 2 1 = ar br 1
2 2 −1 2 4 −2 4 2
x3 y2 1 ar br 2 1
w

21. If A = 3 1 0 and B = 6 2 0 ,then B is

1 1 1 −2 4 2 −2 4 8
ab
= r r 1
2 given by
w

r r2 1
2
[Ans: (4) collinear]
(1) B = 4A (2) B= –4A
(3) B = –A (4) B = 6A
−1 2 4
Hint : A = 3 1 0 and
−2 4 2
 −2 4 2 b + 3d = 1 ⇒ b + 3(–1) = 1 ⇒ b = 4
B = 6 2 0  a b  1 4 
\A =   =  
−2 4 8  c d  0 −1
1 4 
−2 4 8 [Ans: (3)  ]
0 −1

m
Then B = – 6 2 0 R1 ↔ R3
 3 −2
−2 4 2 24. If A + I =   , then (A + I) (A – I) is equal to
4 1
Taking out 2 from R1 and 2 from R2, we get
 −5 −4  −5 4

co
−1 2 4 (1)   (2)  
B = – (2) (2) 3 1 0 = – 4A  8 −9  −8 9 
−2 4 2 5 4   −5 −4
(3)   (4)  −8 −9
[Ans: (2) B = – 4A] 8 9   

.
22. If A is skew-symmetric of order n and C is a  3 −2

ks
Hint : A+I =  
column matrix of order n × 1, then CTAC is 4 1 
(1) an identity matrix of order n 1 0  3 −2
(2) an identity matrix of order 1 A +   =  
0 1 4 1 
(3) a zero matrix of order 1

oo
(4) an identity matrix of order 2  3 −2 1 0
⇒ A =   –  
Hint : C is of order n × 1 ⇒ CT is of order 1× n  4 1  0 1
\ CTA of order 1 × n  2 −2
And CTAC is of order (1 × n ) × ( n × 1) = (1 × 1) ⇒ A =  
4 0 
Since A is a skew-symmetric matrix, CTAC is a
ab
zero matrix of order 1.  2 −2 1 0 1 −2
A–I =  –   =  4 −1
 [Ans: (3) a zero matrix of order 1]  4 0  0 1  
 3 −2 1 −2
 1 3 \ (A + I) (A – I) =   
23. The matrix A satisfying the equation    4 1   4 −1
ur

 1 1 0 1
 3 − 8 −6 + 2  −5 −4
A=   is =   = 
0 −1  4 + 4 −8 − 1   8 −9
 1 4 1 −4  −5 −4
.s

(1)   (2)   [Ans: (1)  ]

 −1 0 1 0  8 −9
 1 4 1 −4 25. Let A and B be two symmetric matrices of same
(3)   (4)   order. Then which one of the following statement
w

0 −1 1 1 is not true?

1 3 1 1  (1) A + B is a symmetric matrix
Hint :
0 1 A = 0 −1
    (2) AB is a symmetric matrix
w

a b  (3) AB = (BA)T
Let A =   (4) AT B = ABT
c d 
1 3  a b  1 1  Hint : For symmetric matrix = AT = A
w

⇒ 0 1  c d  = 0 −1 (BA)T = AT BT = AB
    
AT B = AB = ABT
 a + 3c b + 3d  1 1  Sum of two symmetric matrix is also a symmetric
⇒  c  =  
 d  0 −1 matrix.
⇒ c = 0 and d = – 1 AB is a symmetric matrix is not true.
Also a + 3c = 1 ⇒ a + 0 = 1 ⇒ a = 1 [Ans: (2) AB is a symmetric matrix]
 2. Prove that the relation R defined on the
TEXTUAL QUESTIONS
→ → → →
set V of all vectors by a R b if a = b ’ is an
equivalence relation on V.
EXERCISE 8.1 → → →
Solution : Let a , b , c ∈ V, where V is the set of all
1. Represent graphically the displacement of (i) vectors.

m
→ →
45cm 30° north of east. (ii) 80km, 60° south of
Let R be the relation defined by a = b
west → →
(i) Reflexive: a = a ⇒ aRa ⇒ R is Reflexive.

co
Solution :
→ → → →
(i) 45 cm 30o north of east. (ii) Symmetric: a = b ⇒ b = a
N \ aRb ⇒ bRa ⇒ R is Symmetric.
→ → → → → →
(iii) Transitive: a = b, b = c ⇒ a=c

.
\ aRb, bRc ⇒ aRc

ks
P
\ R is transitive.
W 30o
E Hence, R is an equivalence relation.
O → →
3. Let a and b be the position vectors of the points

oo A and B. Prove that the position vectors of the

points which trisects the line segment AB are
→
a+ 2b
→
b+ 2 a
→ →
ab
S and .
3 3
→ a b
The vector OP represents a displacement of Solution : A B
P Q
45 cm, 30o north of east.
→ →
(ii) 80 km, 60o south of west. Let a and b be the position vectors of the points A and B.
ur

N → → → →
⇒ OA = a and OB = b .
Let P divides the line segment AB in the ratio 1:2 and Q
.s

divides the line segment AB in the ratio 2:1

→ → →
O E
1.(OB) + 2(OA)
w

W \ OP =
60o 1+ 2
→ → → →
1( b ) + 2( a ) b+ 2 a
= =
w

3 3
Q → →
→ 2(OB) + 1(OA)
and OQ =
2 +1
S
w

→ →
2 b+ a
→ →
a+ 2 b
→

The vector OQ represents a displacement of = =

3 3
80 km, 60o south of west. → →
b+ 2 a
Hence, the required position vectors are and
→ → 3
a+ 2 b
.
3
4. If D and E are the midpoints of the sides AB and AC 5. Prove that the line segment joining the midpoints
→ → 3→ of two sides of a triangle is parallel to the third
of a triangle ABC, prove that BE + DC = BC . side whose length is half of the length of the third
2
side.
Solution : Let the position vectors of the vertices of the
→ → → Solution : Let the position vectors of the vertices of the
DABC be a , b and c respectively. → → →

m
triangle be a , b and c respectively.
→ → → → → →
→ → → → → →
⇒ OA = a , OB = b and OC = c OA a , OB = b and OC = c .
=
A

co
a Since D is the mid-point of AB,

→ → → → →
D
OA + OB a+ b
E OD = =
2 2
Also E is the mid-point of AC,

.
→ →

ks
B
c
C → OA + OC a+ c
→ →

b OE = =
2 2
Since D is the mid-point of the side AB,
→ → → a+ c
→
a+ b
→ → →

→ → → DE = OE – OD = –

oo
a+ b 2 2
OD =  ... (1)
2 → → → → → →
and E is the mid-point of the AC a+ c− a− b c− b
= =
2 2
→ → →

⇒ OE =
a+ c
 ... (2) 1 → → 1 →
ab
2 = ( OC − OB ) = ( BC )
2 2
→ → →
a+ c → a+ c − 2 b
→ → → → →
→ → 1
BE = OE – OB = −b = ⇒ DE = l( BC ) where l =
2 2 2
 [From (2)] → → → 1 →
ur

\ DE || BC and DE = ( BC )
2
→ → → a+ b 2 c− a− b→
→ → → → →
→ →
DC = OC – OD = c − = Hence, DE is parallel to BC and whose length is half of
2 2
.s

the length of the third side.

→ →
a+ c − 2 b 2 c − a− b
→ → → → → →

\ BE + DC = +
2 2 6. Prove that the line segments joining the midpoints
w

→ → → → → → of the adjacent sides of a quadrilateral form a

a + c− 2 b+ 2 c− a − b
= parallelogram.
2
→ →
Solution :
3c−3b 3 → →
w

= = (c − b) D R C
2 2 c
d
3 → →
= ( OC − OB )
w

2 S Q
→ → 3→
\ BE + DC = BC
2
Hence proved. A B
a P b
Let the position vectors of the vertices of the
→ → → →
quadrilateral be a , b , c and d .
 Let P, Q, R, S be the mid-points of the adjacent → →
sides of the quadrilateral. 7. If a and b represent a side and a diagonal of a
To prove that PQRS is a parallelogram. parallelogram, find the other sides and the other
A diagonal.
Solution : Let ABCD be the parallelogram.

m
D -a C
D E

co
a−b b
C
b−a
B

→ a+ b → b+ c
→ → → →

.
OP = , OQ = , A
2 2 B

ks
a
→ c+ d → a+ d
→ → → →
→ → → →
OR = , OS = Let AB = a and AC = b .
2 2
→ → → → → →
Now, PQ = OQ − OP In DABC, AC = AB + BC

=
→
b+ c

→
2
→

–
→
a+ b
2
→
→

→
→

oo ⇒

⇒
→
BC = b − a
→

Since ABCD is a parallelogram,

→
b = a + BC
→
→
→
ab
b + c− a− b
= → →
2
→ → CD = – AB
c− a → →
=  ... (1) CD = – a
2
→ → → → →
ur

→ →
SR = OR − OS DA = –( BC ) = − ( b − a )
→ → → →
→ → →
c+ d a+ d ⇒ DA = a – b .
= – → → →
.s

→ → → → →
2 2
→ → → → In D BCD, BD = BC + CD = b – a – a = b – 2 a
c+ d − a− d → →
= Hence, the other sides of the parallelogram are b – a ,
2
w

→ → → → → → →
c− a – a , a – b and the other diagonal is b – 2 a .
=  ... (2)
2 → → → →
→ → 8. If PO + OQ = QO + OR , prove that the points P,
w

From (1) and (2), PQ = SR Q, R are collinear.

→ → → → → → → →
and PQ = 1(SR) ⇒ PQ || SR Solution : Given PO + OQ = QO + OR
→ →
w

Thus, one pair of parallel sides of PQRS are

⇒ PQ = QR
parallel and equal.
 [By triangle law of addition]
\ PQRS is a parallelogram. → →
⇒ PQ = QR
and Q is a common point.
Hence, the points P, Q, R are collinear.
9. If D is the midpoint of the side BC of a triangle 10. If G is the centroid of a triangle ABC, prove that
→ → → → → → →
ABC, prove that AB + AC = 2 AD . GA + GB + GC = 0 . [Hy - 2018]
Solution : Let the position vector of the vertices of the Solution : Let the position vector of the vertices of the
→ → → → → →
triangle be a , b and c respectively. DABC be a , b and c respectively.
→ → → → → →

m
→ → → → → →
\ OA = a , OB = b , OC = c . \ OA = a , OB = b , OC = c .
Since D is the mid-point of BC, Since G is the centroid of DABC, we have
→ →
→

co
b+ c
→ → → →
OD =  ... (1) OA + OB+ OC
2 ⇒ OG =
→ → → 3
To prove that AB + AC = 2 AD → → → →
⇒ 3OG = OA + OB + OC  ... (1)

.
A
a
→ → →
Now, LHS = GA + GB+ GC

ks
→ → → → → →
= OA − OG + OB− OG + OC − OG
→ → → →
= ( OA + OB + OC ) – 3 OG
→ → → → → →

oo
→
B C = ( OA + OB + OC ) – ( OA + OB + OC ) = 0
D c
b
= RHS Hence proved.
→ →
LHS = AB + AC 11. Let A, B, and C be the vertices of a triangle. Let D,E,
ab
→ → → → and F be the midpoints of the sides BC, CA, and
= OB – OA + OC – OA → → → →
→ → → → AB respectively. Show that AD + BE + CF = 0 .
= b–a+c–a
→ → →
Solution : Let the position vector of the vertices of the
ur

= b + c –2 a → → →
DABC be a , b and c respectively.
→ Since D is the mid-point of BC.
RHS = 2 AD → →
→ b+ c
→ →
.s

⇒ OD =
= 2( OD − OA ) 2
 → → → A
b+ c
= 2  − a  [From (1)] a
w

 2 
 
 → → → → → → F E
b+ c− 2 a 
= 2 
w

= b + c –2 a
 2 
 
\ LHS = RHS B C
D c
w

b
Hence proved.
E is the mid-point of AC,
→ →
→ a+ c
⇒ OE = and F is the mid-point of AB
2
→ →
a+ b
→

