Mathematics · Vector Algebra

JEE Main 2025 — 24 January, Morning Shift — Question 1

Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}$ be three vectors such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$ . If the vector

$\overrightarrow{\mathrm{c}}$ is perpendicular to $\vec{b}$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=5$ , then $|\overrightarrow{\mathbf{c}}|$ is equal to

Option A:
$\frac{1}{3 \sqrt{2}}$
Option B:
18
Option C:
16
Option D:
$\sqrt{\frac{11}{6}}$
Correct

Answer: D

Step-by-step solution

$\overrightarrow{\mathrm{c}}=\lambda(\overrightarrow{\mathrm{b}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}))$

$=\lambda((\vec{b} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{b}) \vec{b})$

$=\lambda(11 \vec{a}-2 \vec{b})=\lambda(11 \mathrm{i}+22 \mathrm{j}+33 \mathrm{k}-6 \mathrm{i}-2 \mathrm{j}+2 \mathrm{k})$

$=\lambda(5 \mathrm{i}+20 \mathrm{j}+35 \mathrm{k})$

$=5 \lambda(\mathrm{i}+4 \mathrm{j}+7 \mathrm{k})$

$=$ Given $\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=5$

$=5 \lambda(1+8+21)=5 => \lambda=\frac{1}{30}$

$\Rightarrow \overrightarrow{\mathrm{c}}=\frac{1}{6}(\mathrm{i}+4 \mathrm{j}+7 \mathrm{k})$

$|\overrightarrow{\mathrm{c}}|=\frac{\sqrt{1+16+49}}{6}=\sqrt{\frac{11}{6}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: Vector Algebra
Topic: Collinearity and Coplanarity of Vectors and Points