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MULTIPLE PRODUCTS
OBJECTIVES
1) 8
4.
3) 3
{( ) (
The scalar a. b × c × a + b + c
)} is equal to
1) 0
2) a b c  +  b c a 
3) a b c 
4) 2 a b c 
If a is perpendicular to b and c , a = 2, b = 3, c = 4 and the angle between b and
sh
i
2) 12 3
3)
12
3
4) 12 2
ak
1) 12
The vector a lies in the plane of vectors b and c . Which of the following is
w
.s
correct?
2) a. ( b × c ) = 1
3) a. ( b × c ) = −1
4) a. ( b × c ) = 3
w
1) a. ( b × c ) = 0
For three vectors u, υ, w which of the following expressions is not equal to any
w
6.
4) not defined
n.
2) 1
c is 2 π , then  abc  =


3
5.
5) 2
If a = 3i − j + 2k and b = 3i + j − k , then a. ( a × b ) =
1) 0
3.
3) 4
at
io
2.
2) 6
co
m
If a = i + 2 j, b = j + 2k, c = i + 2k, then a. ( b × c ) =
ed
uc
1.
of the remaining three?
1) u. ( υ× w )
2) ( υ× w ) .u
3) υ. ( u.w )
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4) ( u × υ ) .w
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7.
Volume of the parallelepiped whose coterminous edges are 2i − 3j + 4k,i + 2 j − 2k ,
3i − j + k
2) 6 cu. Units
3) 7 cu. Units
4) 8 cu. Units
If a b c  = 2 , then the volume of the parallelepiped whose coterminous edges


co
m
8.
1) 5 cu. Units
2) 8 cu. Units
3) 18 cu. Units
4) 16 cu. Units
If a b c  = 4 , then the volume of the parallelepiped with a + b, b + c and c + a as


at
io
9.
1) 9 cu. Units
n.
are 2a + b, 2b + c and 2c + a is
coterminous edges is
2) 7
3) 8
4) 5
ed
uc
1) 6
10. If a b c  = 12 , then a + b b + c c + a  =




1) 24
2) 36
3) 48
4) 26
 2a + b 2b + c 2c + a 
=
11. If a, b, c are linearly independent, then 
sh
i
2) 8
3) 7
4) 6
ak
1) 9
a b c 


w
.s
(
12. If a, b, c are linearly independent and 
1) 10
2) 14
3) 18
) (
)(
)
 a + 2b × 2b + c . 5c + a 
 = k , then k is
a.b × c
4) 12
(
) (
)
w
13. If a, b, c are three non-coplanar vectors, then ( a + b + c )  a + b × a + c  is equal


w
to
1) 0
2) a b c 
3) 2 a b c 
4) −  a b c 
14. If a, b, c be three non-coplanar vectors and  a − b b − c c − a  = k a b c  , then k =




1) 1
2) 0
3) -1
4) 2
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15. Let a = i − k, b = xi + j + (1 − x ) k and c = yi + x j + (1 + x + y ) k . Then  a b c  depends
on
1) Only x
2) Only y
3) Neither x nor y
4) Both x and y
co
m
16. The value of p such that the vectors i + 3j − 2k , 2i − j + 4k and 3i + 2 j + pk are
coplanar, is
1) 4
2) 2
3) 8
4) 10
2) 2,1 ± 6
3) −2,1 ± 3
4) −3,1 ± 2
1)
1
6
2) −
1
6
3)
1
5
ed
uc
18. If ( a − λ b ) . ( b − 2c ) × ( c + 3a ) = 0 then λ =
at
io
1) 2,1 ± 5
n.
17. The vectors λi + j + 2k , i + λ j − k and 2i − j + λ k are coplanar if λ =
4) −
1
5
2
19. If a, b, c are unit vectors perpendicular to each other, then a b c  =


