JEE MainMathematics

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Vector Algebra: JEE Main Mathematics Question with Solution

Let

\vec{u} = \hat{i} - \hat{j} - 2\hat{k}, \vec{v} = 2\hat{i} + \hat{j} - \hat{k}, \vec{v}\cdot\vec{w} = 2

and

\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}

. Then

\vec{u}\cdot\vec{w}

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.

2

\frac{3}{2}

1

-\frac{2}{3}

Step-by-step solutionView

Correct answer

Taking the dot product of both sides of $\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}$ with $\vec{v}$ determines $\lambda = -1/2$ , and then taking the dot product with $\vec{w}$ gives $\vec{u}\cdot\vec{w} = 1$ .

Option analysis

Why each option works or fails

A ·

2

The student confuses the target quantity $\vec{u}\cdot\vec{w}$ with the given value $\vec{v}\cdot\vec{w} = 2$ . Distinguish between the given scalar product $\vec{v}\cdot\vec{w}$ and the required dot product $\vec{u}\cdot\vec{w}$ , using the vector equation to relate them.

B ·

\frac{3}{2}

The student makes a sign error when evaluating $\vec{u}\cdot\vec{v}$ or solving for $\lambda$ , obtaining $\lambda = 1/2$ instead of $-1/2$ . Carefully compute $\vec{u}\cdot\vec{v} = (1)(2) + (-1)(1) + (-2)(-1) = 3$ , leading to $0 = 3 + 6\lambda$ , which gives $\lambda = -1/2$ .

C ·

1

This is the correct option. Dotting the equation with $\vec{v}$ gives $\lambda = -1/2$ , and dotting with $\vec{w}$ gives $0 = \vec{u}\cdot\vec{w} + \lambda(\vec{v}\cdot\vec{w})$ , which yields $\vec{u}\cdot\vec{w} = -(-1/2)(2) = 1$ .

D ·

-\frac{2}{3}

The student isolates $\lambda$ incorrectly as $-2/3$ or confuses $\lambda$ with $\vec{u}\cdot\vec{w}$ . Solve $3 + 6\lambda = 0$ to get $\lambda = -1/2$ , then substitute $\lambda$ into the scalar equation $\vec{u}\cdot\vec{w} = -\lambda(\vec{v}\cdot\vec{w})$ .

Reviewed route

Solution

StepWorking

01given

Vectors $\vec{u} = \hat{i} - \hat{j} - 2\hat{k}$ , $\vec{v} = 2\hat{i} + \hat{j} - \hat{k}$ , with $\vec{v}\cdot\vec{w} = 2$ and $\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}$ .

02goal

Find the value of the scalar product $\vec{u}\cdot\vec{w}$ .

03approach

Take the dot product of $\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}$ first with $\vec{v}$ to eliminate the left-hand side and solve for $\lambda$ . Then take the dot product with $\vec{w}$ to relate $\vec{u}\cdot\vec{w}$ directly to $\lambda$ and $\vec{v}\cdot\vec{w}$ .

04execute

Compute $\vec{u}\cdot\vec{v} = (1)(2) + (-1)(1) + (-2)(-1) = 2 - 1 + 2 = 3$ and $|\vec{v}|^2 = 2^2 + 1^2 + (-1)^2 = 6$ . Taking the dot product of both sides with $\vec{v}$ gives $(\vec{v} \times \vec{w})\cdot\vec{v} = \vec{u}\cdot\vec{v} + \lambda|\vec{v}|^2 \implies 0 = 3 + 6\lambda \implies \lambda = -\frac{1}{2}$ .

05execute

Now take the dot product of the relation with $\vec{w}$ : $(\vec{v} \times \vec{w})\cdot\vec{w} = \vec{u}\cdot\vec{w} + \lambda(\vec{v}\cdot\vec{w})$ . Since $(\vec{v} \times \vec{w})\cdot\vec{w} = 0$ , we have $0 = \vec{u}\cdot\vec{w} + \lambda(2) \implies \vec{u}\cdot\vec{w} = -2\lambda = -2\left(-\frac{1}{2}\right) = 1$ .

✓verify

Check consistency: $\vec{u} + \lambda\vec{v} = (\hat{i} - \hat{j} - 2\hat{k}) - \frac{1}{2}(2\hat{i} + \hat{j} - \hat{k}) = -\frac{3}{2}\hat{j} - \frac{3}{2}\hat{k}$ . Its dot product with $\vec{v}$ is $0(2) - \frac{3}{2}(1) - \frac{3}{2}(-1) = 0$ , confirming orthogonality.

