JEE MainMathematics

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Orthogonality of the Cross Product and Vector Dot Products

Let a vector and b vector be two vectors, Let |a vector| = 1, |b vector| = 4 and a vector · b vector = 2. If c vector = (2a vector × b vector) - 3b vector, then the value of b vector · c vector is

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A-24B-84C-48D-60

Step-by-step solutionView

Correct answer

Taking the dot product of $\vec{b}$ with $\vec{c}$ eliminates the cross product term because $\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$ , leaving $- 3 ∣ \vec{b} ∣^{2} = - 3 (16) = - 48$ .

Option analysis

Why each option works or fails

A · -24

Believing that $∣ \vec{b} ∣^{2} = 8$ (forgetting to square $4$ properly and computing $4 \times 2$ or $- 3 \times 8 = - 24$ ). Compute $∣ \vec{b} ∣^{2}$ as $4^{2} = 16$ , giving $- 3 (16) = - 48$ .

B · -84

Attempting to compute $\vec{b} \cdot (2 \vec{a} \times \vec{b})$ non-trivially and introducing extraneous cross product magnitude terms involving $\sin θ$ . Recognize that the scalar triple product with a repeated vector $\vec{b} \cdot (\vec{a} \times \vec{b})$ is identically zero.

C · -48

The cross product $\vec{a} \times \vec{b}$ is orthogonal to $\vec{b}$ , so $\vec{b} \cdot (2 \vec{a} \times \vec{b}) = 0$ , leaving $- 3 (\vec{b} \cdot \vec{b}) = - 3 (16) = - 48$ . This is the correct value.

D · -60

Mistakenly adding or subtracting extraneous multiples of $\vec{a} \cdot \vec{b} = 2$ or using an incorrect expansion for the dot product. Distribute the dot product linearly: $\vec{b} \cdot (2 \vec{a} \times \vec{b}) - 3 \vec{b} \cdot \vec{b} = 0 - 3 ∣ \vec{b} ∣^{2}$ .

Reviewed route

Solution

StepWorking

01Given

$∣ \vec{a} ∣ = 1$ , $∣ \vec{b} ∣ = 4$ , $\vec{a} \cdot \vec{b} = 2$ , and $\vec{c} = 2 (\vec{a} \times \vec{b}) - 3 \vec{b}$ .

02Goal

Find the value of the scalar product $\vec{b} \cdot \vec{c}$ .

03Approach

Take the dot product of $\vec{b}$ with $\vec{c}$ by distributing across the linear combination: $\vec{b} \cdot \vec{c} = 2 (\vec{a} \times \vec{b}) \cdot \vec{b} - 3 (\vec{b} \cdot \vec{b})$ . Note that $(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$ because the cross product is orthogonal to both constituent vectors.

04Execute

Compute $\vec{b} \cdot \vec{c} = 0 - 3 ∣ \vec{b} ∣^{2} = - 3 (4^{2}) = - 48$ .

✓Verify

Check that the extra information $∣ \vec{a} ∣ = 1$ and $\vec{a} \cdot \vec{b} = 2$ is redundant distractor data for this specific dot product, which confirms the computation depends only on $∣ \vec{b} ∣$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why was the value of

\vec{a} \cdot \vec{b} = 2

and

∣ \vec{a} ∣ = 1

given if we never used it?

The examiner provided extraneous parameters as deliberate distractors to make candidates spend time trying to compute the cross product or angle between vectors.

Answer

Taking the dot product of $\vec{b}$ with $\vec{c}$ eliminates the cross product term because $\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$ , leaving $- 3 ∣ \vec{b} ∣^{2} = - 3 (16) = - 48$ .

Why each option works or fails

A: -24 - Believing that $∣ \vec{b} ∣^{2} = 8$ (forgetting to square $4$ properly and computing $4 \times 2$ or $- 3 \times 8 = - 24$ ). Compute $∣ \vec{b} ∣^{2}$ as $4^{2} = 16$ , giving $- 3 (16) = - 48$ .
B: -84 - Attempting to compute $\vec{b} \cdot (2 \vec{a} \times \vec{b})$ non-trivially and introducing extraneous cross product magnitude terms involving $\sin θ$ . Recognize that the scalar triple product with a repeated vector $\vec{b} \cdot (\vec{a} \times \vec{b})$ is identically zero.
C · correct: -48 - The cross product $\vec{a} \times \vec{b}$ is orthogonal to $\vec{b}$ , so $\vec{b} \cdot (2 \vec{a} \times \vec{b}) = 0$ , leaving $- 3 (\vec{b} \cdot \vec{b}) = - 3 (16) = - 48$ . This is the correct value.
D: -60 - Mistakenly adding or subtracting extraneous multiples of $\vec{a} \cdot \vec{b} = 2$ or using an incorrect expansion for the dot product. Distribute the dot product linearly: $\vec{b} \cdot (2 \vec{a} \times \vec{b}) - 3 \vec{b} \cdot \vec{b} = 0 - 3 ∣ \vec{b} ∣^{2}$ .

Step-by-step solution

Given: $∣ \vec{a} ∣ = 1$ , $∣ \vec{b} ∣ = 4$ , $\vec{a} \cdot \vec{b} = 2$ , and $\vec{c} = 2 (\vec{a} \times \vec{b}) - 3 \vec{b}$ .
Goal: Find the value of the scalar product $\vec{b} \cdot \vec{c}$ .
Approach: Take the dot product of $\vec{b}$ with $\vec{c}$ by distributing across the linear combination: $\vec{b} \cdot \vec{c} = 2 (\vec{a} \times \vec{b}) \cdot \vec{b} - 3 (\vec{b} \cdot \vec{b})$ . Note that $(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$ because the cross product is orthogonal to both constituent vectors.
Execute: Compute $\vec{b} \cdot \vec{c} = 0 - 3 ∣ \vec{b} ∣^{2} = - 3 (4^{2}) = - 48$ .
Verify: Check that the extra information $∣ \vec{a} ∣ = 1$ and $\vec{a} \cdot \vec{b} = 2$ is redundant distractor data for this specific dot product, which confirms the computation depends only on $∣ \vec{b} ∣$ .

Shortcut:

When to use it: Recognizing immediately that $\vec{b}$ is perpendicular to any cross product containing $\vec{b}$ .

Quick kill: Since $(\vec{a} \times \vec{b}) ⊥ \vec{b}$ , dotting $\vec{c}$ with $\vec{b}$ gives directly $- 3 ∣ \vec{b} ∣^{2} = - 3 (16) = - 48$ .

Orthogonality of the Cross Product and Vector Dot Products

What feels right?

Taking the dot product of $\vec{b}$ with $\vec{c}$ eliminates the cross product term because $\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$ , leaving $- 3 ∣ \vec{b} ∣^{2} = - 3 (16) = - 48$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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Taking the dot product of $\vec{b}$ with $\vec{c}$ eliminates the cross product term because $\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$ , leaving $- 3 ∣ \vec{b} ∣^{2} = - 3 (16) = - 48$ .

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