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Vector Algebra: JEE Main Mathematics Question with Solution

The vector

\vec{a}=-\hat{i}+2\hat{j}+\hat{k}

is rotated through a right angle, passing through the

y

-axis in its way and the resulting vector is

\vec{b}

. Then the projection of

3\vec{a}+\sqrt{2}\vec{b}

\vec{c}=5\hat{i}+4\hat{j}+3\hat{k}

is :

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\sqrt{3}

1

3\sqrt{2}

\sqrt{6}

Step-by-step solutionView

Correct answer

The rotated vector is $\vec{b} = -\hat{i} - \hat{j} + \hat{k}$ , making the projection of $3\vec{a}+\sqrt{2}\vec{b}$ onto $\vec{c}$ equal to $3\sqrt{2}$ .

Option analysis

Why each option works or fails

A ·

2\sqrt{3}

Believing that the denominator for the projection onto $\vec{c}$ is $|\vec{a}|$ or $\sqrt{6}$ instead of $|\vec{c}| = 5\sqrt{2}$ . The projection of any vector $\vec{v}$ onto $\vec{c}$ is given by $\frac{\vec{v}\cdot\vec{c}}{|\vec{c}|}$ ; always divide by the magnitude of the target vector $\vec{c}$ .

B ·

1

Taking $\vec{b}$ along the positive $y$ -axis direction or computing the dot product without multiplying the scalar coefficient properly. Ensure you correctly express $\vec{b}$ as a linear combination of $\vec{a}$ and $\hat{j}$ , enforcing both $\vec{a}\cdot\vec{b} = 0$ and $|\vec{b}| = |\vec{a}|$ .

C ·

3\sqrt{2}

None. This is the correct option. The rotated vector $\vec{b} = -\hat{i}-\hat{j}+\hat{k}$ gives $(3\vec{a}+\sqrt{2}\vec{b})\cdot\vec{c} = 18 + 12 = 30$ , and dividing by $|\vec{c}| = 5\sqrt{2}$ yields $3\sqrt{2}$ .

D ·

\sqrt{6}

Calculating the magnitude of $\vec{a}$ or $\vec{b}$ ( $|\vec{a}| = \sqrt{6}$ ) and stopping before computing the projection of $3\vec{a}+\sqrt{2}\vec{b}$ on $\vec{c}$ . Read the full question: the magnitude of $\vec{a}$ is intermediate; you need the scalar projection of $3\vec{a}+\sqrt{2}\vec{b}$ on $\vec{c}$ .

Reviewed route

Solution

StepWorking

01given

Given vector $\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$ is rotated by $\frac{\pi}{2}$ passing through the positive $y$ -axis to become $\vec{b}$ . The target vector for projection is $\vec{c} = 5\hat{i} + 4\hat{j} + 3\hat{k}$ .

02goal

Find the projection of $3\vec{a} + \sqrt{2}\vec{b}$ on $\vec{c}$ , which is given by $\frac{(3\vec{a} + \sqrt{2}\vec{b}) \cdot \vec{c}}{|\vec{c}|}$ .

03approach

Since $\vec{a}$ rotates to $\vec{b}$ passing through the $y$ -axis, $\vec{b}$ lies in the plane of $\vec{a}$ and the $y$ -axis ( $\,\hat{j}\,$ , so $\vec{b} = \lambda\vec{a} + \mu\hat{j}$ ). Rotation preserves magnitude, so $|\vec{b}| = |\vec{a}|$ . The rotation angle is $90^\circ$ , which gives $\vec{a} \cdot \vec{b} = 0$ . Passing through the $y$ -axis determines the correct sign branch. The angle between $\vec{a}$ and $\hat{j}$ is acute, and rotating toward $\hat{j}$ gives $\vec{b} \cdot \hat{j} > 0$ .

04execute

Express $\vec{b} = \lambda(-\hat{i} + 2\hat{j} + \hat{k}) + \mu\hat{j} = -\lambda\hat{i} + (2\lambda + \mu)\hat{j} + \lambda\hat{k}$ . Using orthogonality: $\vec{a} \cdot \vec{b} = 0 \implies -1(-\lambda) + 2(2\lambda + \mu) + 1(\lambda) = 0 \implies 6\lambda + 2\mu = 0 \implies \mu = -3\lambda$ .

