Let P be a point in the plane of the vector overrightarrow AB =3 hat i +hat j -hat k and overrightarrow AC =hat i -hat j +3 hat k such that P is equidistant from the lines AB and AC . If overrightarrow AP =frac √(5) 2 , then the area of the triangle ABP is :

Mathematics · Vector Algebra

JEE Main 2026 — 28 January, Evening Shift — Question 17

Let P be a point in the plane of the vector $\overrightarrow{\mathrm{AB}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{AC}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ such that P is equidistant from the lines AB and AC . If $|\overrightarrow{\mathrm{AP}}|=\frac{\sqrt{5}}{2}$ , then the area of the triangle ABP is :

Option A:
2
Option B:
$\frac{3}{2}$
Option C:
$\frac{\sqrt{30}}{4}$
Correct
Option D:
$\frac{\sqrt{26}}{4}$

Answer: C

Step-by-step solution

$\quad \cos 2 \theta=\frac{3-1-3}{\sqrt{11} \cdot \sqrt{11}}=-\frac{1}{11}$

$1-2 \sin ^{2} \theta=-\frac{1}{11} \Rightarrow 2 \sin ^{2} \theta=\frac{12}{11}$

$\Rightarrow \sin \theta=\sqrt{\frac{6}{11}}$

$\therefore \operatorname{Area}(\triangle \mathrm{APB})=\frac{1}{2} \times \sqrt{11} \cdot \frac{\sqrt{5}}{2} \cdot \sqrt{\frac{6}{11}}=\frac{\sqrt{30}}{4}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Vector Algebra
Topic: Introduction to Vectors