OF =
2
 → → → → → → → → →
To prove that AD + BE + CF = 0 = OB – OA + OD – OA + OB
→ → → → → →
LHS = AD + BE + CF – OC + OD – OC
→ → → → → → → → → → → → → →
= OD – OA + OE – OB + OF – OC = b – a + d – a + b – c + d – c

m
→ → → → → → → → → →
b+ c → a+ c → a+ b → = −2 a + 2 b − 2 c + 2 d

= –a + –b+ –c → → → →
2 2 2
= 2[( b + d ) – ( a + c )]

co
→ → → → → → → → →
= b+ c − 2 a+ a+ c − 2 b+ a+ b− 2 c → →
= 2[2 OF – 2 OE ] [From (1)]
2
→
0 → → → →
= = 0 = RHS Hence proved. = 4.[ OF − OE ] = 4. EF = RHS
2

.
Hence proved.

ks
12. If ABCD is a quadrilateral and E and F are the
midpoints of AC and BD respectively, then prove EXERCISE 8.2
→ → → → →
that AB + AD + CB + CD = 4EF . 1. Verify whether the following ratios are direction
cosines of some vector or not.
Solution :

→
Let the position vector of the vertices of
the quadrilateral ABCD be a , b , c and d
respectively.
→ → →
→ → →

oo → (i)
Solution :

(i)
1 3 4
, , (ii)
5 5 5

Given ratios are

1 1 1
2
, ,

1 3
2 2

4
, and .
4 3
(iii) , 0,
3 4
ab
\ OA = a , OB = b, 5 5 5
→ → → → 1 3
Let the ratios are l = , m = , n =
4
OC = c and OD = d . 5 5 5
Since E and F are the mid-points of AC and BD 2 2 2
 1  3  4 1 9 16
ur

respectively, we have \ l2 + m2 + n2 =   +   +   = + +
 5  5  5 25 25 25
→ a+ c
→ →
26
OE = and = ≠1
2 25
.s

→ b+ d
→ →
Hence, the given ratios are not the direction
OF =  ... (1)
2 cosines of any vector.
→ → → → → 1 1 1
w

To prove that AB + AD + CB + CD = 4 EF (ii) Let l = ,m= and n =

2 2 2
C 2 2 2
c  1  1
   1
\ l2 + m2 + n2 =  +   +  
 2 
D
2 2
w

d
1 1 1 2 +1+1 4
= + + = = =1
2 4 4 4 4
w

A
Hence, the given ratios are direction cosines
a of some vector.
B 4 3
b (iii) Let l = , m = 0, n =
3 4
→ → → → 2 2 2  4
2
2  3
2

LHS = AB + AD + CB + CD \ l + m + n =   + 0 + 
 3  4
 16 9 256 + 81 337 ∧ ∧ ∧
= + = = ¹1 (ii) Give vector is 3 i + j + k
9 16 16 × 9 144 ∧ ∧ ∧
Hence, the given ratios are not the direction The direction ratios of 3 i + j + k are 3, 1, 1.
cosines of any vector. r = x2 + y 2 + z 2
= 32 + 12 + 12 = 11

m
2. Find the direction cosines of a vector whose
3 1 1
direction ratios are (i) 1 , 2 , 3 (ii) 3 , – 1 , 3 Hence, its direction cosines are , ,
(iii) 0 , 0 , 7 11 11 11
∧
Solution : (iii) Given vector is j

co
∧
(i) Given direction ratios are 1, 2, 3 The direction ratios of j are 0, 1, 0.
Let x = 1, y = 2, z = 3 x = x2 + y 2 + z 2
r = x 2 + y 2 + z 2 = 1 + 4 + 9 = 14 =0 + 12 + 0 = 1

.
x y z 0 1 0
The direction cosines are , , Hence, its direction cosines are , , ⇒ 0,

ks
r r r 1, 0. 1 1 1
1 2 3
Thus, the direction cosines are , , ∧ ∧ ∧
14 14 14 (iv) The given vector is 5 i − 3 j − 48 k
(ii) Let x = 3, y = – 1, z = 3 The direction ratios are 5, – 3, – 48.
\ r =

2
x +y +z 2 2

Hence, the direction consines are

3
,
−1 3
,
= 9 + 1 + 9 = 19

oo

r =
=

=
x2 + y 2 + z 2
52 + ( −3) + ( −48)

25 + 9 + 2304 =
2 2

2338
ab
19 19 19
Hence, the direction cosines are
(iii) Let x = 0, y = 0, z = 7 5 −3 −48
, ,
\ r = x2 + y 2 + z 2 = 0 + 0 + 72 = 7 2338 2338 2338
∧ ∧ ∧
0 0 7
ur

Hence, the direction consines are , , (v) The given vector is 3 i − 3 k + 4 j

7 7 7 ∧ ∧ ∧
⇒ 0, 0, 1. ⇒ 3 i + 4 j− 3k
The direction ratios are 3, 4, – 3.
.s

3. Find the direction cosines and direction ratios for

r = x2 + y 2 + z 2
the following vectors.
32 + 42 + ( −3)
∧ ∧ ∧ ∧ ∧ ∧ 2
=
(i) 3 i − 4 j + 8 k (ii) 3 i + j + k
w

∧ ∧ ∧ ∧ =9 + 16 + 9 = 34
(iii) j (iv) 5 i − 3 j − 48 k 3 4 −3
∧ ∧ ∧ ∧ ∧ Hence, the direction cosines are , ,
(v) 3 i − 3 k + 4 j (vi) i − k 34 34 34
w

∧ ∧
Solution :
∧ ∧ ∧
(vi) The given vector is i − k
(i) Given vector is 3 i − 4 j + 8 k The direction ratios are 1, 0, – 1.
∧ ∧ ∧
w

The direction ratios of 3 i − 4 j + 8 k are 3, – 4, 8. x = x2 + y 2 + z 2

r = x2 + y 2 + z 2 = 12 + 02 + ( −1)
= 2
2

3 + ( −4) + 8
2 2 2
= 1 0 −1
Hence, the direction cosines are , ,
= 9 + 16 + 64 = 89 1 −1 2 2 2
⇒ , 0,
3 −4 8 2 2
Hence, its direction cosines are , ,
89 89 89
4. A triangle is formed by joining the points (1, 0, 0), → 1 1 1
(0, 1, 0) and (0, 0, 1). Find the direction cosines of The direction cosines of BE are , − ,
6 6 6
the medians. 2× 2×
2 2 2
Solution : Let the vertices of the triangle be A(1, 0, 0), → → →
B(0, 1, 0), C(0, 0, 1). ⇒The median CF = OF − OC
A (1, 0, 0)

m
 1 ∧ 1 ∧ ∧  ∧ ∧ ∧
=  i + j + 0 k  −  0 i + 0 j + k 
2 2   
1 1
∧ ∧ ∧
F E = i + j− k

co
2 2
1 1 6
C r = + +1 =
B 4 4 2
D (0, 0, 1)
(0, 1, 0)
→

.
1 1 −1
Let D, E, F are the mid-point of the sides \ The direction of cosines of CF are , ,

ks
6 6 6
BC, CA and AB respectively. 1 1 −2 2× 2×
⇒ , , 2 2 2
 x1 + x2 y1 + y2 z1 + z2  6 6 6
\ D is  , , 
 2 2 2  1 1
5. If , , a are the direction cosines of some

oo
 1 1 1 1 2 2
⇒ D is  0, ,  and E is  , 0,  ,
 2 2 2 2 vector, then find a.
1 1  Solution : Given direction cosines of some vector are
F is  , , 0
2 2  1 1
→ → → , ,a
ab
Medians AD = OD − OA 2 2
1 1
 ∧ 1 ∧ 1 ∧  ∧ ∧ ∧ Let l = , m = ,n = a
=  0 i + j+ k −  i − 0 j+ 0 k 2 2
 2 2    We know that l2 + m2 + n2 = 1
∧ 1∧ 1∧
2 2
= − i + j + k  1  1 
ur

2
2 2 ⇒   +   +a = 1
2 2 2 1 1 2 2
r = x + y + z = 1+ + 1 1
4 4 ⇒ + + a2 = 1
4 +1+1 6 4 2
.s

= = 1 1 4 −1
4 2 ⇒ a2 = 1 − − =
4 2 4
→ 1 1 4 −1− 2 1
Hence, the direction cosines of AD are, =
1− − = =
4 2 4 4
w

−1 1 1 −2 1 1
, , ⇒ , , 1
6 6 6 6 6 6 a = ±
2× 2× 4
2 2 2
w

1
→ → → ⇒ a = ±
2
The median BE = OE − OB
1∧ ∧ 1 ∧  ∧ ∧ ∧ 6. If (a, a + b , a + b + c) is one set of direction ratios
=  2 i − 0 j + 2 k  –  0 i + j + 0 k 
w

of the line joining (1, 0, 0) and (0, 1, 0), then find a

1∧ ∧ 1∧ set of values of a, b, c.
= i − j+ k
2 2
1 1 1+ 4 +1 6 Solution : Given points are A(1, 0, 0) and B(0, 1,0)
r = +1+ = =
4 4 4 2 and one set of direction ratios are a, a + b,
a + b + c.
 Case (i):
( 35 ) + ( 6 )
2 2 ∧ ∧
= = | a |2 + | c |2
→ → → ∧ ∧ ∧ ∧ ∧ ∧
AB = OB – OA = (0 i + j + 0 k ) − ( i + 0 j + 0 k ) By Pythagoras theorem, the given vectors
∧ ∧
= − i + j form a right angled triangle.
→ 8. Find the value of l for which the vectors

m
\ Direction ratios of the line AB are (–1, 1, 0)
→ ∧ ∧ ∧ → ∧ ∧ ∧
Given (–1, 1, 0) = (a, a + b, a + b + c) a = 3 i + 2 j + 9 k and b = i + λ j + 3 k are parallel.
→ ∧ ∧ ∧ → ∧ ∧ ∧
Equating the like components both sides, we get Solution : Given a = 3 i + 2 j + 9 k , b = i + λ j + 3 k

co
→ →
a = – 1, a + b = 1, a + b + c = 0 Given a || b
→ →
a = – 1, –1 + b = 1 ⇒ b = 2 \ a = (some scalar) b
– 1 + 2 + c = 0 ⇒ c = –1 → ∧ ∧ ∧
a = 3 i + 2 j+ 9 k

.
⇒

\ a = – 1, b = 2, c = –1

ks
∧ 2∧ ∧
Case (ii): = 3( i + j+ 3k )
→ 3
→ → → ∧ ∧ ∧ ∧ ∧ ∧
→
a = 3( b )
BA = OA – OB = ( i + 0 j + 0 k ) − (0 i + j + 0 k ) 2∧ ∧ → ∧
∧ ∧ j+ 3k b = i+

oo
= i − j 3
∧ ∧ ∧
→ Comparing this with i + λ j + 3 k we get
\ Direction ratios of the line BA are (1, –1, 0) 2
l =
Given (1, –1, 0) = (a, a + b, a + b + c) 3
9. Show that the following vectors are coplanar
ab
Equating the like components both sides, we get
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
a = 1, a + b = – 1, a + b + c = 0 (i) i − 2 j + 3 k , − 2 i + 3 j − 4 k , − j + 2 k
a = 1, 1 + b = –1 ⇒ b = – 2 ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
(ii) 2 i + 3 i + k , i − j , 7 i + 3 j + 2 k . [Hy - 2018]
1–2+c =0⇒ c=1 → ∧ ∧ ∧ → ∧ ∧ ∧
ur

\ a = 1, b = – 2, c = 1 Solution : Let a = i − 2 j + 3 k , b = −2 i + 3 j − 4 k ,
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ → ∧ ∧
7. Show that the vectors 2 i − j + k , 3 i − 4 j − 4 k , i − 3 j − 5 k c = − j+ 2 k
∧ ∧ ∧ ∧ ∧ → → →
j − 4 k , i − 3 j − 5 k form a right angled triangle. Let a = s b + t c
.s

∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
Solution : Let the sides of the triangle be
⇒ i − 2 j + 3 k = s( −2 i + 3 j − 4 k ) + t( − j + 2 k )
∧ ∧ ∧ ∧ → ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
a = 2 i − j+ k , b = 3 i − 4 j− 4 k , ⇒ i − 2 j + 3 k = (–2s) i + (3s – t) j + (–4s + 2t) k
w

∧ ∧ ∧ ∧
c = i − 3 j− 5k Equating the like components both sides, we
∧ get
22 + ( −1) + 12
2
|a | =
– 2s = 1  ... (1)
w

= 4 +1+1 = 6 3s – t = – 2  ... (2)

→
|b| = 32 + ( −4) + ( −4)
2 2 – 4s + 2t = 3
 ... (3)
1
w

From (1), s = –
= 9 + 16 + 16 = 41 2
∧ 1
| c | = 1 + ( −3) + ( −5)
2 2 2
Substituting s = – in (2) we get,
2
= 1 + 9 + 25 = 35  −1 3
3  – t = – 2 ⇒ – – t = – 2
( 41) = 41=35 + 6
2
Now | b |2 =
→
 2 2
 3
–t = –2+ 3  1 3 7
2 ⇒ 2 = – +7  ⇒2=– +
2  2 2 2
−4 + 3 −1
–t = =
2 2 −3 + 7 4
1 ⇒ 2 = ⇒ 2= ⇒ 2=2
t = 2 2
2

m
1 1 The value of s and t satisfy equation (1)
Substituting s = – , t = in (3) we get,
2 2 One vector is a linear combination of other two
 −1  1 vectors.
−4   + 2   = 3
 2  2 Hence, the given vectors are co-planar.

co
⇒ 2+1 = 3 10. Show that the points whose position vectors
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
⇒ 3 = 3
4 i + 5 j + k , − j − k , 3 i + 9 j + 4 k and
which satisfies equation (3). ∧ ∧ ∧

Thus, one vector is a linear combination of − 4 i + 4 j + 4 k are coplanar.

