JEE Main 2026 — Jan 28, Shift 2 — Mathematics solution

JEE Main2026Jan 28, Shift 2Mathematics

Let $P$ be a point in the plane of the vector $A B = 3 \hat{i} + \hat{j} - \hat{k}$ and $A C = \hat{i} - \hat{j} + 3 \hat{k}$ such that $P$ is equidistant from the lines $A B$ and $A C$ . If $A P = \frac{5}{2}$ , then the area of the triangle $A B P$ is :

A
2
B
$\frac{3}{2}$
C
$\frac{30}{4}$
D
$\frac{26}{4}$

Solution

$P$ equidistant from lines $A B$ and $A C$ (and lying in their plane) means $A P$ is along the angle bisector at $A$ — the angle $θ = ∠ B A P = ∠ P A C = \frac{1}{2} ∠ B A C$ .

Compute $cos (∠ B A C) = \frac{A B \cdot A C}{∣ A B ∣∣ A C ∣}$ . Dot product $= 3 - 1 - 3 = - 1$ . $∣ A B ∣ = 9 + 1 + 1 = 11$ ; $∣ A C ∣ = 1 + 1 + 9 = 11$ .

$cos (∠ B A C) = - \frac{1}{11}$ . Using $cos 2 θ = 1 - 2 sin^{2} θ$ : $- \frac{1}{11} = 1 - 2 sin^{2} θ \Rightarrow sin^{2} θ = \frac{6}{11} \Rightarrow sin θ = \frac{6}{11}$ .

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

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