2) 3
3) 2
4) 4
sh
i
1) 1
ak
20. If a = 2i + 3 j, b = i + j + k and c = λ i + 4 j + 2k are coterminous edges of a
w
.s
parallelepiped of volume 2 cu. Units, then a value of λ is
1) 1
3) 3
4) 4
(
) + b.( a × c ) is
( c × a ) .b c.( a × b )
a. b × c
w
w
21.
2) 2
1) 0
2) 1
3) -1
4) 2
22. Let p = a × c , q = c × a , r = a × b a b c being any three non-coplanar vectors
a b c 


a b c 


a b c 


Then p ( a + b ) + q. ( b + c ) + r. ( c + a ) is equal to
1) -3
2) 3
3) 0
4) -2
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23. If  2a + 4b c d  = λ ac d  + µ  b c d  , then λ + µ =
1) 6
2) -6
3) 10
4) 8
24. The position vectors of the points A,B,C,D are 3i − 2 j − k , 2i + 3 j − 4 k , −i + j + 2 k
and 4i + 5 j + λ k respectively. If A,B,C,D are coplanar, then λ =
146
17
2)
146
17
3)
146
15
4) −
146
15
co
m
1) −
25. If the points (1,0,3), (-1,3,4), (1,2,1) and (a,2,5) are coplanar, then a =
2) -1
3) 2
4) -2
n.
1) 1
26. Let a be a unit vector and b a non zero vector not parallel to a . Then the angle
π
4
2)
π
3
3)
π
2
4)
π
6
ed
uc
1)
at
io
between the vectors u = 3 ( a × b ) and υ = ( a × b ) × a is
27. The volume of the tetrahedron having vertices i + j − k, 2i + 3j + 2k, −i + j + 3k and
2k is
3)
7
cu. Units
8
2)
7
cu. units
6
4)
5
cu. units
8
sh
i
7
cu. Units
3
ak
1)
28. The volume of the tetrahedron with vertices (2,2,2),(4,3,3),(4,4,4) and (5,5,6) is
w
.s
1
cu. Units
2
2)
w
1)
1
cu. units
4
3)
1
cu. units
3
4)
1
cu. units
5
29. Volume of the tetrahedron with vertices at (0,0,0), (1,0,0), (0,1,0) and (0,0,1) is
1
cu.units
6
2)
1
cu.units
4
3)
1
cu.units
3
4)
1
cu.units
5
w
1)
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30. The volume of the tetrahedron whose vertices are (1,-6,10), (-1,-3,7), (5,-1, λ )
and (7,-4,7) is 11 cu. Units, then the value of λ is
1) 2 or 6
2) 3 or 4 3) 1 or 7
4) 5 or 6
31. If i, j, k are orthonormal unit vectors, then i ( a × i ) + j × ( a × j) + k ( a × k ) is
2) 2a
3) 3a
4) 4a
co
m
1) a
32. If a = i − 2 j − 3k, b = 2i + j − k, and c = i + 3j − 2k and a × ( b × c ) = pi + q j + rk , then
p+q+r =
2) 4
3) 2
4) -2
n.
1) -4
1) 1
2) 0
3) -1
4) 2
at
io
( a × b ) × c = λa + µb, then λ + µ =
33. If a = i + j + k, b = i + j, c = i and
1) a
{ ( )} is equal to
a× a× a×b
1) ( a.a )( b × a )
4) a × b
2) ( a.a )( a × b )
4) a × b
3) a.a
( a × b ) × ( a × c ) .d is equal to
w
.s
1)  a b c 
ak
36.
3) c
sh
i
35.
2) b
ed
uc
34. If a = 2i − 3j + 4k, b = i + j − k and c = i − j + k , then a × ( b × c ) is perpendicular to
4)  b a c  ( b.d )
w
3)  a b c  ( c.d )
2)  a b c  ( c.d )
37. Which of the following statement is not true?
2) ( a × b ) × c = a × ( b × c )
3) a. ( b × c ) = ( a × b ) .c
4) a × ( b + c ) = ( a × b ) + ( a × c )
w
1) a ( b + c ) = a.b + a.c
38.
{a.( b × i )} i + {a.( b × j)} j + {a.( b × k )} k =
1) 2 ( a × b )
2) 3 ( a × b )
3) a × b
4) ( a × b )
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39. The position vectors of three non- collinear points A,B,C are a, b, c respectively.
The distance of the origin from the plane through A,B,C is
3)
1
2)
(a × b) + ( b × c) + (c × a )
(
a b c 


a × b + b×c + c×a
) (
) (
4)
)
2 a b c 
(a × b) + ( b × c) + (c × a )
3  a b c 
(a × b) + ( b × c) + (c × a )
co
m
1)
40. The shortest distance between the lines r = ( 3i + 8 j + 3k ) + s ( 3i − j + k ) and
) (
)
n.
(
1) 30
at
io
r = −3i − 7 j + 6k + t −3i + 2 j + 4k is
2) 2 30
3) 3 30
4) 4 30
ed
uc
41. If the four points a, b, c, d are coplanar, then  b c d  + c a d  + a b d  =

