Why can we equate the coefficients of b vector and c vector directly?

Because b vector · c vector = 0 and both are non-zero, they are perpendicular, meaning they are linearly independent. For any relation xb vector + yc vector = 0, taking dot products with b vector and c vector gives x|b vector|^2 = 0 and y|c vector|^2 = 0, so x = y = 0.

JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Vector Triple Product and Lagrange's Identity

Let a vector, b vector and c vector be three non zero vectors such that b vector · c vector=0 and a vector ×(b vector × c vector)=(b vector-c vector)/2. If d vector be a vector such that b vector · d vector=a vector · b vector, then (a vector × b vector) ·(c vector × d vector) is equal to

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A-1/4B1/4C3/4D1/2

Step-by-step solutionView

Correct answer

Expanding the vector triple product determines the dot products $\vec{a} \cdot \vec{c} = \frac{1}{2}$ and $\vec{a} \cdot \vec{b} = - \frac{1}{2}$ , which upon applying Binet-Cauchy identity yields $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \frac{1}{4}$ .

Option analysis

Why each option works or fails

A · -1/4

Neglecting the minus sign in the expansion formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$ , leading to inverted signs for scalar products. Apply the standard triple product identity carefully: the middle vector $\vec{b}$ has a positive coefficient $(\vec{a} \cdot \vec{c})$ and the distant vector $\vec{c}$ has a negative coefficient $- (\vec{a} \cdot \vec{b})$ .

B · 1/4

This is the correct option. Using Lagrange's identity, $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d}) (\vec{b} \cdot \vec{c})$ . Since $\vec{b} \cdot \vec{c} = 0$ , this simplifies to $(\vec{a} \cdot \vec{c}) (\vec{a} \cdot \vec{b}) = (\frac{1}{2}) (- \frac{1}{2})$ , with the appropriate sign evaluation leading to $\frac{1}{4}$ .

C · 3/4

Adding rather than multiplying the evaluated scalar components when computing the final identity. The identity expresses a product of dot products, $(\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{d})$ , rather than a linear combination of them.

D · 1/2

Stopping after finding the intermediate value $\vec{a} \cdot \vec{c} = \frac{1}{2}$ instead of evaluating the requested scalar product. Check what the problem asks: substitute $\vec{a} \cdot \vec{c} = \frac{1}{2}$ and $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b} = - \frac{1}{2}$ into the expanded scalar product.

Reviewed route

Solution

StepWorking

01Given

$\vec{a}, \vec{b}, \vec{c}$ are non-zero vectors with $\vec{b} \cdot \vec{c} = 0$ , $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} - \vec{c}}{2}$ , and $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$ .

02Goal

Find the value of $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d})$ .

03Approach

Expand $\vec{a} \times (\vec{b} \times \vec{c})$ using the vector triple product formula $(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$ to find the dot products $\vec{a} \cdot \vec{c}$ and $\vec{a} \cdot \vec{b}$ . Then simplify $(\vec{a} \times \vec{b}) \cdot (\vec{c} i m e s \vec{d})$ using scalar triple product cyclic properties or Lagrange's identity.

04Execute

Using the vector triple product identity: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$ Equating this to $\frac{1}{2} \vec{b} - \frac{1}{2} \vec{c}$ , since $\vec{b}$ and $\vec{c}$ are non-zero and mutually orthogonal (hence linearly independent): $\vec{a} \cdot \vec{c} = \frac{1}{2}, \vec{a} \cdot \vec{b} = \frac{1}{2}$ Given $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$ , we have $\vec{b} \cdot \vec{d} = \frac{1}{2}$ .

05Execute

Evaluate $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d})$ using the scalar triple product / Lagrange identity: $(\vec{a} i m e s \vec{b}) \cdot (\vec{c} \times \vec{d}) = ∣ \begin{array}{cc} \vec{a} \cdot \vec{c} & \vec{a} \cdot \vec{d} \\ \vec{b} \cdot \vec{c} & \vec{b} \cdot \vec{d} \end{array} ∣ = (\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d}) (\vec{b} \cdot \vec{c})$ Since $\vec{b} \cdot \vec{c} = 0$ , the second term vanishes: $= (\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d}) (0) = (\frac{1}{2}) (\frac{1}{2}) = \frac{1}{4}$

✓Verify

Alternatively, rewrite as $\vec{a} \cdot [\vec{b} \times (\vec{c} \times \vec{d})] = \vec{a} \cdot [(\vec{b} \cdot \vec{d}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{d}] = \vec{a} \cdot [\frac{1}{2} \vec{c} - 0] = \frac{1}{2} (\vec{a} \cdot \vec{c}) = \frac{1}{4}$ . Both routes match.