.
other two vectors. Solution : Let the position vectors of the given vector

ks
Hence, the given vectors are co-planar. be
→ ∧ ∧ ∧
→ ∧ ∧ ∧ OA = 4 i + 5 j + k
(ii) Let a = −2 i + 3 j + k → ∧ ∧
→ → → OB = − j − k

oo
b = i– j → ∧ ∧ ∧
→ ∧ ∧ ∧ OC = 3 i + 9 j + 4 k and
c = 7 i + 3 j+ 2 k → ∧ ∧ ∧
→ → → OD = −4 i + 4 j + 4 k
Let a = s b + t c → → → →
ab
where s and t are scalars Let a = AB = OB – OA
∧ ∧ ∧ → → ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
⇒ 2 i + 3 j + k = s( i – j ) + t( 7 i + 3 j + 2 k ) = ( − j − k ) – ( 4 i + 5 j + k )
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
⇒ 2i+ 3 j+ k = i (s + 7t) + j (–s + 3t) + k (2t) = −4 i − 6 j − 2 k
→ → → →
ur

Equating the like components both sides, we get b = AC = OC – OA

∧ ∧ ∧ ∧ ∧ ∧
2 = s + 7t  ... (1)
= ( 3 i + 9 j + 4 k ) – ( 4 i + 5 j + k )
3 = – s + 3t ... (2) ∧ ∧ ∧
= − i + 4 j + 3 k and
.s

1 = 2t  ... (3)
Let us solve (2) (3), to get the values of s and t.
→ → → →
c = AD = OD – OA
∧ ∧ ∧ ∧ ∧ ∧
1
= ( −4 i + 4 j + 4 k ) – ( 4 i + 5 j + k )
w

From (3), t =
2 ∧ ∧ ∧
1 = −8 i − j + 3 k
Substituting t = in (2) we get,
2 → → → ∧ ∧ ∧
w

Also, let a = s b + t c = −4 i − 6 j − 2 k
 1
3 = – s + 3  ∧ ∧ ∧ ∧ ∧ ∧
 2 = s( − i + 4 j + 3 k ) + t( −8 i − j + 3 k )
3 ∧ ∧ ∧ ∧ ∧ ∧
⇒ 3 = –s +
w

−4 i − 6 j − 2 k = (– s – 8t) i + (4s – t) j + (3s + 3t) k

2
3 3 − 6 −3 Equating the like components on both sides, we get
⇒ s = –3= =
2 2 2 – 4 = – s – 8t  ... (1)
1 3 – 6 = 4s – t  ... (2)
∴t = ,s =–
2 2 – 2 = 3s + 3t  ... (3)
3 1
Substituting s = – and t = in (1) we get,
2 2
 (1) × 4 ⇒ – 16 = – 4s – 32t → → →
(2) ⇒ – 6 = 4s – t |3 a − 2 b + 5 c | = ( −15)2 + 272 + 132
2
2 − 22
= 225 + 729 + 169 = 1123
Adding, – 22 = – 33t ⇒ t= = → → → −15 27 13
2 3 − 33 Direction cosines of 3 a − 2 b + 5 c is , ,
Substituting t = in (1) we get, 1123 1123 1123
3

m
 2 16 12. The position vectors of the vertices of a triangle
– 4 = – s – 8   ⇒ – 4 = – s – ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
 3 3 are i + 2 j + 3 k , 3 i − 4 j + 5 k and −2 i + 3 j − 7 k .
16 12 − 16 4 Find the perimeter of the triangle.
⇒ = =−
s = 4–

co
3 3 3 Solution : Let the vertices of the triangle be A, B, C.
2 4
Substituting t = , and s = − in (3) we get,
→ ∧ ∧ ∧
3 3 Then, given OA = i + 2 j + 3 k ,
 4  2 → ∧ ∧ ∧
– 2 = 3  −  + 3   OB = 3 i − 4 j + 5 k and
 3 3

.
→ ∧ ∧ ∧

ks
⇒ –2 =– 4 + 2 ⇒ – 2 = – 2 OC = −2 i + 3 j − 7 k
which satisfies equation (3). → → →
Thus, one vector is the linear combination of AB = OB – OA
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
other two vectors. = ( 3 i − 4 j+ 5 k ) – ( i + 2 j+ 3k ) = 2 i − 6 j+ 2 k
Hence, the given points are co-planar. →

∧
11. If
∧ →
→ ∧ ∧ ∧ →
a = 2 i + 3 j − 4 k , b = 3 i − 4 j − 5 k , c =and
∧ ∧ ∧
j − 5 k , c = − 3 i + 2 j + 3 k , find the magnitude and
direction cosines of
∧∧ ∧
−3 i + 2 j + 3k
| AB | = 2 + ( −6) + 2
∧

= 4 + 36 + 4 = 44
→ → →
BC = OC – OB
2 ∧ → 2

oo ∧
2
ab
→ → → → → → ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
(i) a + b + c (ii) 3 a − 2 b + 5 c = ( −2 i + 3 j − 7 k ) – ( 3 i − 4 j + 5 k ) = −5 i + 7 j − 12 k
→
Solution : Given a = 2 i + 3 j − 4 k
∧ ∧ ∧ →
| BC | = ( −5)2 + 72 + ( −12)2
→ ∧ ∧ ∧
ur

b = 3 i − 4 j − 5 k and = 25 + 49 + 144
→ ∧ ∧ ∧
c = −3 i + 2 j + 3 k = 218

→ → → ∧ ∧ ∧ ∧ ∧ ∧ → → →
a + b + c = ( 2 i + 3 j− 4 k ) + ( 3 i − 4 j− 5k ) CA = OA – OC
.s

(i)
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
 + ( −3 i + 2 j + 3 k ) = ( i + 2 j + 3 k ) – ( −2 i + 3 j − 7 k ) = 3 i − j + 10 k
∧ ∧ ∧
= 2 i + j − 6 k →
w

32 + ( −1) + 102
2
→ → → | CA | =
2 + 1 + ( −6)
2 2 2
| a + b + c | =
= 9 + 1 + 100 = 110
= 4 + 1 + 36 = 41 \ Perimeter of DABC,
w

→ → → 2 1 −6 → → →
Direction cosines of ( a + b + c ) is , ,
41 41 41 | AB | + | BC | +| CA | = ( 44 + 218 + 110 units )
→ → →
→ → → ∧ ∧ ∧
13. Find the unit vector parallel to 3 a − 2 b + 4 c , if
w

(ii) 3 a − 2 b + 5 c = 3( 2 i + 3 j − 4 k )–
→ → → → ∧ ∧ ∧ → ∧ ∧ ∧ → ∧
∧ ∧ ∧ ∧
3 i −∧ 4 j − 5 k ) ∧+ → 3∧ a −∧2 b + 4 c , if a = 3 i − j − 4 k , b = − 2 i + 4 j − 3 k , and c = i
2(→ → → → ∧
5( −3 i + ∧2 j + 3∧ k ) ∧ → ∧ ∧ ∧
3 a − 2 b +∧ 4 c , ∧if a =∧ 3 i ∧− j ∧− 4 k ,∧b = ∧− 2 i ∧+ 4 j ∧− 3 k , and c = i + 2 j − k .
= 6 i + 9 j − 12 k − 6 i + 8 j + 10 k − 15 i + 10 j + 15 k → ∧ ∧ ∧
∧ ∧ ∧
Solution : Given a = 3 i − j− 4 k
= −15 i + 27 j + 13 k → ∧ ∧ ∧
b = −2 i + 4 j − 3 k
 → ∧ ∧ ∧ → → →
and c = i + 2 j − k PQ = OQ – OP
→ → → ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
3 a – 2 b + 4 c = 3( 3 i − j − 4 k ) – 2( −2 i + 4 j − 3 k ) + = ( 2 i + 5 j ) – ( i + j + k )
∧ ∧ ∧ ∧ ∧ ∧
4( i + 2 j − k ) = i + 4 j− k

m
∧ ∧ ∧ ∧ ∧ ∧
= 9 i − 3 j − 12 k + 4 i − 8 j + 6 k + 4 i + 8 j − 4 k
∧ ∧ ∧ → → →
RS = OS – OR
∧ ∧ ∧
∧ ∧ ∧ ∧ ∧ ∧
= 17 i − 3 j − 10 k
= ( i − 6 j − k )– ( 3 i + 2 j − 3 k )

co
→ → →
17 2 + ( −3) + ( −10)
2 2 ∧ ∧ ∧
|3 a – 2 b + 4 c | =
= −2 i − 8 j + 2 k
= 289 + 9 + 100 = 398 ∧ ∧ ∧ →
→ → → = – 2( i + 4 j − k ) = – 2 PQ
\ Unit vector parallel to (3 a – 2 b + 4 c ) is → →

.
\ RQ = l PQ where l = – 2

ks
1 ∧ ∧ ∧

398
( 17 i − 3 j − 10 k ) → →
\ RQ || PQ
→ → →
14. The position vectors a , b , c of three points 16. Find the value or values of m for which
→ → → → ∧ ∧ ∧

oo
satisfy the relation 2 a − 7 b + 5 c = 0 . Are these m ( i + j + k ) is a unit vector.
points collinear? → ∧ ∧ ∧
Solution : Let the position vector of three points be Solution : Let a = m( i + j + k )
→ → → →
a, b, c. 2 2 2
| a | = m 1 +1 +1 = m 3
ab
The given relation is → →

→ → → →
To make a as a unit vector, | a | = +1
2 a – 7 b + 5 c = 0 1
→ → → \ m 3 = ± 1 ⇒ m = ± .
⇒ 2a +5c =7b 3
ur

2 → 5 → → 17. Show that the points A (1, 1, 1), B(1, 2, 3) and

⇒ a + c = b
7 7 C(2, –1, 1) are vertices of an isosceles triangle.
→ → → Solution : Let the position vector of the points A, B, C
sa +tc = b
.s

⇒
→ → → be
Thus, b is a linear combination of a and c . → ∧ ∧ ∧
OA = i + j + k
\ The given points are collinear.
w

→ ∧ ∧ ∧
15. The position vectors of the points P, Q, R, S are OB = i + 2 j + 3 k
∧ ∧ ∧ ∧ ∧ ∧ ∧
i + j+ k , 2 i + 5 j , 3 i + 2 j − 3 k , and i − 6 j − k
∧ ∧ ∧ ∧
→ ∧ ∧ ∧
w

OC = 2 i − j + k
respectively. Prove that the line PQ and RS are
parallel. → → →
AB = OB – OA
→ ∧ ∧ ∧
w