 
 

2)  a b c 
2) 2 a b c 
4) 3  a b c 
sh
i
1) 0
42. Let a = a1 i + a 2 j + a 3 k, , b = b1 i + b 2 j + b3 k, and c = c1 i + c2 j + c3 k be three non-zero
ak
and non coplanar vectors such that c is a unit vector perpendicular to both a
w
.s
a1
π
and b . If the angle between a and b is , then a 2
6
a3
2) ±
3) 1
4) -1
w
w
1) 0
1
2
(a
2
1
b1
b2
c1
c2 =
b3
c3
+ a 22 + a 32 )( b12 + b22 + b32 )
43. If a b are non-zero and non-collinear vectors, then a b i  i + a b j j + a b k  k is

 
 

1) a + b
2) a × b
3) a − b
4) b × a
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(
)
44. If three unit vectors a , b , c are such that b × b × c = 1 b , then the vector a
2
makes with b and c respectively the angles
1) 40°, 80°
3) 30°, 60°
4) 90°, 60°
(
)
(
)
1
b + c , then the angle between a and b
2
a × b×c =
1) 60°
co
m
Let a, b, c be three non-coplanar vectors each having a unit magnitude. If
2) 30°
3) 135°
4) 120°
n.
45
2) 45°, 45°
(
If
at
io
46. Let a, b, c be three vectors having magnitude 1, 1 and 2 respectively.
)
a × a × c + b = O , then the acute angle between a and c is
2) 60°
3) 45°
4) 75°
ed
uc
1) 30°
47. Let p, q, r be the three mutually perpendicular vectors of the same magnitude.
{
}
{
}
{
}
sh
i
If a vector x satisfies the equation p × ( x − r ) × p + q × ( x − r ) × q + r × ( x − p ) × r = O
, then x is given by
(
)
(
)
(
)
1
p + q + 2r
2) 2
w
.s
ak
1
1) p + q − 2r
2
(
1
3) p + q + r
3
)
1
2p + q − r
4) 3
w
48. If a ( α ×β ) + b ( β× γ ) + c ( γ × α ) = O and at least one of a a,b and c is non-zero, then
w
vectors α, β γ
1) Parallel
2) Mutually Perpendicular
3) Coplanar
4) None of these
49. If ( b × c ) × ( c × a ) = 3c , then  b × c c × a a × b  =
1) 2
2) 7
3) 9
4) 11
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50. Vector
(
is
c
perpendicular
a a = 2i + 3j − k and b = i − 2 j + 3k .
to
Also
)
c. 2i − j + k = 6 . Then the value of  a b c  is
1) 36
2) -36
3) -63
4) 63
51. If b and c are any two non-collinear unit vectors and a is any vector, then
b×c
2) 3a
)
2
(b × c) =
3) a
4) 4a
n.
1) 2a
(
a. b × c
co
m
( a.b ) b + ( a.c ) c +
52. The position vectors of vertices of ∆ ABC are a, b, c . Given that a.a = b.b = c.c = 9
1) a − b + c
(
) (
)
(
3) a + b + c
)
a × 3b + 2c , b × c − 2c , 2c × a − 3b 