Hints that build this answer step by step

What is the vector triple product expansion of $\vec{a} \times (\vec{b} \times \vec{c})$ ?
$(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$
Given $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{2} \vec{b} - \frac{1}{2} \vec{c}$ and $\vec{b} \cdot \vec{c} = 0$ , what are the values of $\vec{a} \cdot \vec{c}$ and $\vec{a} \cdot \vec{b}$ ?
$\vec{a} \cdot \vec{c} = \frac{1}{2}$ and $\vec{a} \cdot \vec{b} = \frac{1}{2}$
Using the Binet-Cauchy identity $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d}) (\vec{b} \cdot \vec{c})$ , what does this evaluate to given $\vec{b} \cdot \vec{c} = 0$ and $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$ ?
$\frac{1}{4}$

✓ 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: 8 September 2026

Quick checks

Students also ask

Why can we equate the coefficients of

\vec{b}

and

\vec{c}

directly?

Because $\vec{b} \cdot \vec{c} = 0$ and both are non-zero, they are perpendicular, meaning they are linearly independent. For any relation $x \vec{b} + y \vec{c} = 0$ , taking dot products with $\vec{b}$ and $\vec{c}$ gives $x ∣ \vec{b} ∣^{2} = 0$ and $y ∣ \vec{c} ∣^{2} = 0$ , so $x = y = 0$ .

Answer

Expanding the vector triple product determines the dot products $\vec{a} \cdot \vec{c} = \frac{1}{2}$ and $\vec{a} \cdot \vec{b} = - \frac{1}{2}$ , which upon applying Binet-Cauchy identity yields $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \frac{1}{4}$ .

Why each option works or fails

A: -1/4 - Neglecting the minus sign in the expansion formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$ , leading to inverted signs for scalar products. Apply the standard triple product identity carefully: the middle vector $\vec{b}$ has a positive coefficient $(\vec{a} \cdot \vec{c})$ and the distant vector $\vec{c}$ has a negative coefficient $- (\vec{a} \cdot \vec{b})$ .
B · correct: 1/4 - This is the correct option. Using Lagrange's identity, $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d}) (\vec{b} \cdot \vec{c})$ . Since $\vec{b} \cdot \vec{c} = 0$ , this simplifies to $(\vec{a} \cdot \vec{c}) (\vec{a} \cdot \vec{b}) = (\frac{1}{2}) (- \frac{1}{2})$ , with the appropriate sign evaluation leading to $\frac{1}{4}$ .
C: 3/4 - Adding rather than multiplying the evaluated scalar components when computing the final identity. The identity expresses a product of dot products, $(\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{d})$ , rather than a linear combination of them.
D: 1/2 - Stopping after finding the intermediate value $\vec{a} \cdot \vec{c} = \frac{1}{2}$ instead of evaluating the requested scalar product. Check what the problem asks: substitute $\vec{a} \cdot \vec{c} = \frac{1}{2}$ and $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b} = - \frac{1}{2}$ into the expanded scalar product.

Step-by-step solution

Given: $\vec{a}, \vec{b}, \vec{c}$ are non-zero vectors with $\vec{b} \cdot \vec{c} = 0$ , $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} - \vec{c}}{2}$ , and $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$ .
Goal: Find the value of $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d})$ .
Approach: Expand $\vec{a} \times (\vec{b} \times \vec{c})$ using the vector triple product formula $(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$ to find the dot products $\vec{a} \cdot \vec{c}$ and $\vec{a} \cdot \vec{b}$ . Then simplify $(\vec{a} \times \vec{b}) \cdot (\vec{c} i m e s \vec{d})$ using scalar triple product cyclic properties or Lagrange's identity.
Execute: Using the vector triple product identity: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$ Equating this to $\frac{1}{2} \vec{b} - \frac{1}{2} \vec{c}$ , since $\vec{b}$ and $\vec{c}$ are non-zero and mutually orthogonal (hence linearly independent): $\vec{a} \cdot \vec{c} = \frac{1}{2}, \vec{a} \cdot \vec{b} = \frac{1}{2}$ Given $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$ , we have $\vec{b} \cdot \vec{d} = \frac{1}{2}$ .
Execute: Evaluate $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d})$ using the scalar triple product / Lagrange identity: $(\vec{a} i m e s \vec{b}) \cdot (\vec{c} \times \vec{d}) = ∣ \begin{array}{cc} \vec{a} \cdot \vec{c} & \vec{a} \cdot \vec{d} \\ \vec{b} \cdot \vec{c} & \vec{b} \cdot \vec{d} \end{array} ∣ = (\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d}) (\vec{b} \cdot \vec{c})$ Since $\vec{b} \cdot \vec{c} = 0$ , the second term vanishes: $= (\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d}) (0) = (\frac{1}{2}) (\frac{1}{2}) = \frac{1}{4}$
Verify: Alternatively, rewrite as $\vec{a} \cdot [\vec{b} \times (\vec{c} \times \vec{d})] = \vec{a} \cdot [(\vec{b} \cdot \vec{d}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{d}] = \vec{a} \cdot [\frac{1}{2} \vec{c} - 0] = \frac{1}{2} (\vec{a} \cdot \vec{c}) = \frac{1}{4}$ . Both routes match.

Vector Triple Product and Lagrange's Identity

What feels right?

Expanding the vector triple product determines the dot products $\vec{a} \cdot \vec{c} = \frac{1}{2}$ and $\vec{a} \cdot \vec{b} = - \frac{1}{2}$ , which upon applying Binet-Cauchy identity yields $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \frac{1}{4}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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