Vector or Cross Product MCQ Quiz - Objective Question with Answer for Vector or Cross Product - Download Free PDF

Last updated on Jun 14, 2025

Latest Vector or Cross Product MCQ Objective Questions

Vector or Cross Product Question 1:

The position vectors of three points A, B and C respectively, where  a,b and c  respectively, where c=(cos2θ)a+(sin2θ)b. What is (a×b)+(b×c)+(c×a) equal to?

  1. 0
  2. 2c
  3. 3c
  4. Unit vector

Answer (Detailed Solution Below)

Option 1 : 0

Vector or Cross Product Question 1 Detailed Solution

Calculation:

Given,

The position vectors of points A, B, and C are a, b, and c respectively, and c=cos2θa+sin2θb.

The expression to evaluate is: (a×b)+(b×c)+(c×a).

First, substitute c into the equation:

(a×b)+(b×(cos2θa+sin2θb))+(cos2θa+sin2θb)×a.

Using the distributive property of the cross product:

(a×b)+[(b×cos2θa)+(b×sin2θb)]+[(cos2θa×a)+(sin2θb×a)].

Since b×b=0 and a×a=0, we are left with:

(a×b)+cos2θ(b×a)+sin2θ(a×b).

Substitute b×a=(a×b) into the expression:

(a×b)+cos2θ(a×b)+sin2θ(a×b).

Factor out a×b:

a×b[1cos2θsin2θ].

Since cos2θ+sin2θ=1, the expression becomes:

a×b[11]=0.

∴ The final result is 0.

Hence, the correct answer is option 1. 

Vector or Cross Product Question 2:

What is 3α + 2β equal to if (2î + 6ĵ + 27k̂) × (î + αĵ + βk̂) is a null vector?

  1. 36
  2. 33
  3. 30
  4. 27

Answer (Detailed Solution Below)

Option 1 : 36

Vector or Cross Product Question 2 Detailed Solution

Explanation:

Given:

⇒ (2î + 6ĵ + 27k̂) × (î + αĵ + βk̂) = 0

⇒ [i^j^k^ 2627 1αβ]  = 0

⇒ i^(6β27α)j^(2β27)+k^(2α6)=0

Comparing both sides, we get

6β  – 27α  = 0

⇒ 2β  = 9α 

2β - 27 = 0

⇒ β = 27/2

Also

⇒ 2α – 6 = 0 

α =3

Now, 

3α  + 2β = 3×3+2×272 =36

∴ Option (1) is correct

Vector or Cross Product Question 3:

The unit vector perpendicular to each of the vectors a+b and ab, where a=i^+j^+k^ and b=i^+2j^+3k^, is :

  1. 16i^+26j^+16k^
  2. 16i^+16j^16k^
  3. 16i^+26j^+26k^
  4. 16i^+26j^16k^

Answer (Detailed Solution Below)

Option 4 : 16i^+26j^16k^

Vector or Cross Product Question 3 Detailed Solution

Concept:

Vector Perpendicular to Both Vectors:

  • We are given two vectors: a + b and a - b, and we need to find a vector perpendicular to both of them.
  • The method to find a vector perpendicular to both given vectors is by taking their cross product. The result will be a vector perpendicular to both.
  • Once the cross product is found, we normalize the result (i.e., divide it by its magnitude) to find the unit vector perpendicular to both vectors.

 

Calculation:

Given vectors:

a = i + j + k

b = i + 2j + 3k

The vectors we need to take the cross product of are:

a + b = (i + j + k) + (i + 2j + 3k) = 2i + 3j + 4k

a - b = (i + j + k) - (i + 2j + 3k) = 0i - j - 2k

Now, we compute the cross product of (a + b) and (a - b):

(a + b) × (a - b) =

|i^j^k^234012|

Expanding the determinant:

Result of cross product:

(a + b) × (a - b) = -2i + 4j - 2k

Now, let's find the magnitude of the resulting vector:

Magnitude = √((-2)² + 4² + (-2)²) = √(4 + 16 + 4) = √24 = 2√6

To find the unit vector, we divide the result by its magnitude:

Unit vector = (-2i + 4j - 2k) / 2√6

The unit vector is:

Unit vector = (-1/√6)i + (2/√6)j - (1/√6)k

Hence Option 4 is the correct answer. 

Vector or Cross Product Question 4:

If θ is the angle between the vectors 4ij+2k and i+3j2k, then sin2θ=

  1. 395
  2. 395
  3. 28549
  4. 25849

Answer (Detailed Solution Below)

Option 3 : 28549

Vector or Cross Product Question 4 Detailed Solution

Formula Used:

 

Dot product of two vectors: ab=|a||b|cosθ, where θ is the angle between the vectors.