Solution : Given OP = i + j+ k ∧ ∧ ∧ ∧ ∧ ∧
→ ∧ ∧ = ( i + 2 j + 3 k ) – ( i + j + k )
OQ = 2i+5 j ∧ ∧
→ ∧ ∧ ∧ = j + 2 k
OR = 3 i + 2 j− 3k
→
→ ∧ ∧ ∧ | AB | = 12 + 22 = 5
and OS = i − 6 j− k
 → → → Solution :
→ ∧ ∧ ∧
BC = OC – OB (i) Given a = 2 i + λ j + k
→ ∧ ∧ ∧
∧ ∧ ∧ ∧ ∧ ∧
= ( 2 i − j + k ) – ( i + 2 j + 3 k ) and b = i − 2 j + 3 k
→ →
∧ ∧ ∧ Since the vectors are perpendicular, a . b = 0
= i − 3 j − 2 k ∧ ∧ ∧ ∧ ∧ ∧

m
→ ⇒ ( 2 i + λ j + k ).( i − 2 j + 3 k ) = 0
12 + ( −3) + ( −2)
2 2
\ | BC | =
⇒ 2(1) + l(–2) + 1(3) = 0
= 1 + 9 + 4 = 14
→ → → ⇒ 2 – 2l + 3 = 0

co
CA = OA – OC
⇒ 5 – 2l = 0 ⇒ 2l = 5
∧ ∧ ∧ ∧ ∧ ∧
= ( i + j + k ) – ( 2 i − j + k ) 5
⇒ l =
∧ ∧ 2
= − i + 2 j → ∧ ∧ ∧ → ∧ ∧ ∧

.
→ (ii)Given a = 2 i + 4 j − k and b = 3 i − 2 j + λ k
→ →
( −1)2 + 22

ks
\ | CA | = = 5 Since the vectors are perpendicular, a . b = 0
→ → ⇒
∧ ∧ ∧ ∧
( 2 i + 4 j− k ) . ( 3 i − 2 j+ λ k ) = 0
∧ ∧

Since | AB | = | CA |, the given points form an isosceles

triangle. ⇒ 2(3) + 4(–2) – 1(l) = 0
⇒ 6 – 8 – l =0

1.
→
Find a . b when
→
EXERCISE 8.3
oo ⇒
⇒

3. If
→
a and
→
b
–2 – l = 0

are two vectors such that

l =– 2
ab
→ ∧ ∧ ∧ → ∧ ∧ ∧
→ → → →
(i) a = i − 2 j + k and b = 3 i − 4 j − 2 k a = 10, b = 15 and a . b = 75 2 , find the angle
→ ∧ ∧ ∧ → ∧ ∧ ∧ → →
(ii) a = 2 i + 2 j − k and b = 6 i − 3 j + 2 k between a and b .
Solution : → → → →
ur

→ ∧ ∧ ∧ Solution : Given | a | = 10, | b | = 15 and a . b = 75 2

(i) Given a = i − 2 j + k →
→ ∧ ∧ ∧ Let q be the angle between the vectors a and
b = 3 i − 4 j− 2k →
b.
.s

→ → ∧ ∧ ∧ ∧ ∧ ∧
a . b = ( i − 2 j + k ). ( 3 i − 4 j − 2 k ) → → 5

= 1(3) – 2(–4) + 1(–2) = 3 + 8 – 2 = 9

a⋅ b 2 1 75 2
\ cos q = → → = 10 15 = =
→ → ( ) 2 2
\ a.b = 9 a b
w

2 1
→ ∧ ∧ ∧
(ii) Given a = 2 i + 2 j − k π
→ ∧ ∧ ∧
= cos .
4
w

b = 6 i − 3 j+ 2 k π
→ → ∧ ∧ ∧ ∧ ∧ ∧
q = .
a . b = ( 2 i + 2 j − k ).( 6 i − 3 j + 2 k ) 4
4. Find the angle between the vectors
= 12 – 6 – 2 = 12 – 8 = 4
w

∧ ∧ ∧ ∧ ∧ ∧
→ →
(i) 2 i + 3 j − 6 k and 6 i − 3 j + 2 k .
2. Find the value l for which the vectors a and b
∧ ∧ ∧ ∧
are perpendicular, where (ii) i − j and j − k .
→ →∧ ∧∧ ∧∧ →∧ ∧
→ ∧∧ ∧∧ ∧
Solution :
(i) a = a2 i=+2λi j++λkj, and, bi −=2i j−+2 3j k+ 3 k
+bk= → ∧ ∧ ∧ → ∧ ∧ ∧
→ ∧ ∧ ∧ → ∧ ∧ ∧ (i) Let a = 2 i + 3 j − 6 k and b = 6 i − 3 j + 2 k
(ii) a = 2 i + 4 j − k and b = 3 i − 2 j + λ k .
Let q be the angle between the given vectors.
 → → →
→ → ∧ ∧ ∧ ∧ ∧ ∧ ⇒ a + 2b = – c
a . b = ( 2 i + 3 j − 6 k ).( 6 i − 3 j + 2 k ) → → →
⇒ | a + 2 b |2 = |– c |2
= 12 – 9 – 12 = – 9 → → → → →
⇒ | a |2 + 4| b |2 + 4( a . b ) = | c |2
→
22 + 32 + ( −6)
2 → →
|a| = ⇒ 9 + 4(16) + 4( a . b ) = 49

m
= 4 + 9 + 36 = 49 = 7 → →
→ 9 + 64 + 4( a . b ) = 49
6 + ( −3) + 2
2 2 2
and | b | = → →
⇒ 73 + 4( a . b ) = 49

co
= 36 + 9 + 4 = 49 = 7
→ →
→ →
a⋅ b −9 −9 ⇒ 4( a . b ) = 49 – 73
\ cos q = = =
→ →
7 (7 ) 49 → →
| a |⋅| b | ⇒ 4 | a | | b | cos q = – 24

.
 −9 
⇒ q = cos–1   ⇒ 4(3)(4) cos q = – 24
 49 

ks
2
→ ∧ ∧ → ∧ ∧ 6
(ii) Let a = i − j and b = j − k −24
⇒ cos q =
→ → ∧ ∧ ∧ ∧ 4 (3) ( 4)
a . b = ( i − j ). ( j − k ) 2

oo
−1 π
= 1(0) – 1(1) + 0(–1) = – 1 ⇒ cos q = = – cos
→
2 3
| a | = 12 + ( −1) = 1 + 1 =
2
2  π 2π
⇒ cos q = cos  π −  = cos
→  3 3
| b | = 1 + ( −1) = 1 + 1 =
2 2
2 2π
ab
⇒ q =
→ → 3
Let q be the angle between the vectors a and b → ∧ ∧ ∧ → ∧ ∧
→ →
6. Show that the vectors a = 2 i + 3 j + 6 k , b = 6 i + 2
→ ∧ ∧ ∧ → ∧ ∧ ∧ → ∧ ∧ ∧
a⋅ b a = 2 i + 3 j + 6 k , b = 6 i + 2 j − 3 k , and c = 3 i − 6 j + 2 k , are
\ cos q = → →
ur

| a |⋅| b | mutually orthogonal.

→ ∧ ∧ ∧
−1 1  π Solution : Given a = 2 i + 3 j + 6 k ,
⇒ cos q = = – ⇒ cos q = – cos   → ∧ ∧ ∧
2 2 2  3 b = 6 i + 2 j− 3k ,
.s

→ ∧ ∧ ∧
 π  2π  c = 3 i − 6 j+ 2 k
⇒ cos q = cos  π −  = cos  
 3  3 → → ∧ ∧ ∧ ∧ ∧ ∧
2π a . b = ( 2 i + 3 j + 6 k ).( 6 i + 2 j − 3 k )
w

⇒ q = = 2(6) + 3(2) + 6(–3)

3
→ → → = 12 + 6 – 18 = 0
5. If a , b, c are three vectors such that → → ∧ ∧ ∧ ∧ ∧ ∧
w

→ → →
→ → → → b . c = ( 6 i + 2 j − 3 k ).( 3 i − 6 j + 2 k )
a + 2 b + c = 0 and a = 3 , b = 4 , c = 7 , find
→ → = 6(3) + 2(–6) – 3(2)
the angle between a and b . [March - 2019]
→ → → →
= 18 – 12 – 6 = 0
w

Solution : Given a + 2 b + c = 0 → → ∧ ∧ ∧ ∧ ∧ ∧
→ → c . a = ( 3 i − 6 j + 2 k ).( 2 i + 3 j + 6 k )
and | a | = 3, | b | = 4 = 3(2) – 6(3) + 2(6)
→
and | c | = 7 = 6 – 18 + 12 = 0
→ → → → → →
→ →
Let q be the angle between a and b . Since a . b = b . c = c . a = 0 the given
vectors are mutually orthogonal.
 ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ → ∧ ∧ ∧
7. Show that the vectors − i − 2 j − 6 k , 2 i − j + k , and
− i + 3 j + 5 k , OC = 3 i + j + 2 k
∧ ∧ ∧ ∧ ∧
j + k , and − i + 3 j + 5 k ,form a right angled triangle. → → →
Now, AB = OB – OA
Solution : Let the given vectors are ∧ ∧ ∧ ∧ ∧ ∧
→ ∧ ∧ ∧ → ∧ ∧ ∧ = ( 4 i + 3 j + k ) – ( 2 i − j + 3 k )
Given a = − i − 2 j − 6 k , b = 2 i − j + k

m
∧ ∧ ∧
→ ∧ ∧ ∧
= 2 i + 4 j − 2 k
and c = − i + 3 j + 5 k
→ → → →
| a | = ( −1)2 + ( −2)2 + ( −6)2 BC = OC – OB

co
∧ ∧ ∧ ∧ ∧ ∧
= 1 + 4 + 36 = 41
→
= ( 3 i + j + 2 k ) – ( 4 i + 3 j + k )
22 + ( −1) + 12
2 ∧ ∧ ∧
|b| =
= − i − 2 j + k
= 4 +1+1 = 6 → → →

.
→ CA = OA – OC
and | c | = ( −1) 2 2 2
+ 3 + 5 = 1 + 9 + 25

ks
∧ ∧ ∧ ∧ ∧ ∧
= ( 2 i − j + 3 k ) – ( 3 i + j + 2 k )
= 35 ∧ ∧ ∧
= − i − 2 j + k
( 41) = 41 = 35 + 6
→ 2
| a |2 = → ∧ ∧ ∧
Now, AB = 2 i + 4 j − 2 k
= ( 35 ) + ( 6 ) = | b | + | c |
→ →
→

oo
2 2 ∧ ∧ ∧
2 2
= –2( − i − 2 j + k ) = –2 BC
Hence, by Pythagoras theorem, the given
→ →
Thus AB || BC and B is a common points.
vectors form a right angled triangle.
Hence, the given points are collinear.
→ → → → → → → → → → → → →
ab
8. If a = 5, b = 6, c = 7 and a + b + c = 0 , find a ⋅ b + b ⋅ c + c ⋅ a→ →
→ → → → → → → 10. If a , b are unit vectors and q is the angle between
0 , find a ⋅ b + b ⋅ c + c ⋅ a . them, show that
→ →
Solution : Given | a | = 5, | b | = 6, θ 1 θ 1→ →
(i) sin = → a −
→
b (ii) cos = a + b
→ → → → → 2 2 2 2
ur

| c | = 7 and a + b + c = 0 → →
→ → → → → → θ a− b
| a + b + c a 2 b 2 c
| = | | +| | +| | +2 2
(iii) tan =
→ → → → → → 2 → →
( a . b ) + 2( b . c ) + 2( c . a ) a+ b
.s

→ →
= 25 + 36 + 49 + 2 Solution : Let a and b be the unit vectors and q is the
→ → → → → → → → →
(a.b + b.c + c.a) = 0 angle between a and b .
→ → → → → → → → → → → →
w

⇒ –110 = 2( a . b + b . c + c . a ) (i) Consider | a – b |2 = | a |2 +| b |2 – 2( a . b )

→ →
−110 → → → → → → [ | a | = 1; | b | = 1]
⇒ = a.b + b.c + c.a → →
2 = 1 + 1 – 2| a || b |
w

→ → → → → →
⇒ a . b + b . c + c . a = – 55 cos q = 2 – 2 cos q

9. Show that the points (2, – 1, 3), (4, 3, 1) and = 2(1 – cos q)
w

(3, 1, 2) are collinear. θ θ

= 2.2 sin2 = 4 sin2
Solution : Let the given points be A(2, –1, 3), B(4, 3, 1) 2 2
and C(3, 1, 2).
→ → θ
| a – b | = 2 sin
→ ∧ ∧ ∧ 2
Then OA = 2 i − j + 3 k , θ 1 → →
sin = |a–b|
→ ∧ ∧ ∧ 2 2
OB = 4 i + 3 j + k and