1) −1 a × b, b × c, c × a 
2
4) 6  a × b, b × c, c × a 
sh
i
2
2) 18 a b c 
w
w
w
.s
ak
3) −18 a b c 
4) b + c − a
ed
uc
53.
2) a + b − c
at
io
and a b c  = o . Then the position vector of the orthocentre of ∆ ABC is
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MULTIPLE PRODUCTS
HINTS AND SOLUTIONS
Scalar and Vector Triple Products
co
m
1. (2)
1 2 0
a ⋅ (b × c)[a b c |= 0 1 2
n.
1 0 2
at
io
= 1(2 − 0) − 2(0 − 2) = 2 + 4 = 6.
2. (4)
a ⋅ b is a scalar. Hence a × (a ⋅ b) i.e. cross product between a vector and a scalar is not
ed
uc
defined.
3. (1)
G.E. = a ⋅ {(b × c) × (a + b + c)}
sh
i
= a ⋅ {(b × a) + (b × b) + (b × c) + (c × a ) + (c × b) + (c × c)}
= a ⋅ {(b × a) + (b × c) + (c × a ) + (c × b)}
ak
= a ⋅ (b × a ) + a ⋅ (b × c) + a ⋅ (c × a ) + a ⋅ (c × b)
w
.s
= [a b a] + [a b c] + [a c a] + [a c b]
= 0 + [a b c] + 0 − [a b c] = 0.
w
4. (2)
w
a ⊥ b, a ⊥ c ⇒ a //(b × c)
Now | b × c |=| b | ⋅ | c | sin120°
 3
= 3(4) 
 = 6 3.
 2 
Then [a b c] = a ⋅ (b × c)
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=| a | ⋅ | b × c | cos(a, b × c)
= 2(6 3) cos 0° = 12 3.
5. (1)
Given a, b, c are coplanar. But b × c is a vector perpendicular to the plane containing
co
m
b and c .
Hence b × c is also perpendicular to a
n.
∴ a ⋅ (b × c) = 0 .
v ⋅ (u ⋅ w) is not equal to the remaining three.
2 −3
Volume = 1 2
ed
uc
7. (3)
at
io
6. (3)
4
2
3 −1 −1
sh
i
= | 2(2 + 2) + 3(1 − 6) + 4(−1 − 6) |= 7
8. (3)
ak
2 1 0
Volume = 0 2 1 [a b c]
w
.s
1 0 2
= [2(4 − 0) − 0 + 1(1 − 0)](2) = 18 .
w
9. (3)We have
w
[a + b b + c c + a] = 2[a b c] = 2(4) = 8 .
10. (1)
[a + b b + c c + a] = 2[a b c] = 2(12) = 24
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11. (1)
[2a + b 2b + c 2c + a ]
2 1 0
= 0 2 1 [a b c]
1 0 2
G.E. =
co
m
= 9[a b c]
9[a b c]
= 9.
[a b c]
n.
12. (4)
1 0 5
[a b c]
at
io
1 2 0
0 2 1 [a b c]
ed
uc
= k ⇒ 12 = k
13. (4)
G.E. = (a + b + c) ⋅ [(a + b) × (a + c)]
(a + b + c) ⋅ [(a × b) + (a × c) + (b × a) + (b × c)]
sh
i
= a(a × c) + a(b × a) + a(b × c) + b(a + c) + b(b + a) + b(b × c) + c(a × c) + c(b × a) + c(b × c)
ak
(∵ a × a = 0)
= 0 + 0 + [a b c] − [b a c] + 0 + 0 + 0 + [c b a] + 0
w
.s
= [a b c] − [b c a] − [c a b]
= [b c a] − [b c a] − [a b c] = −[a b c]
w
14. (2)
−1
0
0
1
−1 [a b c] = k[a b c]
−1
0
1
w
1
⇒ 0 = k[a b c]
⇒ k = 0 {∵[a b c] ≠ 0}
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15. (3)
1
0
−1
1
0
0
[a b c] = x
1
1− x
= x
1
1
y x 1+ x − y
y x 1+ x
co
m
Applying: C3→C3 + C1
= 1(1 + x) – 1.x = 1 + x – x = 1
(Expanding along R1)
n.
This depends neither on x nor on y.
at
io
16. (2)
Given vectors are coplanar
3
2
ed
uc
1 3 −2
⇒ 2 −1 4 = 0
p
⇒ 1(−p − 8) − 2(3p + 4) + 3(12 − 2) = 0
sh
i
⇒ 7p + 14 = 0 ⇒ p = 2.
17. (3)
1
λ
2
−1 = 0
w
.s
λ
1
ak
Given vectors are coplanar ⇔
2 −1
λ
w
⇒ λ(λ 2 − 1) − 1(λ + 2) + 2(−1 − 2λ) = 0
w
⇒ λ 3 − 6λ − 4 = 0
By inspection one value of λ is –2.
∴ (λ + 2)(λ 2 − 2λ − 2) = 0 .
Now λ 2 − 2λ − 2 = 0 ⇒ λ =
2± 4+8
2
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=
2±2 3
= 1± 3
2
∴λ = −2,1 ± 3
18. (2)
co
m
[a − λ b b − 2c c + 3a] = 0
1 −λ 0
⇒ 0 1 −2 [a b c] = 0
3 0
1
n.
⇒ 1(1 − 0) + λ (0 + 6) = 0
at
io
Since [a b c] ≠ 0 ⇒ λ = −1/ 6.
19. (1)
ed
uc
Given
| a | = | b | = | c | = 1 and a ⊥ b, b ⊥ c, c ⊥ a .
Now b × c =| b | ⋅ | c | sin 90°n
sh
i
= 1 ⋅1 ⋅1 ⋅ n ⇒ n is ⊥ b and c
But a is ⊥ to b and c ⇒ a || n
ak
∴ a ⋅ (b × c) = a ⋅ n =| a | ⋅ | n | cos 0°
or | a || n | cos π = 1 or − 1.
w
.s
⇒ [a b c] = ±1
⇒ [a b c]2 = 1.
w
20. (4)
w
2 3 0
Volume = 1 1 1 = 2(given)
λ 4 2
⇒| 2(2 − 4) − 3(2 − λ) + 0 |= 2
⇒| 3λ − 10 |= 2
3λ − 10 = 2 ⇒ 3λ = 12 ⇒ λ = 4.
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21. (1)
G.E. =
=
[a b c] [b a c]
+
[c a b] [c a b]
[a b c] [a b c]
−
=0
[c a b] [c a b]
G.E. =
co
m
22. (2)
(b × c) ⋅ (a + b)
+
[a b c]
[a b c] + 0 + [b c a] + 0 + [c a b] + 0
[a b c]
=
3[a b c]
= 3.
[a b c]
ak
sh
i
=
23. (1)
at
io
ed
uc
 a ⋅ (a × c) + b ⋅ (b × c) + b ⋅ (c × a ) + 


