Practice Question on Vector for IIT-JEE , NDA and Airforce 31-45 | SAT | CBSE | UPSC | IITJEE | NDA

Vector

---

Question no. 31

If |a.b|=3 and |a✕b|=4 , then the angle between a and b is

looks_one cos^-1(3/4)

looks_two cos^-1(3/5)

looks_3 cos^-1(4/5)

looks_4 45^o

Option looks_two cos^-1(3/5)

Solution :
|a.b|= |a||b|cosθ = 3 ......[1]
|a✕b| = |a||b|sinθ = 4 ......[2]
Divide [2] by [1]
sinθ/cosθ = 4/3
tanθ = 4/3
θ = tan^-1(4/3)
θ = cos^-1(3/5)

Question no. 32

The Vector a= αi+ 2j +βk lies in the plane of the vectors b= i+ j and c = j +k and bisects the angle between b and c. then the value of α and β

looks_one 2 ,2

looks_two 1,1

looks_3 2,1

looks_4 1,2

option looks_two 1,1

Solution :
a= αi+ 2j +βk
a lies in the plane of the vectors b and c
a.(b ✕ c) = 0

Question no. 33

If [a✕b   b✕c   c✕a] = λ[a  b  c ]² then , λ is

looks_one 0

looks_two 1

looks_3 2

looks_4 3

Option looks_two 1

Solution :
[a✕b   b✕c   c✕a]
( a✕b). ( (b✕c) ✕ (c✕a) )
( a✕b ). [((b✕c).a)c - ((b✕c).c)a ]
( a✕b ). [c (a .(b✕c) - 0 ) ]
( a✕b ).c (a .(b✕c) ) = [a  b  c ]²
λ = 1

Question no. 34

The system of vectors a= 12i + 4j + 3k, b= 8i - 12j - 9k, c= 33i - 4j - 24k represents a cube ,then it's volume is

looks_one 616

looks_two 308

looks_3 154

looks_4 none of these

Option looks_4 none of these

Solution :
a= 12i + 4j + 3k
b= 8i - 12j - 9k
c= 33i - 4j - 24k
Volume of cube = a.(b ✕ c)

7692

Question no. 35

If a = i -j , b = i + j , c= i + 3j + 5k , n be a unit vector such that a.n=0 and a.n=0, then c.n is

looks_one 5

looks_two 1

looks_3 2

looks_4 3

Option looks_one 5

Solution :
a = i -j
b = i + j
c= i + 3j + 5k
n = xi + yj + zk
x² + y² + z² = 1
a.n= x -y = 0
b.n = x + y = 0
x = 0 , y = 0
Therefore , c = 1
c.n = 5

Question no. 36

If a= xi + yj +2k , b=i -j +k , c=xi +2j ; angle between a and b is 90^o, a.c=4 , then , then

looks_one [a b c]² =|a|

looks_two [a b c] =|a|

looks_3 [a b c] =0

looks_4 [a b c] =|a|²

Option looks_4 [a b c] =|a|²

Solution :
a= xi + yj +2k
b=i -j +k
c=xi +2j
a.c = x + 2y = 4 ........[1]
Angle between a and b is 90^o
a.b = 0
x - y + 2 = 0 ........[2]
Solve [1] and [2]
x = 0 , y = 2
a= 2j +2k
|a|² = 8
[ a b c ] = a.(b✕c)

-2(-1) + 2(3) = 8

Question no. 37

If the vector b is collinear with the vector a=(2√2 , -1, 4) and |b|=10 , then

looks_one a ∓ b

looks_two a ∓ 2b

looks_3 2a ∓ b

looks_4 none of these

Option looks_3 2a ∓ b

Solution :
|a| = 5
b = λa
|b| = λ|a|
10 = λ(5)
λ = 2

Question no. 38

For Any vector a ,b, c of length 3,4,5 respectively. Let a be perpendicular to b + c , b be perpendicular to c + a ,c be perpendicular to b + a then the value of |a + b + c | =

looks_one 2√5

looks_two 2√2

looks_3 10√5

looks_4 5√2

Option looks_4 5√2

Solution :
|a| = 3
|b| = 4
|c| = 5
a.(b + c) = 0
b.(c + a) = 0
c.(b + a) = 0
2[a.b + b.c + a.c] = 0
|a + b + c|² = |a|² + |b|² + |c|² + 2(a.b + b.c + a.c)
|a + b + c|² = |3|² + |4|² + |5|² + 0
|a + b + c|² = 50
|a + b + c| = 5√2

