Practice question on Vector for NDA , IIT-JEE , and Air force | NDA | IITJEE | CBSE | SAT | UPSC 16-30 (SET 2)

Vector

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Question no. 16

If a , b and c are three coplanar vector , then

looks_one a.(b ✕ c)=1

looks_two a.(b ✕ c)= 3

looks_3 c.(a ✕ b)=0

looks_4 b.(c ✕ a)=1

Option looks_3 c.(a ✕ b)=0

Solution :
If a , b and c are three coplanar vector
c.(a ✕ b)=0

Question no. 17

Given a= i+ j -k and b= -i+ 2j +k , c= -i+ 2j -k. A unit vector prependicular to both a +b and c +b

looks_one i

looks_two j

looks_3 k

looks_4 (i +j +k)/√3

Option looks_3 k

Solution :
a= i+ j -k
b= -i+ 2j +k
c= -i+ 2j -k
a +b = 3j
c +b = -2i + 4j
A unit vector prependicular to both a +b and c +b is cross product of both vector.

Unit Vector = k

Question no. 18

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

looks_one a is parallel to b

looks_two a is prependicular to b

looks_3 either a or b is a null vector

looks_4 none of these

Option looks_3 either a or b is a null vector

Solution :
a ✕ b=0 and a.b=0 , it means that they are neither parallel or prependicular
it means either a or b is a null vector

Question no. 18

The system of vectors a= i - j + k, b= i + 2j - k, c= 3i + pj + 5k are coplanar then the value of p will be is

looks_one -6

looks_two -2

looks_3 2

looks_4 6

Option looks_one -6

Solution :
a= i - j + k
b= i + 2j - k
c= 3i + pj + 5k
a.(b ✕ c ) = 0

Question no. 20

If a = (3i +k)/√10 , b = (2i + 3j -6k)/7 then the value of (2a -b).[(a ✕ b) ✕ (a +2b)] is

looks_one 5

looks_two -5

looks_3 -3

looks_4 3

Option looks_two -5

Solution :
(2a -b).[(a ✕ b)✕ (a +2b)]
(2a -b).[(a ✕ b)✕a) + 2(a ✕ b)✕b ]
-(2a -b).[a✕(a ✕ b) + 2b✕(a ✕ b)]
(2a -b)[(a.b)a - |a|²b + 2a]
(2a -b)(b - 2a)
- |b|² - 4|a|² = -1 -4 = -5

Question no. 21

If a= i -j +2k , b=2i +4j +k , c=xi +j +yk are mutually perpendicular , then (x,y) =

looks_one 2,-3

looks_two -2,3

looks_3 3,-2

looks_4 -3,2

Option looks_4 -3,2

Solution :
a= i -j +2k , b=2i +4j +k , c=xi +j +yk are mutually perpendicular
a.b = 0
b.c = 0
a.c = 0
a.c = x -1 + 2y = 0 ⇒ x + 2y = 1 ......[1]
b.c = 2x+ y + 4 = 0 ⇒ 2x+ y = -4 ........[2]
Solve [1] and [2]
x = -3
y = 2

Question no. 22

Let a = j -k , c= i -j -k. then the vector b satisfying a ✕ b + c=0 and a.b=3

looks_one 2i -j +2k

looks_two i -j -2k

looks_3 i +j -2k

looks_4 -i +j -2k

Option looks_4 -i +j -2k

Solution :
a = j -k
c= i -j -k
b = b₁i + b₂j + b₃k
a.b = 3
b₂ - b₃ = 3
satisfied By Option 3 and 4

Question no. 23

For Any vector r, the value of(r✕i)² + (r✕j)² + (r✕k)² =

looks_one 3r²

looks_two 4r²

looks_3 r²

looks_4 2r²

Option looks_4 2r²

Solution :
r = r₁i + r₂j + r₃k
(r✕i)² = r₃² + r₂²
(r✕j)² = r₁² + r₃²
(r✕k)² = r₁² + r₂²
(r✕i)² + (r✕j)² + (r✕k)² = 2(r₁² + r₂² + r₃² ) = 2r²

Question no. 24

If a = 2i +j -2k , b = i+ j. Let c be a vector such that |c- a| =3 , |(a ✕ b)✕c|=3 , the angle between c and a ✕ b be 30^o , then a.c is

looks_one 1/8

looks_two 25/8

looks_3 2

looks_4 5

Option looks_3 2

Solution :
a = 2i +j -2k
b = i+ j
|(a ✕ b)✕c|=3
(a ✕ b) = 2i -2j + k
|c- a|² = |c|² + |a|² - 2c.a
9 = 4 + 9 - 2c.a
-4 = -2c.a
a.c = 2

Question no. 25

The three non zero vectors a , b and c are related by a=8b and c = -7b , then the angle between a and c is

looks_one 180^o

looks_two 90^o

looks_3 45^o

looks_4 0^o

Option looks_one 180^o

Solution :
a=8b
c = -7b
a = -(8/7) c
a and c are opposite in direction

Question no. 26

If a , b ,c are non coplanar vectors and λ is a real number, then [λ(a + b)    λ²b     λc] = [a     b +c    b] for

looks_one Exactly two values of λ

looks_two Exactly three values of λ

looks_3 No values of λ

looks_4 Exactly one values of λ

Option looks_3 No values of λ

Solution :
a , b ,c are non coplanar vectors
[λ(a + b)    λ²b     λc] = [a     b +c    b]
λ(a + b).λ³(b✕c) = a.(b + c)✕b
λ⁴(a.(b✕c)) = a.( c✕b)
λ⁴ = -1

Question no. 27

Let a= i -k, b= xi +j +(1-x)k and c= yi + xj + (1-x-y)k , then [a b c] depends on

looks_one neither x nor y

looks_two both x and y

looks_3 only x

looks_4 only y

Option looks_3 only x

Solution :
a= i -k
b= xi +j +(1-x)k
c= yi + xj + (1-x-y)k
[a b c]

1(1-x -y -x(1-x)) -1(x² - y)
1 -x -y - x + x² - x² + y = 1 - 2x

Question no. 28

If C is the mid point of AB and P is any point outside AB, then

looks_one PA +PB +PC = 0

looks_two PA +PB +2PC = 0

looks_3 PA +PB = PC

looks_4 PA +PB =2PC

Option looks_4 PA +PB =2PC

Solution :

PC = PB + BC ......[1]
PC = PA + AC ......[2]
Add [1] and [2]
2PC = PB + PA + BC + AC
BC = -AC
2PC = PA +PB

Question no. 29

Let a= i + j + k , b=i -j + 2k and c =xi + (x-2)j - k , if the vector c lies in the plane of a and b , then x=

looks_one 0

looks_two -4

looks_3 -2

looks_4 1

Option looks_3 -2

Solution :
a, b and c are co planar
a.(b ✕ c) = 0

1(1 - 2x + 4) -1(-1 - 2x) + 1 (x-2 +x) = 0
5 - 2x + 1 + 2x + 2x - 2 = 0
2x = -4
x = -2

Question no. 30

If [ a   b   a ✕ b ] =

looks_one | a ✕ b |

looks_two | a ✕ b |²

looks_3 0

looks_4 none of these

Option looks_two | a ✕ b |²

Solution :
[ a   b   a ✕ b ] = a.(b✕(a ✕ b))
= a.((b.b)a - (a.b)b)
= a.(|b|²a -(a.b)b)
= |b|²|a|² - (a.b)²
= | a ✕ b |²

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