c ⋅ (c × a) + c(a × b) + a (a × b) 

=
[a b c]
n.
(c × a ) ⋅ (b + c) (a × b) ⋅ (c + a )
+
[a b c]
[a b c]
(2a + 4b) ⋅ (c × d) = 2(a ⋅ c × d) + 4(b ⋅ c × d)
w
.s
= 2[a c d] + 4[b c d] + λ[a b c] + µ[b c d]
⇒ λ = 2, µ = 4
w
⇒ λ + µ = 6.
w
24. (1)
Given A(3, –2, –1), B(2, 3, –4),
C = (–1, 1, 2), D = (4, 5, λ)
AB = (−1, 5, −3), AC = ( −4,3, 3) , AD = (1, 7, λ + 1)
Given A, B, C, D are coplanar
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−1 5
−3
⇒ [AB AC AD] = 0 ⇒ −4 3
3
1
=0
7 λ +1
⇒ 17λ = −146 ⇒ λ = −146 /17.
co
m
25. (2)
Let A = (1, 0, 3), B = (–1, 3, 4),
C = (1, 2, 1), D = (a, 2, 5)
n.
Then AB = ( −2, 3,1), AC = (0, 2, −2)
at
io
AD = (a − 1, 2, 2)
Given points are coplanar
⇒ 8a = −8 ⇒ a = −1.
26. (3)
ed
uc
⇒ [AB AC AD] = 0
3(a × b) ⋅ {(a × b) × a}
3 | a × b | ⋅ | (a × b) × a |
3(a × b) ⋅ (a × b)a
=0
3 | a × b | ⋅ | (a × b) × a
ak
=
sh
i
If θ is the angle between the given vectors, then cos θ =
w
.s
⇒ θ = 90° = π / 2 .
27. (1)
w
Let A = (1, 1, –1), B = (2, 3, 2),
w
C = (–1, 1, 3), D = (0, 0, 2)
AB = (2 − 1,3 − 1, 2 + 1) = (1, 2,3)
AC = (−2, 0, 4), AD = (−1, −1,3)
1
6
Volume = [AB AC AD]
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1 2 3
1
1
7
= −2 0 4 = (14) = cu.units
6
6
3
−1 −1 3
28. (3)
Let A = (2, 2, 2), B = (4, 3, 3), C = (4, 4, 4) and D = (5, 5, 6), be the vertices. Then
co
m
volume of the tetrahedron
1
= [AB AC AD]
6
n.
1
[(2,1,1), (2, 2, 2), (3,3, 4)]
6
ed
uc
2 1 1
1
1
1
= 2 2 2 = (2) = cu.units.
6
6
3
3 3 4
at
io
=
29. (1)
1
6
sh
i
Volume = [OA OB OC] where O, A, B, C are the given points.
w
.s
30. (3)
ak
1 0 0
1
1
1
⇒ V = 0 1 0 = (1) = cu.units.
6
6
6
0 0 1
Let A = (1, –6, 10), B = (–1, –3, 7),
w
C = (5, –1, λ), D = (7, –4, 7)
w
Then AB = (−2,3, −3), AC = (4,5, λ − 10) and AD = (6, 2, −3)
−2
4
6
1
Volume =
3
5
2 = 11
6
−3 λ − 10 −3
⇒ 22λ − 88 = ±66 ⇒ λ = 7 or 1
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31. (2)
G.E. = ( i ⋅ i )a − ( i ⋅ a ) + ( j ⋅ j)a
−( j ⋅ a ) j + (k ⋅ k)a − (k ⋅ a)k
= 3a − ( i ⋅ a) i + ( j ⋅ a) j + (k ⋅ a)k 
= 3a − a = 2a
co
m
32. (1)
(a ⋅ c)b − (a ⋅ b)c = p i + q i + rk
n.
⇒ (1 − 6 + 6)(2 i + j − k) −
at
io
(2 − 2 + 3)( i + 3 j − 2k) = p i + q j + rk
33. (2)
(c ⋅ a )b − (c ⋅ b)a = λ a + µb
⇒ λ = −(c ⋅ b) = −1
ed
uc
⇒ − i − 8 j + 5k = p i + q j + rk
⇒ p + q + r = −1 − 8 + 5 = −4
sh
i
and µ = c ⋅ a = 1 ⇒ λ + µ = −1 + 1 = 0
34. (1)
ak
a × (b × c) = (a ⋅ c)b − (a ⋅ b)c
= (2 − 3 + 4)( i + j − k) − (2 + 3 − 4)( i − j + k)
w
.s
= 2 i + 4 j − 4k
Now a × (b × c) ⊥ a
w
