Scalar and Vector Product MCQ Quiz - Objective Question with Answer for Scalar and Vector Product - Download Free PDF

Calculation:

Given,

The position vectors of points A, B, and C are , , and respectively, and .

The expression to evaluate is: .

First, substitute into the equation:

.

Using the distributive property of the cross product:

.

Since and , we are left with:

.

Substitute into the expression:

.

Factor out

.

Since , the expression becomes:

.

∴ The final result is .

Hence, the correct answer is option 1. 

Calculation:

Given,

Using the identity , we substitute:

Simplifying the equation:

Rearrange to isolate the sine terms:

Now, calculate the dot product:

∴ The value of is 2.

Hence, the correct answer is Option 4.

Calculation: 

Taking the cross product with 

Hence, the correct answer is Option 2.

Answer (3)

Sol.

dot with 

= 5 × 1 + (–2) (2) + (3) (3)

= 5 – 4 + 9 = 10

Calculation:

The vector is 

⇒

⇒  

⇒  x = 4

Concept:

Dot product of two vectors is defined as:

Cross/Vector product of two vectors is defined as:

where θ is the angle between 

Calculation:

To Find: Value of 

Here angle between them is 0°

Concept:

If  then

Calculation:

Given:  and

The given two vectors are in parallel, therefore the vector product of these two vectors will be zero.

⇒ 9λ – 6 = 0 and 2 - 3λ = 0

∴ λ = 2/3

Given: 

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 

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

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

 = nĵ  + lk̂ + mî +  lk̂ + mî + nĵ 

 = 2(mî + nĵ +lk̂ ) = 2

∴ The correct option is 3

Concept:

Condition for coplanarity:

​Calculation:

Here,  are coplanar

1(λ - 1) + 1(2λ + λ) + 1(-2 - λ) = 0

λ - 1 + 2λ + λ + -2 - λ = 0

3λ - 3 = 0

λ = 1

Hence, option (4) is correct.

Concept:

Let  are two vectors

  is the unit vector perpendicular to both 

Calculation:

Given: 

As we know, 

⇒ 6 = 3 × 4 × cos θ 

⇒ cos θ = 

∴ θ = 60° 

As we know that, If  are two vectors, then

      (∵ Magnitude of a unit vector is one)

 = 3 × 4 × sin 60° 

Concept:

If  is a unit vector, then .

Calculation:

It is given that .

⇒ 

Since  is a unit vector, we get:

⇒ 

⇒ 

⇒  = 4

Concept:

• Unit vector: a vector which has a magnitude of one.

Let

Magnitude of vector of a =

Unit vector =

• Let  and  be the two vectors, then the vector  perpendicular to both  and

 and

• If  then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) - (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) - (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) - (a₂₂ × a₃₁)}

Calculation:

Let vector  and  and vector  perpendicular to both  and 

Unit vector =

Concept:  

Calculation:

Given

Additional Information

Properties of Scalar Product

 (Scalar product is commutative)

 (Distributive of scalar product over addition)

In terms of orthogonal coordinates for mutually perpendicular vectors, it is seen that 

Properties of Vector Product

Concept:

Dot Product: it is also called the inner product or scalar product

• Let the two vectors be  then dot Product of two vector:   Where, || = Magnitude of vectors a and || = Magnitude of vectors b and θ is angle between a and b 

Calculation:

Let

⇒ 1 + 0 + 0 = √3 × 1 × cos θ

⇒ cos θ = 1/√3

∴ θ = cos⁻¹ (1/√3)