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Vector Algebra: JEE Main Mathematics Question with Solution

Q: Why can we equate the coefficients of and directly?

Because and both are non-zero, they are perpendicular, meaning they are linearly independent. For any relation , taking dot products with and gives and , so .

Let

\vec{a}, \vec{b}

and

\vec{c}

be three non zero vectors such that

\vec{b} \cdot \vec{c}=0

and

\vec{a} \times(\vec{b} \times \vec{c})=\frac{\vec{b}-\vec{c}}{2}

. If

\vec{d}

be a vector such that

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

, then

(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})

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.

-\frac{1}{4}

\frac{1}{4}

\frac{3}{4}

\frac{1}{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 ·

-\frac{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 ·

\frac{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 $(\vec{a}\cdot\vec{c})(\vec{a}\cdot\vec{b}) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)$ , with the appropriate sign evaluation leading to $\frac{1}{4}$ .

C ·

\frac{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 ·

\frac{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 $(\vec{a} \times \vec{b}) \cdot (\vec{c} imes \vec{d})$ 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): $\vec{a} \cdot \vec{c} = \frac{1}{2}, \quad \vec{a} \cdot \vec{b} = \frac{1}{2}$ 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: $(\vec{a} imes \vec{b}) \cdot (\vec{c} \times \vec{d}) = \begin{vmatrix} \vec{a} \cdot \vec{c} & \vec{a} \cdot \vec{d} \\ \vec{b} \cdot \vec{c} & \vec{b} \cdot \vec{d} \end{vmatrix} = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})$ Since $\vec{b} \cdot \vec{c} = 0$ , the second term vanishes: $= (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(0) = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{4}$

✓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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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: $-\frac{1}{4}$ - 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: $\frac{1}{4}$ - 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 $(\vec{a}\cdot\vec{c})(\vec{a}\cdot\vec{b}) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)$ , with the appropriate sign evaluation leading to $\frac{1}{4}$ .
C: $\frac{3}{4}$ - 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: $\frac{1}{2}$ - 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 $\frac{1}{4}$ and $\vec{a} \cdot \vec{b}$ . Then simplify $(\vec{a} \times \vec{b}) \cdot (\vec{c} imes \vec{d})$ 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): $\vec{a} \cdot \vec{c} = \frac{1}{2}, \quad \vec{a} \cdot \vec{b} = \frac{1}{2}$ 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: $(\vec{a} imes \vec{b}) \cdot (\vec{c} \times \vec{d}) = \begin{vmatrix} \vec{a} \cdot \vec{c} & \vec{a} \cdot \vec{d} \\ \vec{b} \cdot \vec{c} & \vec{b} \cdot \vec{d} \end{vmatrix} = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})$ Since $\vec{b} \cdot \vec{c} = 0$ , the second term vanishes: $= (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(0) = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{4}$
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 Algebra: JEE Main Mathematics Question with Solution

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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