The problem states that is perpendicular to . The dot product of any two mutually perpendicular vectors is zero.

Why does completely disappear?

Because is a linear combination of and . Any component parallel to vanishes when taking the cross product with .

JEE MainMathematics

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Vector Algebra: JEE Main Mathematics Question with Solution

\vec{a}, \vec{b}, \vec{c}

are three non-zero vectors and

\hat{n}

is a unit vector perpendicular to

\vec{c}

such that

\vec{a}=\alpha \vec{b}-\hat{n},(\alpha \neq 0)

and

\vec{b} \cdot \vec{c}=12

, then

|\vec{c} \times(\vec{a} \times \vec{b})|

is equal to :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

9

15

6

12

Step-by-step solutionView

Correct answer

Using $\vec{a} \times \vec{b} = -\hat{n} \times \vec{b}$ and expanding the vector triple product or magnitude simplifies $|\vec{c} \times (\vec{a} \times \vec{b})|$ to $|\vec{b} \cdot \vec{c}| = 12$ .

Option analysis

Why each option works or fails

A ·

9

Subtracting component magnitudes or assuming a 3-4-5 geometric triangle relationship involving the given dot product. Express $\vec{a} \times \vec{b}$ directly in terms of $\hat{n} \times \vec{b}$ and apply the BAC-CAB expansion or magnitude identity with orthogonal vectors.

B ·

15

Adding components or unit vector terms to the scalar product rather than evaluating the exact vector triple product. Notice that $\hat{n} \cdot \vec{c} = 0$ , so in the BAC-CAB expansion $(\vec{c} \cdot \vec{b})\hat{n} - (\vec{c} \cdot \hat{n})\vec{b}$ , the second term vanishes completely.

C ·

6

Halving the result by confusing the magnitude formula with the area of a triangle $\frac{1}{2}|\vec{u} \times \vec{v}|$ . The expression asks for the vector magnitude $|\vec{c} \times (\vec{a} \times \vec{b})|$ , not the area of a triangle formed by the vectors.

D ·

12

None. The student correctly evaluates $\vec{a} \times \vec{b} = -\hat{n} \times \vec{b}$ and uses $\hat{n} \perp \vec{c}$ to find the magnitude is $|(\vec{b} \cdot \vec{c})\hat{n}| = 12$ . Keep using direct substitution of linear vector relations into cross products to annihilate collinear terms.

Reviewed route

Solution

StepWorking

01given

Vectors $\vec{a}, \vec{b}, \vec{c}$ are non-zero. $\hat{n}$ is a unit vector perpendicular to $\vec{c}$ (hence $\hat{n} \cdot \vec{c} = 0$ and $|\hat{n}| = 1$ ). We are given $\vec{a} = \alpha \vec{b} - \hat{n}$ with $\alpha \neq 0$ , and $\vec{b} \cdot \vec{c} = 12$ .

02goal

Evaluate the magnitude of the vector triple product $|\vec{c} \times (\vec{a} \times \vec{b})|$ .

03approach

Expand $\vec{c} \times (\vec{a} \times \vec{b})$ using the vector triple product expansion $(\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}$ . Then determine $\vec{c} \cdot \vec{a}$ by taking the dot product of the given relation $\vec{a} = \alpha \vec{b} - \hat{n}$ with $\vec{c}$ .

04execute

Using the BAC-CAB rule: $\vec{c} \times (\vec{a} \times \vec{b}) = (\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b} = 12\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}$ Now, compute $\vec{c} \cdot \vec{a}$ from $\vec{a} = \alpha \vec{b} - \hat{n}$ : $\vec{c} \cdot \vec{a} = \alpha (\vec{c} \cdot \vec{b}) - \vec{c} \cdot \hat{n} = \alpha(12) - 0 = 12\alpha$ Substitute $\vec{c} \cdot \vec{a} = 12\alpha$ back into the expansion: $\vec{c} \times (\vec{a} \times \vec{b}) = 12\vec{a} - 12\alpha \vec{b} = 12(\vec{a} - \alpha \vec{b})$ From the given relation, $\vec{a} - \alpha \vec{b} = -\hat{n}$ . Taking the magnitude: $|\vec{c} \times (\vec{a} \times \vec{b})| = |12(-\hat{n})| = 12 |\hat{n}| = 12 \times 1 = 12$