 → → → → → → ∧ ∧ ∧
(ii) Consider | a + b |2 = | a |2 +| b |2 + 2( a . b ) 12. Find the projection of the vector i + 3 j + 7 k on
→ → ∧ ∧ ∧
= 1 + 1 + 2| a || b | the vector 2 i + 6 j + 3 k . [Hy - 2018]
→ ∧ ∧ ∧ → ∧ ∧ ∧
cos q = 2 + 2 cos q Solution : Let a = i + 3 j + 7 k and b = 2 i + 6 j + 3 k
→ → → → ∧ ∧ ∧ ∧ ∧ ∧

m
[ | a | = 1; | b | = 1] a . b = ( i + 3 j+ 7 k ) . ( 2 i + 6 j+ 3k )
θ
= 2(1 + cos q) = 2.2 cos2 = 1(2) + 3(6) + 7(3)
2
θ = 2 + 18 + 21 = 41
= 4 cos2

co
→
2 2 2 2
| b | = 2 + 6 + 3 = 4 + 36 + 9
→ → θ
\ | a + b | = 2 cos = 49 = 7
2
→ →
θ 1 → → → → a⋅ b 41
⇒ cos = |a+b|

.
2 2 Now, projection of a on b = → =
|b|
7

ks
θ 1 → →
θ sin | a − b | → ∧ ∧ ∧ →
(iii) tan = 2 = 2 13. Find l, when the projection of a = λ i + j + 4 k on b
2 θ 1 → →→ ∧ ∧ ∧ → ∧ ∧ ∧
cos | a +a b=| λ i + j + 4 k on b = 2 i + 6 j + 3 k is 4 units.
2 2
→ ∧ ∧ ∧
 [From (i) and (ii)] Solution :

oo
→ →
Given a = λ i + j + 4 k and
→ ∧ ∧ ∧
θ | a− b |
tan = → → Hence, proved. b = 2 i + 6 j+ 3k
2 →
| a+ b | 2 2 2
→ → → → | b | = 2 +6 +3
→ →
11. Let a , b , c be three vectors such that a = 3, b = 4, c = 5
ab
→ → → = 4 + 36 + 9 = 49 = 7
a = 3, b = 4, c = 5 and each one of them being → → ∧ ∧ ∧ ∧ ∧ ∧
a . b = ( λ i + j+ 4 k ) . ( 2 i + 6 j+ 3k )
perpendicular to the sum of the other two, find
→ → →
a+ b+ c . = 2l + 6 + 12 = 2l + 18
→ →
ur

→ → →
Solution : Given | a | = 3, | b | = 4, | c | = 5.
Also projection of a on b = 4 units
→ →
→ → → → → →
a⋅b
Also a . ( b + c ) = 0 We know that, projection of a on b is →
→ → → →
b.(c+a) = 0 2 λ + 18 |b|
.s

→ → → → 4 =
7
and c . ( a + b ) = 0
28 = 2l + 18
Since they are perpendicular
28 – 18 = 2l
w

→ → → → → → → → → → → →
⇒ a . b + a . c = 0, b . c + b . a = 0 and c . a + c . b = 0 10 = 2l
Adding all the above we get, 10
l = =5
→ → → → → → 2
w

2( a . b + b . c + c . a ) = 0
→ → → → → → \l = 5
⇒ a . b + b . c + c . a = 0 ...(1)
→ → →
→ → → → → →
14. Three vectors a , b and c are such that
w

Consider| a + b + c |2 =|a |2 + |b |2 + |c |2 +
→ → → → → → →
→ → → → → →
2( a . b + b . c + c . a ) a = 2, b = 3, c = 4 , and a + b + c = 0 .
→ → → → → →
=9 + 16 + 25 + 2 (0) = 50 [From (1)] Find 4 a ⋅ b + 3 b ⋅ c + 3 c ⋅ a .
→ → → → → →
\ | a + b + c | = 50 = 25 × 2 = 5 2 Solution : Given | a | = 2, | b | = 3, | c | = 4 and
→ → → → → → →
a+b+c = 0 ⇒ a+b =–c
 → → → 42
\ | a + b |2 = |– c |2 12 − 63 − 33 12 − 96 − 84
= = = = – 42
→ → → → → 2 2 2
⇒ | a |2 + | b |2 + 2( a . b ) = | c |2 → → → → → →
→ → \ 4 a . b + 3 b . c + 3 c . a = – 42
⇒ 4 + 9 + 2( a . b ) = 16
EXERCISE 8.4

m
→ →
⇒ 13 + 2( a . b ) = 16 → → → ∧ ∧ ∧
→ →
1. Find the magnitude of a × b if a = 2 i + j + 3 k
⇒ 2( a . b ) = 16 – 13 = 3 → ∧ ∧ ∧

co
and b = 3 i + 5 j − 2 k .
→ → 3 → ∧ ∧ ∧
⇒ a.b = Solution : Given a = 2 i + j + 3 k ,
2
→ ∧ ∧ ∧
→ → 2 3 b = 3 i + 5 j− 2 k
⇒ 4( a . b ) = 4 × = 6 ... (1)

.
2 i j k
→ → → → →

ks
Also, b + c = – a a×b = 2 1 3
→ → → 3 5 −2
| b + c |2 = |– a |2 Expanding along R1 we get,
→ → → → → ∧ ∧ ∧
| b |2 + | c |2 + 2( b . c ) = | a |2 = i (–2 – 15) – j (–4 – 9) + k (10 – 3)

oo
→ → ∧ ∧ ∧
→ → a × b = –17 i + 13 j + 7 k
9 + 16 + 2( b . c ) = 4 → →
→ → |a×b| = ( −17)2 + 132 + 72
25 + 2( b . c ) = 4
= 289 + 169 + 49 = 507
ab
→ →
2( b . c ) = 4 – 25 = – 21 → → → → → → → → → →
2. Show that a × ( b + c ) + b × ( c + a ) + c × ( a + b ) = 0 .
→ → −21
(b.c) = → → → → → → → → →
2 Solution : LHS = a × ( b + c ) + b × ( c + a ) + c × ( a + b )
→ →
 −21 −63  (By associative property)
ur

3( b . c ) = 3  =  ... (2)
 2  2
→ → → → → → → → → → → →

→ → →
=a×b+a×c+b×c+b×a+c×a+c×b
Also, c + a = – b → → → →
→ → → [ b × a = – a × b
| c + a | = |– b |
.s

→ → → →
c×a =–a×c
→ → → → → → →
| c + a |2 = |– b |2 c×b=–b×c]
→ → → → →
w

| c |2 + | a |2 + 2( c . a ) = | b |2 → → → → → → → → → → → → →
→ →
16 + 4 + 2( c . a ) = 9 = a × b + a × c + b × c – a × b – a × c – b × c = 0 = RHS
→ →
Hence proved.
20 + 2( c . a ) = 9
w

→ → 3. Find the vectors of magnitude 10 3 that are

⇒ 2( c . a ) = 9 – 20 = – 11 perpendicular to the plane which contains
→ → −11 ∧ ∧ ∧ ∧ ∧ ∧
(c.a) = i + 2 j + k and i + 3 j + 4 k ,
w

2 → ∧ ∧ ∧ → ∧ ∧ ∧
→ →
 −11 −33 Solution : Let a = i + 2 j + k and b = i + 3 j + 4 k
\ 3( c . a ) = 3   =  ... (3)
 2  2 A unit vector which is perpendicular to the
Adding (1), (2) and (3) we get, → →
→ → a× b
→ → → → → → 63 33 vector a and b is → →
4a.b +3b.c +3c.a = 6– − | a× b |
2 2
 ∧ ∧ ∧ ∧ ∧ ∧
i j k i j k
→ → ∧ ∧ ∧
3 = i ( 2 + 6) − j (1 − 9) + k ( −2 − 6)
→ → ∧ ∧ ∧
a × b = 1 2 1 = i (8 – 3) – j (4 – 1) + k (3 – 2) a×b=1 2
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
1 3 4 3 −2 1 = 8 i + 8 j − 8 k = 8( i + j − k )

=5i –3 j + k

m
→ →
52 + ( −3) + 12 =
2 → →
|a×b|= 25 + 9 + 1 = 35
|a×b|= 8 1 + 1 + ( −1) = 8 3
2 2 2

→
\ A unit vector which is perpendicular to the vector a \ Area of the parallelogram = 8 3 sq. units.

co
∧ ∧ ∧
→ 5 i − 3 j+ k 6. Find the area of the triangle whose vertices are
and b is A(3, – 1, 2), B(1, – 1, – 3) and C(4, – 3, 1).
35
Solution : Given that the vertices of the DABC as
Hence, a vector of magnitude 10 3 , which is perpendicular
A(3, – 1, 2), B(1, –1, –3) and C(4, –3, 1)

.
→ 10 3  ∧ ∧ ∧ 
→
→ → →

ks
∧ ∧ ∧ ∧ ∧ ∧
to the vectors a and b is ±  5 i − 3 j + k  AB = OB− OA = ( i − j − 3 k ) − (3 i − j + 2 k )
35 
∧ ∧
4. Find the unit vectors perpendicular to each of the = −2 i − 5 k
vectors a + b and a − b , where a = i + j + k and b = i + 2 j +→ → →
→ → → → → ∧ ∧ ∧ → ∧ ∧ ∧
3k ∧ ∧ ∧ ∧ ∧ ∧
AC = OC − OA = ( 4 i − 3 j + k ) − (3 i − j + 2 k )

oo
∧ → ∧ ∧ ∧
+ k and b = i + 2 j + 3 k . [March - 2019] ∧ ∧ ∧
→ ∧ ∧ ∧ → ∧ ∧ ∧ = i − 2 j − k
Solution : Given a = i + j + k and b = i + 2 j + 3 k ∧ ∧ ∧
→ → ∧ ∧ ∧ i j k
\ a + b = 2 i + 3 j+ 4 k → →
ab
→ → ∧ ∧
AB × AC = −2 0 −5
a – b = − j− 2k
→ → 1 −2 −1
A unit vector which is perpendicular to ( + ) and a b ∧ ∧ ∧
→ → = i (0 − 10) − j ( 2 + 5) + k ( 4)
ur

( a – b ) is ∧ ∧ ∧

∧ ∧ ∧
= − 10 i − 7 j + 4 k
i j k → →
∧ ∧ ∧ | AB × AC | = ( −10)2 + ( −7)2 + 42
.s

2 3 4 = i ( −6 + 4) − j ( −4 + 0) + k ( −2 + 0)
= 100 + 49 + 16 = 165
∧ ∧ ∧
0 −1 −2 = −2 i + 4 j − 2 k 1
Hence the required area of DABC = 165 sq. units.
w

2
Its magnitude is ( −2) + 4 + ( −2) = 4 + 16 + 4 = 24
2 2 2 → → →
7. If a , b , c are position vectors of the verticesA, B, C
= 4 × 6 =2 6 of a triangle ABC, show that the area of the triangle
w

→ → 1→ → → → → →
\ The unit vector which is perpendicular to ( a + b ) and ABC is a × b + b × c + c × a . Also deduce the
∧ ∧ ∧
2
∧ ∧ ∧
→ → ( −2 i + 4 j − 2 k ) (− i + 2 j − k ) condition for collinearity of the points A, B, and C.
( a – b ) is ± = ±
w

2 6 6 Solution : Given that the position vector of the vertices

→ → →
5. Find the area of the parallelogram whose two
of the DABC is a , b and c .
adjacent sides are determined by the vectors
∧ ∧ ∧ ∧ ∧ ∧ → → → → → →
i + 2 j + 3 k and 3 i − 2 j + k . \ OA = a , OB = b and OC = c
Solution : Let the adjacent sides of the parallelogram → → → → →
→ ∧ ∧ ∧ → ∧ ∧ ∧ AB = OB− OA = b – a
are a = i + 2 j + 3 k and b = 3 i − 2 j + k
 → → → → → → ∧
AC = OC – OA = c – a | a ×k | = ( − a1 )2 + a22 = a12 + a22
→ → → → → → → ∧
2 2
\ | a × k |2 = a1 + a2  ... (3)
\ AB × AC = ( b – a ) × ( c – a )
→ ∧ → ∧ → ∧
→ → → → → → → → Adding (1), (2) and (3) we get, | a × i |2 + | a × j |2 + | a × k |2
= b×c–b×a–a×c+a×a

m
→ → → → → → → 2 2 2 2 2 2
b×c+a×b+c×a+ 0 = a3 + a2 + a1 + a3
+ a1 + a2
=
→ → → → → → →
[ a × a = 0 , b × a = – ( a × b ), a × c = –( c × a ) ]
→ → → → 2 2 2
= 2 a1 + a2 + a3 ( )
( ) = 2| a |

co
→ → → → → → 2 →
= a×b+b×c+c×a = 2 a12 + a22 + a32 2
→ → → → → → → →
| AB × AC | = | a × b + b × c + c × a | Hence proved.
→ → → → → → →
1 → → → → → → 9. Let a , b , c be unit vectors such that a . b = a . c = 0
\ Area of DABC = |a×b+b×c+c×a|