Question no. 39

Consider the points A,B,C and D with position vectors 7i +4j +7k , i- 6j +10k , -i -3j +4k , 5i -j +5k respectively. The ABCD is a

looks_one square

looks_two rhombus

looks_3 rectangle

looks_4 none of these

Option looks_4 none of these

Solution :
A = 7i +4j +7k
B = i- 6j +10k
C = -i -3j +4k
D = 5i -j +5k
AB = B - A = -6i - 10j + 3k
BC = -2i + 3j - 6k
CD = 6i + 2j + k
AD = -2i - 5j - 2k

Question no. 40

The value of a , for which A, B, C with positional vectors 2i - j - k , i - 3j - 5k , ai -3j - k respectively are the vertices of right angled triangle with C=90^o

looks_one -1,-2

looks_two -1,2

looks_3 2,1

looks_4 -2,1

Option looks_3 2,1

Solution :
A = 2i - j - k
B = i - 3j - 5k
C = ai -3j - k
AB = -i + 2j + 4k
BC = (1-a)i - 4k
AC = (2-a)i + 2j
According to Pythagoras Theorem
|AB|² = |BC|² + |AC|²
21 = (1-a)² + 16 + (2-a)² + 16
a² - 3a - 2 = 0
(a-2)(a-1) = 0
a = 2 , 1

Question no. 41

If |a|=5, | a-b |=8 and | a + b |=10 , then |b| is equal to

looks_one1

looks_two 3

looks_3 √57

looks_4 none of these

Option looks_3 √57

Solution :
| a-b |² = |a|² + |b|² - 2a.b ..........[1]
| a + b |² = |a|² + |b|² + 2a.b .........[2]
Add [1] and [2]
| a-b |² + | a + b |² = 2|a|² + 2|b|²
64 + 100 = 50 + 2|b|²
|b|² = 57
|b| = √57

Question no. 42

Angle between diagonals of a parallelogram whose side are represented by a= 2i + j + k, and b= i - j +k

looks_one cos^-1(1/3)

looks_two cos^-1(1/2)

looks_3 cos^-1(4/9)

looks_4 cos^-1(5/9)

Option looks_one cos^-1(1/3)

Solution :
a= 2i + j + k
b= i - j +k
Diagonals :
a + b = 3i
a - b = i + 2j + 2k
(a + b).(a - b) = |a + b||a - b| cosθ
3 = 3 *3 cosθ
θ = cos^-1(1/3)

Question no. 43

If u and v are unit vector and α is the acute angle between them , then 2u ✕ 3v is a unit vector for

looks_one exactly two values of α

looks_two more than two values of α

looks_3 no values of α

looks_4 exactly one value of α

Option looks_4 exactly one value of α

Solution :
|u| = 1
|v| = 1
|2u ✕ 3v| = |2u||3v|sinα
Scaling the vector does not change the angle between them
therefore , α is the angle between them
exactly one value of α for 2u ✕ 3v

Question no. 44

Vector of length 3 unit which is perpendicular to i + j + k ,and lies in plane of i + j + k and 2i -3j

looks_one (3/√6)(i - 2j + k)

looks_two (3/√6)(2i - j - k)

looks_3(3/√114)(8i - 7j - k)

looks_4 (3/√114)(-7i + 8j - k)

Option looks_4 (3/√114)(-7i + 8j - k)

Solution :
A = xi + yj + zk
B = i + j + k
A and B are prependicular to each other
A.B = x + y+ z = 0
C = i + j + k
D = 2i -3j
A , C , D are lie in plane
A.(C ✕ D) = 3x - 2y -5z = 0
|A|² = x² + y² + z² = 9
-7i + 8j - k satisfy the equation
Unit Vector = (3/√114)(-7i + 8j - k)

Question no. 45

If a ,b,c be the unit vectors such that b is not parallel to c and a✕(2b✕c) =b, then the angle that a makes with b and c are

looks_one π/3, π/4

looks_two π/3 ,2π/3

looks_3 π/2 ,2π/3

looks_4 π/2 ,π/3

Option looks_4 π/2 ,π/3

Solution :
|a| =|b| = |c| = 1
a✕(2b✕c) =b
b.c ≠ 0
(a.c)2b - 2(a.b)c = b
a.b = 0 ⇒ θ = π/2
(a.c)2b = b
a.c = 1/2 ⇒ θ = π/3

Previous Next