sin ce[a × (b × c)] ⋅ a = 4 + 12 − 16 = 0
w
35. (1)
a × {a × (a × b)} = a{(a ⋅ b)a − (a ⋅ a)b}
(a × a )(a ⋅ b) − (a ⋅ a )(a × b)
= 0 + (a ⋅ a)(b × a ) = (a ⋅ a)(b × a )
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36. (3)
(a × b) × (a × c) ⋅ d
co
m
= {(a × b)c}a − {(a × b)a}c  d
= {[a b c]a − [a b a ]c}d = [a b c](a ⋅ d).
37. (2)
n.
(a × b) × c ≠ a × (b × c) .
at
io
Hence (2) is not true.
38. (3)
Interchanging ⋅ and × , we get
ed
uc
G.E. = {a ⋅ (b × i )}i + {a ⋅ (b × j)} j + {a ⋅ (b × k)}k
= {(a × b) ⋅ i } i + {(a × b) ⋅ j} j + {(a × b) ⋅ k} k
sh
i
Let a × b = x i + yj + zk
Hence
ak
⇒ (a × b) ⋅ i = x, (a × b) ⋅ j = y, (a × b) ⋅ k = z
w
.s
G.E. = [a b i ]i + [a b j] j + [a b k]k
= x i + yj + zk = a × b.
w
39. (3)
w
The equation of the plane passing through the points A(a), B(b), C(c) is
r ⋅{(a × b) + (b × c) + (c × a)} = [a b c]
∴ Distance from the origin to the plane
=
[a b c]
.
| (a × b) + (b × c) + (c × a )
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40. (3)
Here a = 3 i + 8 j + 3k, b = 3 i − j + k,
c = −3i − 7 j + 6k, d = −3 i + 2 j + 4k
b×d = 3
−3
j
k
−1 1
2
co
m
i
4
= i (−4 − 2) − j(12 + 3) + k(6 − 3)
n.
= −6 i − 15 j + 3k
Points A(a), B(b), C(c), D(d) are coplanar.
⇒ [AB, AC, AD] = 0
⇒ [b − a c − a d − a] = 0
sh
i
⇒ (b − a) × (c − a) ⋅ (d − a) = 0
ed
uc
⇒ AB, AC, AD are coplanar
at
io
41. (2)
⇒ {(b × c) − (b × a) − (a × c)} ⋅ (d − a) = 0
ak
⇒ {(b × c) − (a × b) + (c × a)} ⋅ (d − a) = 0
w
.s
⇒ (b × c) ⋅ d − (b × c) ⋅ a + (a × b) ⋅ d − (a × b) ⋅ a + (c × a) ⋅ d − (c × a) ⋅ a = 0
⇒ [b c d] + [a b d] + [c a d] = [a b c]
since [a b c] = 0 and [c a a] = 0 and
w
(b × c) ⋅ a = a ⋅ (b × c) .
w
42. (2)
c || (a × b) and | c | = 1 . Also (a, b) = 30°
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a1
a2
b1 b 2
c1 c2
a3
b3 = [a b c] = c ⋅ (a × b)
c3
= | c | . | a × b | cos 0° or cos180°
= ±1⋅ | a | ⋅ | b | sin 30°
1 2 2
a1 + a 2 + a 32 b12 + b 22 + b32
2
n.
=±
1
2
co
m
= ± a12 + a 22 + a 32 b12 + b 22 + b32
Let a × b = x i + yj + zk
∴ (a × b) ⋅ i = x, (a × b) ⋅ j = y, (a × b) ⋅ k = z
ed
uc
∴[a b i ]i + [a b j] j + [a b k]k
at
io
43. (2)
= x i + yj + zk = a × b
44. (4)
b
2
b
2
ak
⇒ (a ⋅ c) − (a ⋅ b)c =
sh
i
Given a × (b × c) =
… (1)
w
.s
Taking dot product with b ,
w
1
(a ⋅ c)(b ⋅ b) − (a ⋅ b)(c ⋅ b) = (b ⋅ b)
2
w
⇒ a ⋅ c − (a ⋅ b)(c ⋅ b) =
1
2
… (2)
In (1) take dot product with c :
(a ⋅ c)(b ⋅ c) − (a ⋅ b)(c ⋅ c) =
c⋅b
2
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⇒ (a ⋅ c)(b ⋅ c) − (a ⋅ b) =
c ⋅b
2
c⋅b
1

