Practice question on Vector for IIT-JEE , NDA and Airforce (Set 1)

Vector

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Question no. 1

For any vector a and b , (a ✕ b)²

looks_one a² - b²

looks_two a² + b²

looks_3 a²b² -(a.b)²

looks_4 none of these

Option looks_3 a²b² -(a.b)²

Solution :
(a ✕ b) = |a||b|sinθn
(a ✕ b)² = |a|²|b|²sin²θ
= |a|²|b|²( 1 - cos²θ)
= |a|²|b|² - |a|²|b|²cos²θ
= |a|²|b|² -(a.b)²

Question no. 2

A unit vector prependicular to 3i+ 2j -k and 12i+ 5j -5k , is

looks_one (5i- 3j -9k)/√115

looks_two (5i+ 3j -9k)/√115

looks_3 (-5i+ 3j -9k)/√115

looks_4 (5i+ 3j +9k)/√115

option looks_3 (-5i+ 3j -9k)/√115

Solution :

Question no. 3

For any vector a and b ,if a ✕ b=0 , then

looks_one a=0

looks_two b=0

looks_3 not parallel

looks_4 none of these

option looks_4 none of these

They (non-zero) vector must be parallel

Question no. 4

The system of vectors i , j, k is

looks_one orthogonal

looks_two coplanar

looks_3 collinear

looks_4 none of these

option looks_one orthogonal

i , j, k are mutually prependicular to each other , therefore they are orthogonal.

Question no. 5

If | a |= 3 , | b |= 4 and | a + b |=5 , then | a - b | =

looks_one 6

looks_two 5

looks_3 4

looks_4 3

option looks_two 5

Solution :
| a + b |² = |a|² + |b|² + 2|a||b|cosθ
25 = 9 + 16 + 2|a||b|cosθ ⇒ a.b = 0
| a - b | = |a|² + |b|² - 2|a||b|cosθ = 16 + 9 + 0 = 25
| a - b | = 5

Question no. 6

If a , b , c are mutually perpendicular unit vectors , then |a + b + c| =

looks_one 3

looks_two 0

looks_3 1

looks_4 √3

option looks_4 √3

Solution :
a , b , c are mutually perpendicular unit vectors
a.b = b.c = a.c = 0
|a + b + c|² = |a|² + |b|² + |c|² + 2a.b + c.a + b.c
|a + b + c|² = 1+ 1 + 1
|a + b + c| = √3

Question no. 7

If | a |= 3 , | b |= 1 , | c |= 4 and a + b + c=0, then a.b + c.a + b.c

looks_one -13

looks_two 13

looks_3 10

looks_4 -10

option looks_one -13

Solution :
| a |= 3
| b |= 1
| c |= 4
a + b + c = 0
(a + b + c)² = |a|² + |b|² + |c|² + 2a.b + c.a + b.c = 0
9 + 1 + 16 + 2(a.b + c.a + b.c) = 0
a.b + c.a + b.c = -26/2 = -13

Question no. 8

(r.i)² + (r.j)² + (r.k)² =

looks_one 3r²

looks_two r²

looks_3 0

looks_4 none of these

option looks_two r²

Solution :
r = xi + yj + zk
r.i = x
r.j = y
r.k = z
(r.i)² + (r.j)² + (r.k)² = x² + y² + z² = r²

Question no. 9

If | a |= 3 , | b |= 4 and the angle between a and b be 120^o , then |4a + 3b|

looks_one 25

looks_two 12

looks_3 7

looks_4 13

option looks_two 12

Solution :
|4a + 3b|² = 16|a|² + 9| b |² +24a.b
=16|a|² + 9| b |² +24|a||b|cos120^o
= 16.9 + 9.16 + 24.3.4.cos120^o
= 288 - 144 = 144
|4a + 3b|² = 144
|4a + 3b| = 12

Question no. 10

If θ be the angle between a and b and |a ✕ b| = a . b then, θ=

looks_one 180^o

looks_two 90^o

looks_3 45^o

looks_4 0^o

option looks_3 45^o

Solution :
|a ✕ b| = a . b
|a||b|sinθ = |a||b|cosθ
sinθ = cosθ
θ = 45^o

Question no. 11

If |a| + |b|= |c| and a + b = c , then the angle between a and b is

looks_one 90^o

looks_two 180^o

looks_3 0^o

looks_4 none of these

option looks_3 0^o

Solution :
( |a| + |b| )²= |c|²
|a|² + |b|² + 2|a||b| = |c|²
a + b = c
|a|² + |b|² + 2a.b = |c|²
2|a||b| = 2a.b
= 2|a||b|cosθ
cosθ = 1
θ = 0^o

Question no. 12

Let α , β , γ be distinct real numbers. The points with position vectors αi +βj +γk , γi +αj +βk , βi +γj +αk

looks_one collinear

looks_two form an equilateral triangle

looks_3 form an scalene triangle

looks_4 form an right angled triangle

option looks_two

Solution :
OA = αi +βj +γk
OB = γi +αj +βk
OC = βi +γj +αk
Side (AB) = OB - OA = (γ - α)i + (α - β)j +(β - γ )k
Side (BC) = OC - OB = (β- γ )i + (γ - α )j + (α - β)k
Side (AC) = OC - OA = (β - α )i + (γ - β )j +( α - γ )k
|AB|= |AC| = |BC|
Equilateral triangle

Question no. 13

Volume of parallelopiped whose edges are 2i - 3j + 4k , i + 2j - 2k and 3i - j + k

looks_one 5 cubic unit

looks_two 6 cubic unit

looks_3 7 cubic unit

looks_4 8 cubic unit

option looks_3 7 cubic unit

Solution :

Question no. 14

If vectors 2i - 3j + 4k , i + 2j - k and xi - j + 2k are colpanar , then x=

looks_one 8/5

looks_two 5/8

looks_3 0

looks_4 1

option looks_one 8/5

solution :

Question no. 15

If a.i = 4, then (a ✕ j).(2j - 3k)=

looks_one 12

looks_two 2

looks_3 0

looks_4 -12

option looks_4 -12

Solution :

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