.
2 → → π
Condition for the points A, B, C to be collinear is area of and the angle between b and c is = . Prove that

ks
3
DABC = 0
→ 2 → →
a =± (b× c ) .
1 → → → → → → 3 → → →
⇒ |a×b+b×c+c×a|=0 Solution : Given a , b and c are unit vectors.
2 → → →
→ → → → → →
⇒ |a| = |b| =|c|=1

oo
⇒| a × b + b × c + c × a | = 0 which is the required condition. → → → → →
a .2b = a . c = 0, and angle between b and
→ → ∧ 2 → ∧2 → ∧ 2 →
+a × j +a ×k =2 a → π
8. For any vector a prove that a × i  c is
∧2 → ∧ 2 → 2
3
=2 a . → → → → → →
× j +a ×k
→
a . b = a . c = 0 ⇒ a is ⊥ r to both b and c .
ab
→ ∧ ∧ ∧ → → → → → →
Solution : Let the components of a = a1 i + a2 j + a3 k a is ⊥ r to b × c ⇒ a = l ( b × c ) for some scalar l.
→ ∧ ∧ ∧ ∧ ∧ → → →
a×i = ( a1 i + a2 j + a3 k )× i \ | a |2 = l2 | b × c |2
∧ ∧ ∧ ∧ → → → →
= a2 ( j × i ) + a3 (k × i ) ⇒ 1 = l2 [| b |2 | c |2 – ( b . c )2]
ur

∧ ∧ ∧ ∧ → → → → → → →
= a2( − k ) + a3( j ) = a3 j – a2 k [ | a | = 1 and | a × b |2 = | a |2 | b |2 – ( a . b )2]
→ ∧ → → π
a32 + ( − a2 ) =
2
|a×i | = a32 + a22 ⇒ 1 = l2 [(1) (1) – | b |2| c |2 cos2 ]
.s

3
∧
→
2 2
→ → π
\ | a × i |2 = a3 + a2  ... (1) [ angle between b and c is ]
→ ∧ ∧ ∧ ∧ ∧ π → → 3
w

a× j = ( a1 i + a2 j + a3 k )× j ⇒ 1 = l2 [1 – cos2 ] [ | b | = | c | = 1]
∧ ∧ ∧ ∧
3
= a1 ( i × j ) + a3( k × j ) 1  3
⇒ 1 = l2 [1 – ] ⇒ 1 = l2  
∧ ∧ 4  4
w

= a1 k – a3 i 4 2
⇒ 2
l = ⇒l= ±
→ ∧ 3 3
a12 + ( − a3 ) =
2
| a × j| = a12 + a32 2
Substituting l = ± in (1) we get,
w

→ ∧
2 2
3
\ | a × j |2 = a1 + a3  ... (2) → 2 → →
→ ∧ ∧ ∧ ∧ ∧ a = ± (b×c)
a ×k = ( a1 i + a2 j + a3 k ) × k 3
∧ ∧ ∧ ∧
= a1 ( i × k ) + a2( j × k )
∧ ∧
= – a1 j + a2 i
 ∧ ∧ ∧ 3. The unit vector parallel to the resultant of the
10. Find the angle between the vectors 2 i + j − k and ∧ ∧ ∧ → → →
∧ ∧ ∧ vectors i + j − k and i − 2 j + k is [March - 2019]
i + 2 j + k using vector product. ∧ ∧ ∧ ∧ ∧
→ ∧ ∧ ∧ → ∧ ∧ ∧
i − j+ k 2i+ j
(1) (2)
Solution : Let a = 2 i + j − k and b = i + 2 j + k 5 5
→ →

m
Let q be the angle between the vectors a and b ∧ ∧ ∧ ∧ ∧
2 i − j+ k 2i− j
∧ ∧ ∧ (3) (4)
i j k 5 5
→ → ∧ ∧ ∧ ∧ ∧ ∧
a × b = 2 1 −1 Hint : Resultant vector of i + j − k and i − 2 j + k is

co
∧ ∧
1 2 1 2i− j
∧ ∧ ∧
22 + ( −1) =
2
= i (1 + 2) − j ( 2 + 1) + k ( 4 − 1) Its magnitude is 4 +1 = 5
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
2i− j

.
= 3 i − 3 j + 3 k = 3( i − j + k ) 2i− j
\ Required unit vector = [Ans: (4) ]

ks
→ → 5 5
| a × b | = 3 1 + 1 + ( −1) = 3 3
2
2 2
→
→ 4. A vector OP makes 60° and 45° with the positive
22 + 12 + ( −1) =
2
|a| = 6 direction of the x and y axes respectively. Then the
→
|b| =
2
1 + 2 +1 = 2 2
6 →
angle between OP and the z-axis is

oo
→ →
| a× b | 3 3 3 3 3 π (1) 45° (2) 60° (3) 90° (4) 30°
= → → = = 6 = = sin
6 6 2 3
| a || b | 2 Hint : Given a = 60o, b = 45o
π cos2 a + cos2 b + cos2 g = 1
q= ⇒ cos2 60 + cos2 45 + cos2 g = 1
ab
3 2 2
 1  1 
EXERCISE 8.5 ⇒ 2
  +   + cos g = 1
2 2
CHOOSE THE CORRECT OR THE ⇒
1 1 3
+ + cos2 g = 1 ⇒ + cos2 g = 1
ur

MOST SUITABLE ANSWER FROM 4 2 4

3
THE GIVEN FOUR ALTERNATIVES. ⇒ cos2 g = 1 –
4
→ → → → 1  1
2
.s

1. The value of AB + BC + DA + CD is = (
=   == (cos 60)2
cos 60)
2

→ → → → 4  2
(1) AD (2) CA (3) 0 (4) − AD ⇒ cos g = cos 60
→ → → → → → \ g = 60o [Ans: (2) 60°]
w

Hint : AB+ BC+ DA + CD = AB + BC + → ∧ ∧ ∧

→ → → → → 5. If BA = 3 i + 2 j + k and the position vector of B is
CD + DA = AA = 0 [Ans : (3) 0 ] ∧ ∧ ∧
i + 3 j − k then the position vector A is
w

→ → → → ∧ ∧ ∧ ∧ ∧
2. If a + 2 b and 3 a + m b are parallel, then the (1) 4 i + 2 j + k (2) 4i+ 5 j
value of m is ∧ ∧
1 1 (3) 4 i (4) −4 i
w

(1) 3 (2) (3) 6 (4)

→ →
3 → →
6 → ∧ ∧ ∧
Hint : BA = 3 i + 2 j + k
Hint : a + 2 b = 3( a + 2 b ) → →
→ → → → ∧ ∧ ∧
→ → → →
OA − OB = 3 i + 2 j + k
= 33aa++66bb ==33aa++m
mbb
∧ ∧ ∧ → ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
m = 6 [Ans: (3) 6] 3 i + 2 j + k + OA = 3 i + 2 j + k + i + 3 j − k = 4 i + 5 j

∧ ∧
 [Ans: (2) 4 i + 5 j ]
 6. A vector makes equal angle with the positive D C
direction of the coordinate axes. Then each angle
is equal to
 1  2 b
(1) cos–1   (2) cos–1   a+
 3  3 b
 1  2 

m
(3) cos–1   (4) cos–1  
 3  3
A a B
Hint : Given a = b = g
→ → → → →

co
cos2 a + cos2 a+ cos2 a = 1
1 Hint : In D BCD, BD = BC + CD = b − a → →
⇒ 3cos2 a = 1 ⇒ cos2 a =
3  [Ans: (2) b − a ]
1
⇒ cos a = → →
3 10. If a , b are the position vectors A and B, then
 1 
⇒ a = cos–1   which one of the following points whose position

.
 3
 1  vector lies on AB, is → →[March - 2019]

ks
[Ans: (3) cos–1  ] → → 2a − b
 3  (1) a + b (2)
→ → → → → → 2
→ → → →
7. The vectors a − b , b − c , c − a are
2a + b a −b
(1) parallel to each other (3) (4)
3 3

oo
(2) unit vectors a P b
(3) mutually perpendicular vectors Hint : A 1
B
2
(4) coplanar vectors. [Ans: (4) coplanar vectors] →
→ 2a+b
→ → → →
→ → → → 1( b ) + 2( a )
OP = ⇒ OP =
8. If ABCD is a parallelogram, then AB + AD + CB + CD 1+ 2 3 → →
ab
→ → → 2a + b
+ AD + CB + CD is equal to [Ans: (3) ]
3
→ → → → → →
(1) 2(AB + AD) (2) 4 AC 11. If a , b , c are the position vectors of three
→ → collinear points, then which of the following is
(3) 4 BD (4) 0
ur

true?
D C
→ → → → → →
(1) a = b + c (2) 2a = b + c
→ → → → → → →
(3) b = c + a (4) 4 a + b + c = 0
.s

A B Hint : Since the points are collinear.

→ → → → → → → → → a b c
Hint : AB + AD + CB + CD = AB + AD − AD − AB = 0
w

A 1 2 C
→ B
 [Ans: (4) 0 ] → → → → → →
AB = CA ⇒ OB− OA = OA − OC
9. One of the diagonals of parallelogram ABCD with
w

→ → → → → → →
→ → → →
⇒ b −a = a − c ⇒ b+ c = 2 a
a and b as adjacent sides is a + b . The other
→ → →
→  [Ans: (2) 2 a = b + c ]
diagonal BD is
w

→ → → →
(1) a − b (2) b −a
→ →
→ → a +b
(3) a + b (4)
2
 → → ∧ ∧ ∧ ∧ ∧ → ∧ ∧ ∧ →
9a+ 7 b
→
( 3 i + 6 j + 9 k ) – ( 5 i + 7 j ) = OC ; −2 i − j + 9 k = OC
12. If r = then the point P whose position ∧ ∧ ∧
16 [Ans: (1) −2 i − j + 9 k ]
→
→ → → → → →
vector r divides the line joining the points with 15. If a + b = 60, a − b = 40 and b = 46 , then a is
→ →
position vectors a and b in the ratio. (1) 42 (2) 12 (3) 22 (4) 32

m
→ → → → → →
(1) 7 : 9 internally (2) 9 : 7 internally Hint : We know | a + b |2 + | a − b |2 = 2[| a |2 + | b |2 ]
(3) 9 : 7 externally (4) 7 : 9 externally →
602 + 402 = 2(| a |2+ 462)

co
a r b →
Hint : 3600 + 1600 = 2(| a |2 + 2116)
A 7 9 C
→ →
5200 →
9 a+ 7 b
→
= | a |2 + 2116
Given r =  [Ans: (1) 7 : 9 internally] 2
9+7 →
2600 – 2116 = | a |2

.
∧ ∧ ∧
13. If λ i + 2λ j + 2λ k is a unit vector, then the value →

ks
of l is 484 = | a |2
→
1 1 1 1
(1) (2) (3) (4) |a| = 484 = 22 [Ans: (3) 22]
3 4 9 2 → →
∧ ∧ ∧
16. If a and b having same magnitude and angle
Hint : | λ i + 2λ j + 2λ k | = 1

oo
between them is 60° and their scalar product is
λ 2 + ( 2λ ) + ( 2λ )
2 2 1 →
= 1 then a is
2
⇒ λ 2 + 4 λ 2 + 4 λ 2 = 1⇒ 9 λ 2 = 1 (1) 2 (2) 3 (3) 7 (4) 1
1 → → → → 1
ab
⇒ 3l = 1 ⇒ l = Hint : | a | = | b |, q = 60o, a . b =
3 2
1 → → → →
[Ans: (1) ] a . b = | a || b | cos q
3
14. Two vertices of a triangle have position vectors 1 → → 1 → 1
= | a || a | cos 60 ⇒ = | a |2 .
∧ ∧ ∧ ∧ ∧ ∧ 2 2 2
ur

