Mathematics · Vector Algebra

JEE Main 2026 — 6 April, Evening Shift — Question 34

Let $\vec{a} = 2\hat{i} +3\hat{j} +3\hat{k}$ and $\vec{b} = 6\hat{i} +3\hat{j} +3\hat{k}$ . Then the square of the area of the triangle with adjacent sides determined by the vectors $(2\vec{a}+3\vec{b})$ and $(\vec{a}-\vec{b})$ is :

Option A:
$450$
Option B:
$900$
Option C:
$1800$
Correct
Option D:
$2400$

Answer: C

Step-by-step solution

$2 \vec{a}+3 \vec{b}=2(2 \hat{i}+3 \hat{j}+3 \hat{k})+3(6 \hat{i}+3 \hat{j}+3 \hat{k})$ $\Rightarrow 22 \hat{\mathrm{i}}+15 \hat{\mathrm{j}}+15 \hat{\mathrm{k}}$ $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=-4 \hat{\mathrm{i}}$ Area $=\frac{1}{2}|(2 \vec{a}+3 \vec{b}) \times(\vec{a}-\vec{b})|$ $=\frac{1}{2}|-60 \hat{\mathrm{j}}+60 \hat{\mathrm{k}}|=\frac{1}{2} \sqrt{(60)^{2} \times 2}$ $\mathrm{A}=\frac{60}{\sqrt{2}}$ Square of area is $\mathrm{A}^{2}=1800$

Answer key and solution verified before publishing.

Practise Vector Algebra

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Mathematics
Chapter: Vector Algebra
Topic: Vector or Cross Product of Two Vectors