⇒  + (a ⋅ b)(b ⋅ c)  (b ⋅ c) − a ⋅ b =
2
2

Using (2)
b⋅c
b⋅c
− (a ⋅ c)(b ⋅ c)2 − a ⋅ b =
2
2
co
m
⇒
⇒ (b ⋅ c) 2 = 1
1
1
⇒ 1 ⋅1cos(a, c) =
2
2
at
io
∴a ⋅ c =
n.
⇒ b ⋅ c = 1 or a ⋅ b = 0
⇒ (a ⋅ c) = 60° and a ⋅ b = 0
45. (3)
1
(b + c)
2
sh
i
Given a × (b × c) =
ed
uc
⇒ (a, b) = 90°
1
(b + c)
2
ak
⇒ (a ⋅ c)b − (a ⋅ b)c =
Taking cross product with b :
w
.s
(a ⋅ c)(b × b) − (a ⋅ b)(b × c)
=
1
(b × b + b × c)
2
1
(b × c)
2
w
w
⇒ −(a ⋅ b)(b × c) =
1 

⇒ (b × c)  a ⋅ b +
=0
2

⇒ a ⋅b = −
1
(∵ c × b ≠ 0)
2
⇒ 1 ⋅1cos θ = −
1
⇒ θ = 135°
2
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46. (1)
Given | a |= 1,| b |= 1 and | c |= 2
Also given a × (a × c) + b = 0
⇒ (a ⋅ c)a − (a ⋅ a)c + b = 0
⇒ (a ⋅ c)a − c = − b
co
m
…(1)
⇒ a × (a × c) = − b
⇒| a × (a × c) |=| − b |= 1
⇒| a | ⋅ | c | sin θ = 1
1
⇒ θ = 30°
2
ed
uc
⇒ sin θ =
n.
π
=1
2
at
io
But (a, a × c) = π / 2 | a | ⋅ | a × c | sin
⇒ (a ⋅ c)a − c = − b [From (1)]
⇒ {(a ⋅ c)a − c} ⋅{(a ⋅ c)a − c} = b ⋅ b
sh
i
⇒ (a ⋅ c) 2 (a ⋅ c) + c ⋅ c − 2(a ⋅ c)(a ⋅ c) = 1
⇒ (a ⋅ c)2 + 4 − 2(a ⋅ c)2 = 1
ak
⇒ (a ⋅ c)2 = 3 ⇒ a ⋅ c = 3
⇒| a | ⋅ | c | cos(a, c) = 3
w
.s
3
π
= cos
1.2
6
⇒ (a, c) = 30°
w
⇒ cos(a, c) =
w
47. (2)
Given | p | = | q | = | r | = λ (say) and
p ⋅ q = 0, p ⋅ r = 0, q ⋅ r = 0
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p × {(x − q) × p} + q × {(x − r ) × q} + r × {(x − p) × r} = 0
⇒ (p ⋅ p)(x − q) − {p ⋅ (x − q)} ⋅ p + (q ⋅ q)(x − r ) − {q ⋅ (x − r )} + ( r ⋅ r )(x − p) − { r ⋅ (x − p)} ⋅ r = 0
⇒ λ 2 (x − q + x − r + x − p) − (p ⋅ x)p + (p ⋅ q)p − (q ⋅ x)q + (q ⋅ r )q − ( r ⋅ x) r + ( r ⋅ p) r = 0
⇒ λ 2 {3x − (p + q + r )} − [(p ⋅ x)p + (q ⋅ x)q + ( r ⋅ x) r ] = 0
1
2
co
m
Clearly this is satisfied by x = (p + q + r )
48. (3)
n.
Given a(α × β) + b(β × γ ) + c( γ × α) = O
a[α β γ ] = 0, b[α β γ ] = 0, c[α β γ ] = 0
at
io
Taking dot product with α, β, γ respectively
Hence α, β, γ are coplanar.
49. (3)
sh
i
Given (b × c) × (c × a) = 3c
ed
uc
Also given that at least one of a, b, c is non-zero.
⇒ {(b × c) ⋅ a} c − {(b × c) ⋅ c} a = 3c