Hints that build this answer step by step

What operation allows us to eliminate $\vec{w}$ from the left-hand side of $\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}$ to solve for $\lambda$ ?
Take the dot product of both sides with $\vec{v}$ , since $\vec{v} \cdot (\vec{v} \times \vec{w}) = 0$ .
What is the value of $\lambda$ obtained from $\vec{v} \cdot (\vec{u} + \lambda\vec{v}) = 0$ ?
$\lambda = -\frac{1}{2}$
Now, how do we find $\vec{u}\cdot\vec{w}$ using $\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}$ and $\vec{v}\cdot\vec{w} = 2$ ?
Take the dot product of both sides with $\vec{w}$ , giving $0 = \vec{u}\cdot\vec{w} + \lambda(\vec{v}\cdot\vec{w})$ , which yields $\vec{u}\cdot\vec{w} = 1$ .

✓ 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 (v x w) . v = 0?

The cross product of any two vectors is perpendicular to both constituent vectors. Hence, (v x w) is orthogonal to v, making their scalar dot product identically zero.

Answer

Taking the dot product of both sides of $\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}$ with $\vec{v}$ determines $\lambda = -1/2$ , and then taking the dot product with $\vec{w}$ gives $\vec{u}\cdot\vec{w} = 1$ .

Why each option works or fails

A: $2$ - The student confuses the target quantity $\vec{u}\cdot\vec{w}$ with the given value $\vec{v}\cdot\vec{w} = 2$ . Distinguish between the given scalar product $\vec{v}\cdot\vec{w}$ and the required dot product $\vec{u}\cdot\vec{w}$ , using the vector equation to relate them.
B: $\frac{3}{2}$ - The student makes a sign error when evaluating $\vec{u}\cdot\vec{v}$ or solving for $\lambda$ , obtaining $\lambda = 1/2$ instead of $-1/2$ . Carefully compute $\vec{u}\cdot\vec{v} = (1)(2) + (-1)(1) + (-2)(-1) = 3$ , leading to $0 = 3 + 6\lambda$ , which gives $\lambda = -1/2$ .
C · correct: $1$ - This is the correct option. Dotting the equation with $\vec{v}$ gives $\lambda = -1/2$ , and dotting with $-\frac{2}{3}$ gives $0 = \vec{u}\cdot\vec{w} + \lambda(\vec{v}\cdot\vec{w})$ , which yields $\vec{u}\cdot\vec{w} = -(-1/2)(2) = 1$ .
D: $-\frac{2}{3}$ - The student isolates $\lambda$ incorrectly as $-2/3$ or confuses $\lambda$ with $\vec{u}\cdot\vec{w}$ . Solve $3 + 6\lambda = 0$ to get $\lambda = -1/2$ , then substitute $\lambda$ into the scalar equation $\vec{u}\cdot\vec{w} = -\lambda(\vec{v}\cdot\vec{w})$ .

Step-by-step solution

given: Vectors $\vec{u} = \hat{i} - \hat{j} - 2\hat{k}$ , $\vec{v} = 2\hat{i} + \hat{j} - \hat{k}$ , with $\vec{v}\cdot\vec{w} = 2$ and $\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}$ .
goal: Find the value of the scalar product $\vec{u}\cdot\vec{w}$ .
approach: Take the dot product of $\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}$ first with $\vec{v}$ to eliminate the left-hand side and solve for $\lambda$ . Then take the dot product with $\vec{w}$ to relate $\vec{u}\cdot\vec{w}$ directly to $\lambda$ and $\vec{v}\cdot\vec{w}$ .
execute: Compute $\vec{u}\cdot\vec{v} = (1)(2) + (-1)(1) + (-2)(-1) = 2 - 1 + 2 = 3$ and $|\vec{v}|^2 = 2^2 + 1^2 + (-1)^2 = 6$ . Taking the dot product of both sides with $\vec{v}$ gives $(\vec{v} \times \vec{w})\cdot\vec{v} = \vec{u}\cdot\vec{v} + \lambda|\vec{v}|^2 \implies 0 = 3 + 6\lambda \implies \lambda = -\frac{1}{2}$ .
execute: Now take the dot product of the relation with $\vec{w}$ : $(\vec{v} \times \vec{w})\cdot\vec{w} = \vec{u}\cdot\vec{w} + \lambda(\vec{v}\cdot\vec{w})$ . Since $(\vec{v} \times \vec{w})\cdot\vec{w} = 0$ , we have $0 = \vec{u}\cdot\vec{w} + \lambda(2) \implies \vec{u}\cdot\vec{w} = -2\lambda = -2\left(-\frac{1}{2}\right) = 1$ .
verify: Check consistency: $\vec{u} + \lambda\vec{v} = (\hat{i} - \hat{j} - 2\hat{k}) - \frac{1}{2}(2\hat{i} + \hat{j} - \hat{k}) = -\frac{3}{2}\hat{j} - \frac{3}{2}\hat{k}$ . Its dot product with $\vec{v}$ is $0(2) - \frac{3}{2}(1) - \frac{3}{2}(-1) = 0$ , confirming orthogonality.

Vector Algebra: JEE Main Mathematics Question with Solution

What feels right?

Taking the dot product of both sides of $\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}$ with $\vec{v}$ determines $\lambda = -1/2$ , and then taking the dot product with $\vec{w}$ gives $\vec{u}\cdot\vec{w} = 1$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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