✓verify

Check dimensions and properties: $\vec{a} \times \vec{b} = (\alpha \vec{b} - \hat{n}) \times \vec{b} = -\hat{n} \times \vec{b} = \vec{b} \times \hat{n}$ . Then $\vec{c} \times (\vec{b} \times \hat{n}) = (\vec{c} \cdot \hat{n})\vec{b} - (\vec{c} \cdot \vec{b})\hat{n} = 0 - 12\hat{n} = -12\hat{n}$ , whose magnitude is $|-12\hat{n}| = 12$ . Both paths match cleanly.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is

\vec{c} \cdot \hat{n} = 0

The problem states that $\hat{n}$ is perpendicular to $\vec{c}$ . The dot product of any two mutually perpendicular vectors is zero.

Why does

\alpha

completely disappear?

Because $\vec{a}$ is a linear combination of $\vec{b}$ and $\hat{n}$ . Any component parallel to $\vec{b}$ vanishes when taking the cross product with $\vec{b}$ .

Answer

Using $\vec{a} \times \vec{b} = -\hat{n} \times \vec{b}$ and expanding the vector triple product or magnitude simplifies $|\vec{c} \times (\vec{a} \times \vec{b})|$ to $|\vec{b} \cdot \vec{c}| = 12$ .

Why each option works or fails

A: $9$ - Subtracting component magnitudes or assuming a 3-4-5 geometric triangle relationship involving the given dot product. Express $\vec{a} \times \vec{b}$ directly in terms of $\hat{n} \times \vec{b}$ and apply the BAC-CAB expansion or magnitude identity with orthogonal vectors.
B: $15$ - Adding components or unit vector terms to the scalar product rather than evaluating the exact vector triple product. Notice that $\hat{n} \cdot \vec{c} = 0$ , so in the BAC-CAB expansion $(\vec{c} \cdot \vec{b})\hat{n} - (\vec{c} \cdot \hat{n})\vec{b}$ , the second term vanishes completely.
C: $6$ - Halving the result by confusing the magnitude formula with the area of a triangle $\frac{1}{2}|\vec{u} \times \vec{v}|$ . The expression asks for the vector magnitude $|\vec{c} \times (\vec{a} \times \vec{b})|$ , not the area of a triangle formed by the vectors.
D · correct: $12$ - None. The student correctly evaluates $\vec{a} \times \vec{b} = -\hat{n} \times \vec{b}$ and uses $\hat{n} \perp \vec{c}$ to find the magnitude is $|(\vec{b} \cdot \vec{c})\hat{n}| = 12$ . Keep using direct substitution of linear vector relations into cross products to annihilate collinear terms.