05execute

Substitute $\mu = -3\lambda$ into $\vec{b}$ : $\vec{b} = -\lambda\hat{i} - \lambda\hat{j} + \lambda\hat{k}$ . Now equate magnitudes: $|\vec{b}|^2 = |\vec{a}|^2 \implies (-\lambda)^2 + (-\lambda)^2 + \lambda^2 = (-1)^2 + 2^2 + 1^2 \implies 3\lambda^2 = 6 \implies \lambda^2 = 2 \implies \lambda = \pm\sqrt{2}$ . Since the rotation passes through the $y$ -axis, $\vec{a}$ rotates toward the positive $y$ -axis, so $\vec{b}$ has a positive $y$ -component: $-(-\sqrt{2}) = \sqrt{2} > 0$ , giving $\lambda = -\sqrt{2}$ . Thus $\vec{b} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} - \sqrt{2}\hat{k}$ .

06execute

Compute $\vec{a} \cdot \vec{c} = (-1)(5) + 2(4) + 1(3) = 6$ . Compute $\vec{b} \cdot \vec{c} = (\sqrt{2})(5) + (\sqrt{2})(4) + (-\sqrt{2})(3) = 6\sqrt{2}$ . Compute $|\vec{c}| = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{50} = 5\sqrt{2}$ . Then the projection is $\frac{3(\vec{a}\cdot\vec{c}) + \sqrt{2}(\vec{b}\cdot\vec{c})}{|\vec{c}|} = \frac{3(6) + \sqrt{2}(6\sqrt{2})}{5\sqrt{2}} = \frac{18 + 12}{5\sqrt{2}} = \frac{30}{5\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$ .

✓verify

$3\sqrt{2} \approx 4.24$ . Check $|\vec{b}| = \sqrt{2 + 2 + 2} = \sqrt{6} = |\vec{a}|$ . $\vec{a} \cdot \vec{b} = -\sqrt{2} + 2\sqrt{2} - \sqrt{2} = 0$ . Both magnitude and orthogonality conditions are strictly satisfied.

Hints that build this answer step by step

Because $\vec{a}$ passes through the $y$ -axis to reach $\vec{b}$ , the vectors $\vec{a}$ , $\hat{j}$ , and $\vec{b}$ are coplanar. How can $\vec{b}$ be represented in terms of $\vec{a}$ and $\hat{j}$ ?
$\vec{b} = \alpha\vec{a} + \beta\hat{j}$ with $\vec{a}\cdot\vec{b} = 0$ and $|\vec{b}| = |\vec{a}| = \sqrt{6}$
Using $\vec{a} = -\hat{i}+2\hat{j}+\hat{k}$ , $\vec{a}\cdot\vec{b}=0$ , and $|\vec{b}|=\sqrt{6}$ with $\vec{b}$ passing through $\hat{j}$ , what is $\vec{b}$ ?
$\vec{b} = -\hat{i} - \hat{j} + \hat{k}$
What is the projection of the vector $\vec{v} = 3\vec{a} + \sqrt{2}\vec{b}$ on $\vec{c} = 5\hat{i}+4\hat{j}+3\hat{k}$ ?
$3\sqrt{2}$

✓ 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 is b coplanar with a and the y-axis?

The problem states that 'a is rotated... passing through the y-axis'. A vector rotating continuously towards and through another vector sweeps out a single plane containing both vectors.

Answer

The rotated vector is $\vec{b} = -\hat{i} - \hat{j} + \hat{k}$ , making the projection of $3\vec{a}+\sqrt{2}\vec{b}$ onto $\vec{c}$ equal to $3\sqrt{2}$ .

Why each option works or fails

A: $2\sqrt{3}$ - Believing that the denominator for the projection onto $\vec{c}$ is $|\vec{a}|$ or $\sqrt{6}$ instead of $|\vec{c}| = 5\sqrt{2}$ . The projection of any vector $\vec{v}$ onto $\vec{c}$ is given by $\frac{\vec{v}\cdot\vec{c}}{|\vec{c}|}$ ; always divide by the magnitude of the target vector $\vec{c}$ .
B: $1$ - Taking $\vec{b}$ along the positive $y$ -axis direction or computing the dot product without multiplying the scalar coefficient properly. Ensure you correctly express $\vec{b}$ as a linear combination of $\vec{a}$ and $\hat{j}$ , enforcing both $\vec{a}\cdot\vec{b} = 0$ and $|\vec{b}| = |\vec{a}|$ .
C · correct: $3\sqrt{2}$ - None. This is the correct option. The rotated vector $\vec{b} = -\hat{i}-\hat{j}+\hat{k}$ gives $(3\vec{a}+\sqrt{2}\vec{b})\cdot\vec{c} = 18 + 12 = 30$ , and dividing by $|\vec{c}| = 5\sqrt{2}$ yields $3\sqrt{2}$ .
D: $\sqrt{6}$ - Calculating the magnitude of $\vec{a}$ or $\vec{b}$ ( $|\vec{a}| = \sqrt{6}$ ) and stopping before computing the projection of $3\vec{a}+\sqrt{2}\vec{b}$ on $\vec{c}$ . Read the full question: the magnitude of $\vec{a}$ is intermediate; you need the scalar projection of $3\vec{a}+\sqrt{2}\vec{b}$ on $\vec{c}$ .

