Why is c vector · n hat = 0?

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

JEE MainMathematics

Answer verified against the official NTA key

+4 marks−1 if incorrectSingle correctPrevious-year question

Magnitude of Vector Triple Product with Linear Relation

Q: Why does α completely disappear?

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

If a vector, b vector, c vector are three non-zero vectors and n hat is a unit vector perpendicular to c vector such that a vector=α b vector-n hat,(α ≠ 0) and b vector · c vector=12, then |c vector ×(a vector × b vector)| 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.

A9B15C6D12

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

The student correctly evaluates $\vec{a} \times \vec{b} = - \hat{n} \times \vec{b}$ and uses $\hat{n} ⊥ \vec{c}$ 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 $\vec{a} = α \vec{b} - \hat{n}$ with $α \neq 0$ , 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 $\vec{a} = α \vec{b} - \hat{n}$ 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 $\vec{a} = α \vec{b} - \hat{n}$ : $\vec{c} \cdot \vec{a} = α (\vec{c} \cdot \vec{b}) - \vec{c} \cdot \hat{n} = α (12) - 0 = 12 α$ Substitute $\vec{c} \cdot \vec{a} = 12 α$ back into the expansion: $\vec{c} \times (\vec{a} \times \vec{b}) = 12 \vec{a} - 12 α \vec{b} = 12 (\vec{a} - α \vec{b})$ From the given relation, $\vec{a} - α \vec{b} = - \hat{n}$ . 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: $\vec{a} \times \vec{b} = (α \vec{b} - \hat{n}) \times \vec{b} = - \hat{n} \times \vec{b} = \vec{b} \times \hat{n}$ . 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
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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

α

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 - The student correctly evaluates $\vec{a} \times \vec{b} = - \hat{n} \times \vec{b}$ and uses $\hat{n} ⊥ \vec{c}$ 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 $∣ \hat{n} ∣ = 1$ ). We are given $\vec{a} = α \vec{b} - \hat{n}$ with $α \neq 0$ , 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 $\vec{a} = α \vec{b} - \hat{n}$ 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 $\vec{a} = α \vec{b} - \hat{n}$ : $\vec{c} \cdot \vec{a} = α (\vec{c} \cdot \vec{b}) - \vec{c} \cdot \hat{n} = α (12) - 0 = 12 α$ Substitute $\vec{c} \cdot \vec{a} = 12 α$ back into the expansion: $\vec{c} \times (\vec{a} \times \vec{b}) = 12 \vec{a} - 12 α \vec{b} = 12 (\vec{a} - α \vec{b})$ From the given relation, $\vec{a} - α \vec{b} = - \hat{n}$ . 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: $\vec{a} \times \vec{b} = (α \vec{b} - \hat{n}) \times \vec{b} = - \hat{n} \times \vec{b} = \vec{b} \times \hat{n}$ . 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: $\vec{a} = α \vec{b} - \hat{n}$ , $\vec{c} \cdot \hat{n} = 0$ , $∣ \hat{n} ∣ = 1$ , and $\vec{b} \cdot \vec{c} = 12$ .

Goal: Find $∣ \vec{c} \times (\vec{a} \times \vec{b}) ∣$ .

Approach: Directly substitute $\vec{a} = α \vec{b} - \hat{n}$ 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: $\vec{a} \times \vec{b} = (α \vec{b} - \hat{n}) \times \vec{b} = α (\vec{b} \times \vec{b}) - \hat{n} \times \vec{b} = \vec{0} + \vec{b} \times \hat{n} = \vec{b} \times \hat{n}$ 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 $\hat{n} ⊥ \vec{c}$ , 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 $α$ , which perfectly aligns with $α$ canceling out in the standard route.

Magnitude of Vector Triple Product with Linear Relation

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

We think you should solve this next.

Let $\vec{a}$ and $\vec{b}$ be two vectors, Let…

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non zero vectors such that…

Suppose $f : R \to (0, \infty)$ be a differentiable function such that…

Students also ask

Magnitude of Vector Triple Product with Linear Relation

Using a⃗×b⃗=−n^×b⃗a×b=−n^×b and expanding the vector triple product or magnitude simplifies ∣c⃗×(a⃗×b⃗)∣∣c×(a×b)∣ to ∣b⃗⋅c⃗∣=12∣b⋅c∣=12.

01Given

02Goal

03Approach

04Execute

✓Verify

Let a⃗a and b⃗b be two vectors, Let…

Let a⃗,b⃗a,b and c⃗c be three non zero vectors such that…

Suppose f:R→(0,∞)f:R→(0,∞) be a differentiable function such that…

Students also ask

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 $∣ \vec{b} \cdot \vec{c} ∣ = 12$ .

Let $\vec{a}$ and $\vec{b}$ be two vectors, Let…

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non zero vectors such that…

Suppose $f : R \to (0, \infty)$ be a differentiable function such that…