Step-by-step solution

given: Vectors $\vec{a}, \vec{b}, \vec{c}$ are non-zero. $\hat{n}$ is a unit vector perpendicular to $\vec{c}$ (hence $\hat{n} \cdot \vec{c} = 0$ and $6$ ). We are given $\vec{a} = \alpha \vec{b} - \hat{n}$ with $\alpha \neq 0$ , and $\vec{b} \cdot \vec{c} = 12$ .
goal: Evaluate the magnitude of the vector triple product $|\vec{c} \times (\vec{a} \times \vec{b})|$ .
approach: Expand $\vec{c} \times (\vec{a} \times \vec{b})$ using the vector triple product expansion $(\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}$ . Then determine $\vec{c} \cdot \vec{a}$ by taking the dot product of the given relation $\vec{a} = \alpha \vec{b} - \hat{n}$ with $\vec{c}$ .
execute: Using the BAC-CAB rule: $\vec{c} \times (\vec{a} \times \vec{b}) = (\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b} = 12\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}$ Now, compute $\vec{c} \cdot \vec{a}$ from $\vec{a} = \alpha \vec{b} - \hat{n}$ : $\vec{c} \cdot \vec{a} = \alpha (\vec{c} \cdot \vec{b}) - \vec{c} \cdot \hat{n} = \alpha(12) - 0 = 12\alpha$ Substitute $\vec{c} \cdot \vec{a} = 12\alpha$ back into the expansion: $\vec{c} \times (\vec{a} \times \vec{b}) = 12\vec{a} - 12\alpha \vec{b} = 12(\vec{a} - \alpha \vec{b})$ From the given relation, $\vec{a} - \alpha \vec{b} = -\hat{n}$ . Taking the magnitude: $|\vec{c} \times (\vec{a} \times \vec{b})| = |12(-\hat{n})| = 12 |\hat{n}| = 12 \times 1 = 12$
verify: Check dimensions and properties: $\vec{a} \times \vec{b} = (\alpha \vec{b} - \hat{n}) \times \vec{b} = -\hat{n} \times \vec{b} = \vec{b} \times \hat{n}$ . Then $\vec{c} \times (\vec{b} \times \hat{n}) = (\vec{c} \cdot \hat{n})\vec{b} - (\vec{c} \cdot \vec{b})\hat{n} = 0 - 12\hat{n} = -12\hat{n}$ , whose magnitude is $|-12\hat{n}| = 12$ . Both paths match cleanly.

Shortcut:

When to use it: Fastest route when the linear combination involves one of the cross-product terms directly.

given: $\vec{a} = \alpha \vec{b} - \hat{n}$ , $\vec{c} \cdot \hat{n} = 0$ , $|\hat{n}| = 1$ , and $\vec{b} \cdot \vec{c} = 12$ .

goal: Find $|\vec{c} \times (\vec{a} \times \vec{b})|$ .

approach: Directly substitute $\vec{a} = \alpha \vec{b} - \hat{n}$ into the inner cross product $\vec{a} \times \vec{b}$ using the property $\vec{b} \times \vec{b} = \vec{0}$ .

execute: Compute the inner cross product: $\vec{a} \times \vec{b} = (\alpha \vec{b} - \hat{n}) \times \vec{b} = \alpha(\vec{b} \times \vec{b}) - \hat{n} \times \vec{b} = \vec{0} + \vec{b} \times \hat{n} = \vec{b} \times \hat{n}$ Now substitute into the full expression: $\vec{c} \times (\vec{a} \times \vec{b}) = \vec{c} \times (\vec{b} \times \hat{n})$ Apply the vector triple product expansion: $\vec{c} \times (\vec{b} \times \hat{n}) = (\vec{c} \cdot \hat{n})\vec{b} - (\vec{c} \cdot \vec{b})\hat{n}$ Since $\hat{n} \perp \vec{c}$ , we have $\vec{c} \cdot \hat{n} = 0$ , and $\vec{c} \cdot \vec{b} = 12$ : $\vec{c} \times (\vec{a} \times \vec{b}) = 0\vec{b} - 12\hat{n} = -12\hat{n}$ Take the magnitude: $|-12\hat{n}| = 12|\hat{n}| = 12(1) = 12$

verify: Since $|\vec{c} \times (\vec{b} \times \hat{n})| = |-12\hat{n}| = 12$ , the result is independent of $\alpha$ , which perfectly aligns with $\alpha$ canceling out in the standard route.

Vector Algebra: JEE Main Mathematics Question with Solution

What feels right?

Using $\vec{a} \times \vec{b} = -\hat{n} \times \vec{b}$ and expanding the vector triple product or magnitude simplifies $|\vec{c} \times (\vec{a} \times \vec{b})|$ to $|\vec{b} \cdot \vec{c}| = 12$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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