3 i + 4 j − 4 k and 2 i + 3 j + 4 k . If the position → →

∧ ∧ ∧ ⇒ | a | = 1 ⇒ | a | = 1
2
[Ans: (4) 1]
vector of the centroid is i + 2 j + 3 k , then the
 π
position vector of the third vertex is 17. The value of θ ∈ 0,  for which the vectors
 2
.s

∧ ∧ ∧ ∧ ∧ ∧ → ∧ ∧ → ∧ ∧ ∧
(1) −2 i − j + 9 k (2) −2 i − j − 6 k a = (sinθ) i + (cosθ) j and b = i − 3 j + 2 k are
∧ ∧ ∧ ∧ ∧ ∧ perpendicular, is equal to
(3) 2 i − j + 6 k (4) −2 i + j + 6 k
π π π π
w

→ ∧ ∧ ∧ (1)
3
(2)
6
(3)
4
(4)
2
Hint : OA = 3 i + 4 j− 4k
→ ∧ ∧ ∧
→ → → →
Hint : a ⊥ b ⇒ a . b = 0
OB = 2 i + 3 j + 4 k ∧ ∧ ∧ ∧ ∧
w

→ ∧ ∧ ∧ [sin θ i + (cos θ) j ] . [ i − 3 j + 2 k ] = 0
OG = i + 2 j+ 3k
sin q (1) − 3 cosθ + 2 (0) = 0
→ → →
→ = OA + OB+ OC ⇒ sin q = 3 cosθ
w

OG sin θ
3 ⇒ = 3 ⇒ tanq = 3
→ → → → cos θ π π
⇒ 3 OG = OA + OB + OC ⇒ q = [Ans: (1) ]
3 3
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ →
⇒3 ( i + 2 j + 3 k ) = (3 i + 4 j − 4 k ) + (2 i + 3 j + 4 k ) + OC
∧ ∧ ∧ ∧ ∧ →
3 i + 6 j + 9 k = ( 5 i + 7 j ) + OC
 → →  1 3
( )
→ → → →
18. If a = 13, b = 5 and a . b = 60 then | a × b | is 4+2 2   6
 2 4+2
 [Hy- 2018] = = = 4 3
4 3 4 3
(1) 15 (2) 35 (3) 45 (4) 25 2
3 3
→ → → → → → = = ⇒ a = 30o
Hint : | a × b |2 + ( a . b )2 = [| a |2 | b |2]  2 3 2
[Ans: (1) 30o]

m
→ →
| a × b |2 + 602 = [132 . 52] ∧ ∧ ∧
→ → 21. If the projection of 5 i − j − 3 k on the vector
| a × b |2 + 3600 = 169(25) ∧ ∧ ∧ ∧ ∧ ∧
→ →
i + 3 j + λ k is same as the projection of i + 3 j + λ k

co
| a × b |2 = 4225 – 3600 = 625
∧ ∧ ∧
→ → on 5 i − j − 3 k then l is equal to
| a × b | = 625 = 25[Ans: (4) 25]
→ →
(1) ± 4 (2) ± 3 (3) ± 5 (4) ±1
19. Vectors a and b are inclined at an angle q =120°. → ∧ ∧ ∧ → ∧ ∧ ∧ → ∧ ∧ ∧
Hint : Let a = 5 i − j − 3 k , b = i + 3 j + λ k , c = i + 3 j + λ k ,

.
→ → → → → →
2
If | a | = 1, | b |= 2 , then [( a + 3 b ) × (3 a − b )] is → ∧ ∧ ∧

ks
equal to d = 5 i − j− 3k
→ → → →
(1) 225 (2) 275 (3) 325 (4) 300 Given projection of a on b = projection of c on d
→ → → →
Hint : a⋅ b c⋅ d
→ → → → ⇒ =

oo
→ → → → → → → → → →
[( a + 3 b ) × (3 a − b ) ]2 = [a× 3 a − a× b+ 9 b× a − 3 b× b] 2
|b| |d |
→ → → → → → → → →
5 (1) − 1(3) − 3 ( λ )
2
= [0 − a × b − 9 a × b − 0] [ [∵ a × a = b × b = 0 ] 5 − 3 − 3λ
→
2
→
2
→ → ⇒ =
12 + 32 + λ 2 52 + ( −1) + ( −3)
2 2
= [ −10 a × b ] = 100 | a × b | = 100. [| a |2 | b |2 sin2 q]
ab
= 100[(1)2 (2)2 sin2 120] =100 × 4 × [sin (180 – 60)]2 2 − 3λ 2 − 3λ
2 ⇒ =
 3 100
3 10 + λ 2 25 + 1 + 9
= 400 [sin 60]2 = 400 ×   = 400 × = 300
 2  4 10 + λ 2 = 35
 [Ans: (4) 300]  [Equating the denominator]
ur

→ →
20. If a and b are two vectors of magnitude 2 and Squaring, 10 + l2 = 35 ⇒ l2 = 25
inclined at an angle 60°, then the angle between
→ → → ⇒ l =± 5 [Ans: (3) ± 5]
a and a + b is 22. If (1, 2, 4) and (2, – 3λ, – 3) are the initial and
.s

(1) 30° (2) 60° (3) 45° (4) 90° ∧ ∧ ∧

terminal points of the vector i + 5 j − 7 k , then the
→ → → → → → → →
value of λ is equal to
Hint : | a + b |2= | a |2 + | b |2 + 2 a . b = 22 + 22 + 2| a |.| b |
w

7 7 5 5
cos q = 4 + 4 + 2(2) (2) (cos 60) (1) (2) − (3) − (4)
3 3 3 3
4
 1 → ∧ ∧ ∧ → ∧ ∧ ∧
= 8 + 8   = 8 + 4 = 12 Given OA = i + 2 j + 4 k and OB = 2 i − 3λ j − 3 k
Hint : 
w

 2
→ → → ∧ ∧ ∧
\ | a + b | = 12 = 2 3
2
and AB = i + 5 j − 7 k
→ → → → → →
Let a be the angle between a and a + b But AB = OB – OA .
w

→ → → → → → → ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
a⋅ ( a + b ) a⋅ a + a⋅ b i + 5 j − 7 k = ( 2 i − 3λ j − 3 k ) – ( i + 2 j + 4 k )
\ cos a = → → → = → → →
∧ ∧ ∧ ∧ ∧ ∧
| a || a + b | | a || a + b | i + 5 j − 7 k = i + ( −3λ − 2) j − 7 k
→ → → → →
| a |2 + a ⋅ b 22 + | a || b | cos θ Equating the like components both sides, we get
= =
→ →
| a || a + b |
→
( )
2 2 3 5 = – 3l – 2 ⇒ 7 = – 3l
 −7 −7
⇒ l = [Ans: (2) l = ] → ∧ ∧ ∧ → →
3 3 Hint : Given a = i + 2 j + 2 k , | b | = 5 angle between a
23. If the points whose position vectors → π
∧ ∧ ∧ ∧ ∧ ∧ and b is .
10 i + 3 j , 12 i − 5 j and a i + 11 j are collinear 6
→
then a is equal to |a|= 12 + 22 + 22 = 9 =3

m
(1) 6 (2) 3 (3) 5 (4) 8 → →

→ ∧ ∧ → ∧ ∧ → ∧ ∧
Area of the triangle formed by a and b
Hint : OA = 10 i + 3 j ; OB = 12 i − 5 j and OC = a i +11 j 1 → → 1 → →
= | a × b | = | a || b | sin q
→ → → 2 2

co
AB = OB – OA 1 π 1  1  15
∧ ∧ ∧ ∧ ∧ ∧ = 3 (5) sin  = 15 ×  =
= ( 12 i − 5 j ) – ( 10 i + 3 j ) = 2 i − 8 j 2 6 2  2 4
→ ∧ ∧
BC = ( a − 12) i + 16 j 15
 [Ans: (2) ]
→ ∧ ∧ 4

.
CA = (10 − a ) i − 8 j
→ → ADDITIONAL PROBLEMS

ks
∧ ∧ ∧ ∧
AB = CA ⇒ 2 i − 8 j = (10 − a ) i − 8 j
∧ SECTION - A (1 MARK)
⇒
⇒
2 = 10 – a 
a = 10 – 2 = 8
[Equating i components]
[Ans: (4) 8]
1. ( → → →
)
If m 2 + j + k is a unit vector then the value of m
is

oo
[Hy - 2018]
→ ∧ ∧ ∧ → ∧ ∧ ∧ → ∧ ∧ ∧
24. If a = i + j + k , b = 2 i + x j + k , c = i − j + 4 k 1 1 1 1
→ → →
(1) ± (2) ± (3) ± (4) ±
3 5 6 2
and a ⋅ ( b × c ) = 70 , then x is equal to
(1) 5 (2) 7 (3) 26 (4) 10 Hint : ( → →
m 2 + j + k is a unit vector
→
)
ab
Hint :
→ ∧ ∧ ∧ →
a = i + j+ k , b = 2 i + x j+ k , c = i − j+ 4 k
∧
∧

∧
∧ ∧

∧
→ ∧ ∧ ∧
(
m 2+ j + k
→ → →
) =1
→ → →
i j k m 2+ j + k =1
→ →
2 2 2
b×c = 2 x 1 m 2 + 1 + ( −1) = 1
ur

|m| 6 = 1
1 −1 4
∧ ∧ ∧ 1
= i ( 4 x + 1) − j (8 − 1) + k ( −2 − x ) |m| =±
6
.s

∧ ∧ ∧ 1 1
= i ( 4 x + 1) + j ( −7 ) + k ( −2 − x )  [Ans:(3) ±
|m| = ± ]
→ → →   6 6
Given a .( b × c ) = 70 2. If a , b are the position vectors of A and B, then
w

⇒ 1(4x + 1) + 1(–7) + 1(–2 – x) = 70 which one of the following points whose position
vector lies on AB? [March - 2019]
⇒ 4x + 1 – 7 –2 – x = 70    
⇒ 3x – 8 = 70 2a + b a−b
w

(1) (2)
⇒ 3x = 78 3 3
26  
  2a − b
⇒ x = 78 = 26[Ans: (3) 26] (3) a + b (4)
3 2
w

→ ∧ ∧ ∧ → →
→ ∧ ∧
25. If a = i + 2 j + 2 k , | b | = 5 and the angle between a Hint : OA = 2 i + 5 j
→ π → ∧ ∧ ∧
and b is , then the area of the triangle formed OB = 5 i + 7 j + 4 k
6
by these two vectors as two sides, is → → → ∧ ∧ ∧
AB = OB – OA = 5 i + 2 j + 4 k
7 15 3 17 ∧ ∧ ∧
(1) (2) (3) (4)  [Ans: (3) −5 i + 2 j + 4 k ]
4 4 4 4
3. The vector having initial and terminal points as The Correct match is
(2, 5, 0) and (–3, 7, 4) respectively is (i) (ii) (iii) (iv)
∧ ∧ ∧ ∧ ∧ ∧
(1) − i + 12 j + 4 k (2) 5 i + 2 j − 4 k (1) b c d a
∧ ∧ ∧ ∧ ∧ ∧ (2) c a d b
(3) −5 i + 2 j + 4 k (4) i + j+ k
(3) d b a c

m
→ ∧ ∧
Hint : OA = 2 i + 5 j (4) d c b a
→ ∧ ∧ ∧  [Ans : (2) i – c ii – a iii – d iv – b]
OB = 5 i + 7 j + 4 k

co
→ → 7. Assertion (A) : If ABCD is a prallelogram,
→ ∧ ∧ ∧
→ → → →
AB = OB – OA = 5 i + 2 j + 4 k
AB + AD + CB + CD then is equal to zero.
∧ ∧ ∧
D C
 [Ans: (3) −5 i + 2 j + 4 k ]

.
→ → → →
4. The value of λ when the vectors a = 2 i + λ j + k

ks
→ → → →
and b = i + 2 j + 3 k are orthogonal is A B
3 5 → →
(1) 0 (2) 1 (3) (4) − Reason (R) :  AB and CD are equal in magnitude
2 2 →

oo
→ → and opposite in direction. Also AD
Hint : a⋅ b = 2(1) + λ(2) + (1)3 = 0 →
and CB are equal in magnitude and
⇒
2+ 2λ+ 3 = 0
opposite in direction
−5 5
λ =  [Ans: (4) − ] (1) Both A and R are true and R is the correct
2 2 explanation of A
ab
∧ ∧ ∧
5. The value of m for which the vectors 3 i − 6 j + k (2) Both A and R are true and R is not a correct
∧ ∧ ∧ explantion of A
and 2 i − 4 j + λ k are parallel is (3) A is true but R is false
2 3 5 2 (4) A is false but R is true
(1) (2) (3) (4) [Ans: (1) Both A and R are true and R is the correct
ur