Step-by-step solution

given: Given vector $\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$ is rotated by $\frac{\pi}{2}$ passing through the positive $y$ -axis to become $\vec{b}$ . The target vector for projection is $\vec{c} = 5\hat{i} + 4\hat{j} + 3\hat{k}$ .
goal: Find the projection of $3\vec{a} + \sqrt{2}\vec{b}$ on $\vec{c}$ , which is given by $\frac{(3\vec{a} + \sqrt{2}\vec{b}) \cdot \vec{c}}{|\vec{c}|}$ .
approach: Since $\vec{a}$ rotates to $\vec{b}$ passing through the $y$ -axis, $\vec{b}$ lies in the plane of $\vec{a}$ and the $y$ -axis ( $\,\hat{j}\,$ , so $\vec{b} = \lambda\vec{a} + \mu\hat{j}$ ). Rotation preserves magnitude, so $|\vec{b}| = |\vec{a}|$ . The rotation angle is $90^\circ$ , which gives $\vec{a} \cdot \vec{b} = 0$ . Passing through the $y$ -axis determines the correct sign branch. The angle between $\vec{a}$ and $\hat{j}$ is acute, and rotating toward $\hat{j}$ gives $\vec{b} \cdot \hat{j} > 0$ .
execute: Express $\vec{b} = \lambda(-\hat{i} + 2\hat{j} + \hat{k}) + \mu\hat{j} = -\lambda\hat{i} + (2\lambda + \mu)\hat{j} + \lambda\hat{k}$ . Using orthogonality: $\vec{b}$ .
execute: Substitute $\frac{(3\vec{a} + \sqrt{2}\vec{b}) \cdot \vec{c}}{|\vec{c}|}$ into $\vec{b}$ : $\vec{b} = -\lambda\hat{i} - \lambda\hat{j} + \lambda\hat{k}$ . Now equate magnitudes: $|\vec{b}|^2 = |\vec{a}|^2 \implies (-\lambda)^2 + (-\lambda)^2 + \lambda^2 = (-1)^2 + 2^2 + 1^2 \implies 3\lambda^2 = 6 \implies \lambda^2 = 2 \implies \lambda = \pm\sqrt{2}$ . Since the rotation passes through the $y$ -axis, $\vec{a}$ rotates toward the positive $y$ -axis, so $\vec{b}$ has a positive $y$ -component: $-(-\sqrt{2}) = \sqrt{2} > 0$ , giving $\lambda = -\sqrt{2}$ . Thus $\vec{b} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} - \sqrt{2}\hat{k}$ .
execute: Compute $\vec{a} \cdot \vec{c} = (-1)(5) + 2(4) + 1(3) = 6$ . Compute $\vec{b} \cdot \vec{c} = (\sqrt{2})(5) + (\sqrt{2})(4) + (-\sqrt{2})(3) = 6\sqrt{2}$ . Compute $|\vec{c}| = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{50} = 5\sqrt{2}$ . Then the projection is $\frac{3(\vec{a}\cdot\vec{c}) + \sqrt{2}(\vec{b}\cdot\vec{c})}{|\vec{c}|} = \frac{3(6) + \sqrt{2}(6\sqrt{2})}{5\sqrt{2}} = \frac{18 + 12}{5\sqrt{2}} = \frac{30}{5\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$ .
verify: $3\sqrt{2} \approx 4.24$ . Check $|\vec{b}| = \sqrt{2 + 2 + 2} = \sqrt{6} = |\vec{a}|$ . $\vec{a} \cdot \vec{b} = -\sqrt{2} + 2\sqrt{2} - \sqrt{2} = 0$ . Both magnitude and orthogonality conditions are strictly satisfied.

Vector Algebra: JEE Main Mathematics Question with Solution

What feels right?

The rotated vector is $\vec{b} = -\hat{i} - \hat{j} + \hat{k}$ , making the projection of $3\vec{a}+\sqrt{2}\vec{b}$ onto $\vec{c}$ equal to $3\sqrt{2}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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