ak
⇒ [b c a ]c − [b c c]a = 3c
⇒ [b c a ]c = 3c
w
.s
{∴[b c c] = 0} ⇒ [b c a ] = 3
w
[b × c, c × a, a × b] = [a b c]2 = [b c a ]2 = 9
w
50. (4)
Given c is ⊥ to a and b
⇒ c is parallel to a × b
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i
j
k
a×b = 2
3
−1
1 −2
3
= i (9 − 2) − j(6 + 1) + k(−4 − 3)
= 7 i − 7 j − 7k = 7( i − j − k)
co
m
∴ c is parallel to i − j − k .
Let c = λ( i − j − k)
n.
Also given c ⋅ (2 i − j + k) = 6
at
io
⇒ λ( i − j − k) ⋅ (2 i − j + k) = 6
⇒ λ(2 + 1 − 1) = 6 ⇒ 2λ = 6
Hence c = 3( i − j − k)
Now [a b c] = (a × b) ⋅ c
= 7( i − j − k) ⋅ 3( i − j − k)
sh
i
= 21(1 + 1 + 1) = 63
ed
uc
∴λ = 3
51. (3)
ak
We know that b, c and b × c are mutually ⊥ vectors.
w
.s
∴ Any vector a can be expressed in terms of b, c , b × c
⇒ a = xb + yc + z(b × c)
…(1)
w
Taking dot product on (1) with b × c , we get
w
a ⋅ (b × c) = x{b ⋅ (b × c)} + y{c ⋅ (b × c)} + z(b × c)2
= x(0) + y(0) + z(b × c)2 = z(b × c)2
⇒z=
a ⋅ (b × c)
| b × c |2
Given | b | = 1,| c | = 1
Again taking dot product on (1) with b and c
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⇒ a ⋅ b = x(b ⋅ b) + y(c ⋅ b) + z(b × c) ⋅ c
= x(1) + y(0) + z[b c c] = x + y(0) + z(0) = x
Also a ⋅ c = x(b ⋅ c) + y(c ⋅ c) + z(b × c) ⋅ c
= x(0) + y(1) + z[b c c] = 0 + y + 0 = y
a ⋅ (b × c)
.
| b × c |2
co
m
∴ a = (a ⋅ b)b + (a ⋅ c)b +
52. (3)O, A, B, C are coplanar since [a b c] = 0
Hence origin O is the circumcentre.
a+b+c
3
ed
uc
P.V. op G i.e., centroid =
at
io
(∵| a |2 =| b |2 =| c |2 = 9 given)
n.
OA = OB = OC ⇒| a | = | b | = | c |= 3
3(a × b × c)
− 2(O)
3
∴ P.V.of H =
= a +b+c
3− 2
ak
sh
i
We know that orthocenter H divides GO in the ratio 3 : 2 externally
53. (3)
w
.s
G.E. = [3(a × b) + 2(a × c), b × c − 2(b × a), 2(c × a) − 6(c × b)]
Let a × b = p, b × c = q, c × a = r
w
∴ G.E. = [3p − 2 r , q + 2p, 2 r + 6q]
w
3 0 −2
= 2 1 0 [p q r ]
0 6
2
= [3(2 − 0) − 0 − 2(12 − 0)][p q r ]
= −18[p q r ] = −18[a × b, b × c, c × a]
= −18[a b c]2
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