Magnitude of a vector: |a|=a12+a22+a32

Sine of double angle: sin2θ=2sinθcosθ

Relation between sine and cosine: sin2θ+cos2θ=1

Calculation:

Given:

Vector 1: a=4i^j^+2k^

Vector 2: b=i^+3j^2k^

ab=(4)(1)+(1)(3)+(2)(2)=434=3

|a|=42+(1)2+22=16+1+4=21

|b|=12+32+(2)2=1+9+4=14

cosθ=ab|a||b|=32114=33727=376

sin2θ=1cos2θ=19496=1398=9598

sinθ=9598=9572

sin2θ=2sinθcosθ=29572376=6954912=6954923=395493

sin2θ=3953493=28549

∴ The value of sin2θ is 28549

Hence option 3 is correct

Vector or Cross Product Question 5:

A vector of magnitude 2 units along the internal bisector of the angle between the vectors 2i2j+kandi+2j+2k is

  1. j+k
  2. ij
  3. ik
  4. i+k

Answer (Detailed Solution Below)

Option 4 : i+k

Vector or Cross Product Question 5 Detailed Solution

Formula Used:

Unit vector along the internal bisector of two vectors a and b is given by:

c^=a|a|+b|b|

Resultant vector R=|R|c^ where c^ is the unit vector along the bisector and |R| is the magnitude of the resultant vector.

Calculation:

Given:

Vector 1: a=2i^2j^+k^

Vector 2: b=i^+2j^+2k^

Magnitude of resultant vector: 2

|a|=22+(2)2+12=4+4+1=9=3

|b|=12+22+22=1+4+4=9=3

⇒ Unit vector along bisector: c^=2i^2j^+k^3+i^+2j^+2k^3

c^=3i^+3k^3=i^+k^

 

Hence option 4 is correct

Top Vector or Cross Product MCQ Objective Questions

Find the value of a×a

  1. 1
  2. 0
  3. |a|
  4. |a|2

Answer (Detailed Solution Below)

Option 2 : 0

Vector or Cross Product Question 6 Detailed Solution

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Concept:

Dot product of two vectors is defined as:

A.B=|A|×|B|×cosθ

Cross/Vector product of two vectors is defined as:

A×B=|A|×|B|×sinθ×n^

where θ is the angle between AandB

Calculation:

To Find: Value of a×a

Here angle between them is 0°

a×a=|a|×|a|×sin0×n^=0

If u=i^×(a×i^)+j^×(a×j^)+k^×(a×k^) then u is equal to

  1. 0
  2. a
  3. 2a
  4. 3a

Answer (Detailed Solution Below)

Option 3 : 2a

Vector or Cross Product Question 7 Detailed Solution

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Given: 

u=i^×(a×i^)+j^×(a×j^)+k^×(a×k^)

Concept:

î × î  = ĵ × ĵ = k̂ × k̂  = 0 

î × ĵ = k̂ , ĵ × k̂ = î , k̂ × î = ĵ 

Calculation:

Let a = mî + nĵ +lk̂ 

According to the Question 

u=i^×(a×i^)+j^×(a×j^)+k^×(a×k^)

u = î  × (mî + nĵ +lk̂  × î) + ĵ ×  (mî + nĵ +lk̂  × ĵ) + k̂ × (mî + nĵ +lk̂  × k̂)

u =  î  × (-nk̂ + lĵ) + ĵ × (mk̂ -lî  ) + k̂ × (-mĵ + nî) 

u = nĵ  + lk̂ + mî +  lk̂ + mî + nĵ 

u = 2(mî + nĵ +lk̂ ) = 2a

∴ The correct option is 3

If |a|=3,|b|=4andab=6, then find the value of |a×b|

  1. √3
  2. 8√3 
  3. 6√3 
  4. 4√3 

Answer (Detailed Solution Below)

Option 3 : 6√3 

Vector or Cross Product Question 8 Detailed Solution

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Concept:

Let aandb are two vectors

ab=|a|×|b|×cosθ

 a×b=|a|×|b|×sinθ×n^,wheren^ is the unit vector perpendicular to both aandb

 

Calculation:

Given: |a|=3,|b|=4andab=6

As we know, ab=|a|×|b|×cosθ

⇒ 6 = 3 × 4 × cos θ 

⇒ cos θ = 612=12

∴ θ = 60° 

As we know that, If aandb are two vectors, then

a×b=|a|×|b|×sinθ×n^

|a×b|=|a|×|b|×|sinθ|×|n^|=|a|×|b|×sinθ      (∵ Magnitude of a unit vector is one)

|a×b| = 3 × 4 × sin 60° 

|a×b|=3×4×32=63

What is (2a3b)×(2a+3b)equal to?

  1. 0
  2. a×b
  3. 12(a×b)
  4. 4|a|29|b|2

Answer (Detailed Solution Below)

Option 3 : 12(a×b)

Vector or Cross Product Question 9 Detailed Solution

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Concept:  

a×a=0

a×b=b×a

Calculation:

Given

=(2a3b)×(2a+3b)

=2a×2a+2a×3b3b×2a3b×3b

=0+2a×3b3b×2a0

=6(a×b)+6(a×b)

=12(a×b)

 

Additional Information

Properties of Scalar Product

a.a=|a|2

a.b=b.a (Scalar product is commutative)

a.0=0

a.(b+c)=a.b+a.c (Distributive of scalar product over addition)

In terms of orthogonal coordinates for mutually perpendicular vectors, it is seen that i.i=j.j=k.k=1

Properties of Vector Product

a×a=0

a×b=b×a (non-commutative)

 a×(b+c)=a×b+a×c (Distributive of vector product over addition)

i×i=j×j=k×k=0

i×j=k,j×k=i,k×i=j

What is the vector perpendicular to both the vectors î - ĵ and î ?