3 2 2 5
( )
∧ ∧ ∧ 3 ∧ ∧ ∧  explanation of A]
Hint : 3 i − 6 j + k =
2 2 i − 4 j+ λ k 8. Find the odd one out of the following
∧ ∧ 3λ ∧ ∧ ∧ ∧ ∧ ∧ ∧
= 3 i − 6 j + k
.s

2 (1) i + 2 j + 3 k (2) 2 i + 4 j+ 6 k
∧ ∧ ∧ ∧ ∧ ∧
3λ 2 2 (3) 7 i + 14 j + 21 k (4) i + 3 j+ 2 k
= 1 ⇒ λ =  [Ans: (1) ]
2 3 3 Hint : (1), (2), (3) are parallel vectors
w

∧ ∧ ∧
6. Match List - I with List II  [Ans: (4) i + 3 j + 2 k ]
List I List II → → →
∧ ∧
9. Assertion (A) : a , b , c are the position vector of
w

i. i .i (a) 0 → → →
three collinear points then 2 a = b + c
∧ ∧ ∧
ii. i. j (b) Reason (R) : Collinear points, have same direction
k
(1) Both A and R are true and R is the correct
w

iii.
∧ ∧ (c) 1 explanation of A
i× i
(2) Both A and R are true and R is not a correct
∧ ∧ (d) 0 explantion of A
iv. i× j (3) A is true but R is false
(4) A is false but R is true
[Ans: (1) Both A and R are true and R is the correct
 explanation of A]
10. Find the odd one out of the following 5. Find the scalar and vector components of the
(1) matrix multiplication vector with initial point (2, 1) and terminal point
(2) vector cross product (–5, 7).
(3) Subtraction Solution : Let A(2, 1) be initial point and B(–5, 7) be
(4) Matrix Addition terminal point of given vector.
→ ∧ ∧ ∧ ∧

m
Hint : Only (4) is commutative Then, AB = ( −5 − 2) i + (7 − 1) j = −7 i + 6 j
 [Ans: (4)Matrix Addition] →
\ The scalar components of AB are –7 and 6.
SECTION - B (2 MARKS) →

co
∧ ∧
The vector components of AB are −7 i and 6 j .
1. Define diagonal and scalar matrices.[March - 2019]
Solution : Diagonal; In a square matrix A = [aij]n × r ∧ ∧ ∧ ∧ ∧ ∧
6. Show that the vectors 2 i − 3 j + 4 k are −4 i + 6 j − 8 k
Of order n, the elements a11, a22 , a33..... ann are
are collinear.→

.
called the principal diagonal or simply the diagonal ∧ ∧ ∧ → ∧ ∧ ∧
Scalar matrix: Solution : Let a = 2 i − 3 j + 4 k and b = −4 i + 6 j − 8 k

ks
→
A diagonal matrix whose entries along the principal Then | a | = 22 + ( −3) + 42
2

diagonal are equal is called a scalar matrix.

= 4 + 9 + 16 = 29
2. Find a unit vector along the direction of the vector →
5i − 3 j +4k  and | b | = ( −4) 2
+ 62 + ( −8)
2

oo
[March - 2019]
Solution : a = 5i − 3 j + 4k = 16 + 36 + 64 = 116

\a = )  = ±

a (
5i − 3 j + 4k ) =
→
4 × 29 = 2 29
→
a 5 2 \|b| = 2| a |
ab
→ ∧ ∧ ∧ → ∧ ∧ ∧ → →
3. If a = 3 i − 2 j + k and b = 2 i − 4 j + k then find Thus, a and b are collinear.
→ → → ∧ ∧ ∧ → ∧ ∧ ∧
| a – 2 b |. 7. If a = i + 2 j + 3 k and b = 2 i + 3 j − 5 k then find
Solution : → → → → →
ur

→ ∧ ∧ ∧ a × b . Verify that a and a × b are perpendicular

Given a = 3 i − 2 j+ k
→ ∧ ∧ ∧ to each other.
b = 2 i − 4 j+ k → ∧ ∧ ∧ → ∧ ∧ ∧
→ → ∧ ∧ ∧ ∧ ∧ ∧ Solution : Given a = i + 2 j + 3 k and b = 2 i + 3 j − 5 k
\ a – 2b = ( 3 i − 2 j + k ) – 2( 2 i − 4 j + k )
.s

∧ ∧ ∧
∧ ∧ ∧
→ → i j k
= − i + 6 j − k \ a×b = 1 2
→ → 3
| a –2b | = ( −1) 2
+ 6 + ( −1)
2 2
2 3 −5
w

∧ ∧ ∧
= 1 + 36 + 1 = 38 = i ( −10 − 9) − j ( −5 − 6) + k (3 − 4)
∧ ∧ ∧
4. Write two different vectors having same
w

= −19 i + 11 j − k
magnitude. → ∧ ∧ ∧ → ∧ ∧ ∧ → → → ∧ ∧ ∧ ∧ ∧ ∧
Solution : Let a = 2 i − j + 3 k and b = i + 2 j − 3 k be Now, a . ( a × b ) = ( i + 2 j + 3 k ) . ( −19 i + 11 j − k )
two vectors.
w

→ = 1(–19) + 2(11) + 3(–1)

Then, | a | = 22 + ( −1) + 32 = 14
2

→ = – 19 + 22 – 3 = – 22 + 22 = 0
and | b | = 1 + 2 + ( −3) = 14
2 2 2
→ → →
∧ ∧ ∧ ∧ ∧ ∧
Hence the required vectors are 2 i − j + 3 k and i + 2 j − 3 k . This shows that a and ( a × b ) are perpendicular to each
other.
 SECTION - C (3 MARKS) → → → → → →
⇒ | x |2 – a . x + a . x –| a |2 = 12
1. Find the unit vector in the direction of the vector → → →
→ → → → ∧ ∧ → ∧ ∧ → ∧ ∧ ⇒ | x |2 – | a |2 = 12 [ | a | = 1]
a − 2 b + 3 c if a = i + j , b = j + k and c = i + k . →
→ ∧ ∧ → ∧ ∧ → ∧ ∧ ⇒ | x |2 = 13
Given now, a = i + j ; b = j + k ; c = i + k →

m
Solution :
→ → → ∧ ∧ ∧ ∧ ∧ ∧ ⇒ |x| = 13
\ a – 2 b +3 c = ( i + j ) – 2( j + k ) + 3( i + k ) → → →
∧ ∧ ∧ 4. Let a , b and c be non-coplanar vectors. Let
= 4 i − j + k A, B and C be the points whose position vectors

co
→ → →
→ → →
with respect to the origin O are a + 2 b + 3 c ,
42 + ( −1) + 12 = 16 + 1 + 1
2
\ | a – 2 b +3 c | = → → → → →
−2 a + 3 b + 5 c and 7 a − c respectively. Then
= 18 = 9×2 = 3 2
prove that A, B and C are collinear.
→ → →

.
Thus, the unit vector in the direction of a – 2 b +3 c is → → → →
Solution : Given OA = a + 2 b + 3 c

ks
→ → →
a− 2b+3c 1 ∧ ∧ ∧ → → → → → → →
→ → → = ( 4 i − j+ k ) OB = −2 a + 3 b + 5 c and OC = 7 a − c
| a− 2b+3c | 3 2 → → →
Then AB = OB – OA
2. Find the direction cosines of the vector joining the

oo
→ → → → → →
points A(1, 2, –3) and B(–1, –2, 1) directed from = ( −2 a + 3 b + 5 c ) – ( a + 2 b + 3 c )
A to B. → → →
= −3 a + b + 2 c
Solution : Given points are A (1, 2, –3) and B (–1, – 2, 1). → → → → → → → →
→ → → AC = OC – OA = ( 7 a − c ) – ( a + 2 b + 3 c )
ab
Then AB = OB – OA → → →
∧ ∧ ∧ ∧ ∧ ∧ = 6 a − 2 b − 4 c
= ( − i − 2 j + k ) – ( i + 2 j − 3 k ) → → → →
∧ ∧ ∧ = – 2( −3 a + b + 2 c ) = – 2 AB
= −2 i − 4 j + 4 k → →
 \ AC || AB and A is a common points. Hence, the
ur

→ points A, B and C are collinear.

AB = ( −2)2 + ( −4)2 + 42
5. If ABCDE is a pentagaon then prove that
= 4 + 16 + 16 = 36 = 6 → → → → → → →
AB + AE + BC + DC + ED + AC = 3 AC
.s

x −2 −1 → → → → → →
Now, l = → = = Solution : AB + AE + BC + DC + ED + AC 
6 3
| AB | 3 → → → → → →
w

2 = (AB + BC) + (AE + ED + DC) + AC

y −4 −2 D
m = → = 6
=
| AB | 3
3
w

2
z 4 2
n = → = 6
= E C
3
| AB | 3
w

→  1 2 2
Thus, the direction cosines of AB are  − , − , 
 3 3 3
→ → → → → → A B
3. Find | x | if for a unit vector a , ( x − a ) ⋅ ( x + a ) = 12
→ → → → →
→ → →
Solution : Given | a | = 1 and ( x − a )( x + a ) = 12 = AC + AC + AC

⇒
→ → → →
x.x –a.x + x.a –a.a
→ → → →
= 12
→ (Using triangle law of addition)
= 3 AC Hence proved.
 SECTION - D (5 MARKS) → → → ∧ → → ∧ ∧ ∧
and GB = GO + OB = − k + OA + AB = i + j − k
→ → → → → → → →
1. Let a = 2 j + j − 2 k ; b = 2 i + j . If c is a vector Let q be the smaller angle between the diagonals OE and
→ → → → →
GB, then
such that a . c = c , c − a = 2 2 and the →→
→ → → OE⋅ GB 1(1) + 1(1) + 1( −1)
cos q =
→ → = 12 + 12 + 12 ⋅ 12 + 12 + ( −1)2

m
angle between a × b and c is 30º. Find the value
( )
→ → → | OE || GB |
of a × b × c  [Hy - 2018]
2 −1 1
= =

co
→ → → 3 3 3
Solution : a . b = 3 ⇒ | c |= 3
    1
i j k Thus q = cos–1  
→ →     3
a× b = 2 1 -2 = 2i − 2 j + k

.
→ → → →
1 1 0
3. If a , b and c are three vectors such that | a | = 3,

ks
→ → → →
c× i = 4 + 4 +1 = 9 =3 | b | = 4 and | c | = 24 and sum of any two vectors
→ → →

(a× i ) × c
→ → →
(
= a× i
→
×→ →
)
c sin 30º
is orthogonal to the third vector, then find | a + b + c |.
→ → →

oo
Solution : Given ( a + b ) . c = 0
1 9
= 3 × 3 × 2
= 2
→ → → →
⇒ a.c+ b.c = 0
2. Prove that the smaller angle between any two → → →
 1 (b+c).a = 0
diagonals of a cube is cos–1   .
ab
→ → → →
 3
⇒ b.a+c.a = 0
Solution : Let OABCDEFG be a unit cube.
→ → →
G D (c+a).b = 0
→ → → →
⇒ c.b +a.b = 0
ur

F k → → → → → →
E Adding, 2( a . b + b . c + c . a ) = 0
→ → → → → →
a . b + b . c + c . a = 0  ... (1)
.s

j → → → → → →
i C | a + b + c |2 = | a |2 +| b |2 + | c |2 +
O
→ → → → → →
2( a . b + b . c + c . a )
w

A B
= 9 + 16 + 24 + 2(0)
Keeping O as origin,
w

= 49
→ ∧ → ∧ → ∧
→ → →
Let OA = i , OC = j and OG = k ⇒ |a+b+c| = 7
Consider the diagonals OE and BG.
w

→ → → → → →
OE = OB + BE = OA + AB + BE
→ → → ∧ ∧ ∧
= OA + OC + OG = i + j + k
→ → → →
[ AB = OC, BE = OG ]