Mathematics · Vector Algebra

JEE Main 2025 — 8 April, Evening Shift — Question 28

Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$ . Let $\hat{c}$ be a unit vector in the plane of the vector $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$ . Then such a vector $\hat{c}$ is:

Option A:
$\frac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})$
Option B:
$\frac{1}{\sqrt{5}}(\hat{j}-2 \hat{k})$
Option C:
$\frac{1}{\sqrt{2}}(-\hat{i}+\hat{k})$
Correct
Option D:
$\frac{1}{\sqrt{3}}(-\hat{i}+\hat{j}-\hat{k})$

Answer: C

Step-by-step solution

$\vec{c}=x \vec{a}+y \vec{b}$

\begin{aligned} & \vec{c}=x(\hat{i}+2 \hat{j}+\hat{k})+y(2 \hat{i}+\hat{j}-\hat{k}) \\& \vec{a} \cdot \vec{c}=(\hat{i}+2 \hat{j}+\hat{k}) \cdot(x(\hat{i}+2 \hat{j}+\hat{k})+y(2 \hat{i}+\hat{j}-\hat{k})) \\& (\hat{i}+2 \hat{j}+\hat{k}) \cdot(x \hat{i}+2 x \hat{j}+x \hat{k})+2 y \hat{i}+y \hat{j}-y \hat{k}=0 \\& \Rightarrow \quad(x+2 y)+2(x+9)+(x-y)=0 \\& \Rightarrow \quad y=-2 x \\& \therefore \quad \vec{c}=x(-3 \hat{i}+3 \hat{k}) \\& |\vec{c}|=|x| \sqrt{9+9}=3|x| \sqrt{2} \\& \therefore \quad|\vec{c}|=1 \\& 3|x| \sqrt{2}=1 \\& |x|=\frac{1}{3 \sqrt{2}} \end{aligned}

Let $x=\frac{1}{3 \sqrt{2}}$

$\vec{c}=\frac{1}{3 \sqrt{2}}(-3 \hat{i}+3 \hat{k})$

or $\vec{c}=\frac{1}{\sqrt{2}}(-\hat{i}+\hat{k})$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 8 April, Evening Shift
Subject: Mathematics
Chapter: Vector Algebra
Topic: Applications of Vectors