  1. -ĵ

Answer (Detailed Solution Below)

Option 4 : k̂

Vector or Cross Product Question 10 Detailed Solution

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Concept:

Let a and b be the two vectors and the vector c perpendicular to both a and b

Hence  c=a×b

Calculation:

Let vector c is perpendicular to both the vectors î - ĵ and î

Therefore, c = (î - ĵ) × î

= (î × î) -  (ĵ × î)

= 0 - (-k̂)

= k̂

If (a×b)×c=a×(b×c), then which of the following is true?

  1. b×(c×a)=0
  2. a=b=c=0
  3. a+b+c=0
  4. None of these.

Answer (Detailed Solution Below)

Option 1 : b×(c×a)=0

Vector or Cross Product Question 11 Detailed Solution

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Concept:

For three vectors AB and C:

  • Triple Cross Product: is defined as: A×(B×C)=(A.C)B(A.B)C

Calculation:

Given that: (a×b)×c=a×(b×c)

⇒ c×(a×b)=a×(b×c)

⇒ (c.b)a+(c.a)b=(a.c)b(a.b)c

⇒ (a.b)c(c.b)a=0

⇒ b×(c×a)=0, which is the required answer.

Find sin θ if theta is the angle between the vectors a=3j^+4k^andb=6i^+8k^

  1. 1/2
  2. 1/12
  3. 34125
  4. 33125

Answer (Detailed Solution Below)

Option 3 : 34125

Vector or Cross Product Question 12 Detailed Solution

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Concept:

Cross product of vectors:

  •  |a×b|=|a||b|sinθ
  • If a=a1i^+a2j^+a3k^andb=b1i^+b2j^+b3k^ then a×b=|i^j^k^a1a2a3b1b2b3|
Calculation:

Given: a=3j^+4k^andb=6i^+8k^

⇒ a×b=|i^j^k^034608|

⇒ a×b=i^(240)j^(024)+k^(018)=24i^+24j^18k^

⇒ |a×b|=242+242+182=1476=641

|a|=5 and |b|=10

As we know that, |a×b|=|a||b|sinθ

⇒ sinθ=|a×b||a||b|=64150=34125

Hence, the correct option is 3.

A vector is perpendicular to both the vectors i^+j^ and j^+k^ is

  1. î - 2ĵ + k̂
  2. î + ĵ + k̂
  3. î - ĵ + k̂
  4. None of these

Answer (Detailed Solution Below)

Option 3 : î - ĵ + k̂

Vector or Cross Product Question 13 Detailed Solution

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Concept:

Let a and b be the two vectors and the vector c perpendicular to both a and b

Hence  c=a×b

 

Calculation:

Let vector c is perpendicular to both the vectors î + ĵ and ĵ + k̂

Let a = î + ĵ and b = ĵ + k̂

Therefore, c=a×b

= (î + ĵ) × (ĵ + k̂)

= î × ĵ + î × k̂ + ĵ × ĵ + ĵ × k̂ 

= k̂ - ĵ + 0 + î 

= î - ĵ + k̂  

If |a×b|=|ab| ,then angle between aandb is

  1. 90° 
  2. 60° 
  3. 45°
  4. 30° 

Answer (Detailed Solution Below)

Option 3 : 45°

Vector or Cross Product Question 14 Detailed Solution

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Concept:

 

Let a and b be the two vectors,

Dot product of two vectors  is given by:  a.b=|a||b|cosθ

Cross product of two vectors  is given by: a×b=|a||b|sinθn^, Where n^ is a unit vector

 

Calculation:

Given: |a×b|=|ab|

To Find: Angle between aandb

|a×b|=|ab|

|a||b|sinθ|n^|=|a||b|cosθ

⇒ sin θ = cos θ                       (∵ |n^| = 1)

⇒ tan θ = 1

∴ θ = 45° 

What is (a+b)×(ab)equal to?

  1. 0
  2. 2(a×b)
  3. 2(a×b)
  4. |a|2|b|2

Answer (Detailed Solution Below)

Option 2 : 2(a×b)

Vector or Cross Product Question 15 Detailed Solution

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Concept:

Properties of vectors:

If a and b are two vectors parallel to each other then  a×b=0

Cross product of parallel vectors are zero ⇔  a×a=0,b×b=0andc×c=0

A cross or vector product is not commutative ⇔ a×b=b×a


Calculation:

We have to find the value of (a+b)×(ab)

(a+b)×(ab)=a×aa×b+b×ab×b

We know that a×b=b×a

(a+b)×(ab)=0a×ba×b0                (a×a=b×b=0) 

(a+b)×(ab)=2(a×b)

∴ Option 2 is correct.

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