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 :

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Vector Algebra

Let P be a point in the plane of the vector

AB = 3 \hat{i} + \hat{j} - \hat{k}

and

AC = \hat{i} - \hat{j} + 3 \hat{k}

such that P is equidistant from the lines AB and AC . If

∣ AP ∣ = \frac{5}{2}

, then the area of the triangle ABP is :

Answer: (C)

Since

P

is equidistant from the lines

A B

and

A C

, it lies on a bisector of the angle at

A

. With

A B = 3 \hat{i} + \hat{j} - \hat{k}

and

A C = \hat{i} - \hat{j} + 3 \hat{k}

∣ A B ∣ = ∣ A C ∣ = 11

and

A B \cdot A C = 3 - 1 - 3 = - 1

, so the full angle

2 θ

between them satisfies

cos 2 θ = - \frac{1}{11}

. Hence

sin^{2} θ = \frac{1 - cos 2 θ}{2} = \frac{6}{11}

, and

A P

makes angle

θ

with

A B

. Therefore

[△ A B P] = \frac{1}{2} ∣ A B ∣ ∣ A P ∣ sin θ = \frac{1}{2} \cdot 11 \cdot \frac{5}{2} \cdot \frac{6}{11} = \frac{30}{4} .

The area of triangle

A B P

\frac{30}{4}

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option