JEE Main 2026 Jan 28 Shift 2, Mathematics Q17: Applications of Vectors

$P$ equidistant from lines $AB$ and $AC$ (and lying in their plane) means $AP$ is along the angle bisector at $A$ — the angle $\theta =\angle BAP=\angle PAC=\frac{1}{2}\angle BAC$.

Compute $\cos (\angle BAC)=\frac{\overrightarrow{AB}\cdot \overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|}$. Dot product $=3-1-3=-1$. $|\overrightarrow{AB}|=\sqrt{9+1+1}=\sqrt{11}$; $|\overrightarrow{AC}|=\sqrt{1+1+9}=\sqrt{11}$.

$\cos (\angle BAC)=-\frac{1}{11}$. Using $\cos 2\theta =1-2{\sin }^2\theta$: $-\frac{1}{11}=1-2{\sin }^2\theta \Rightarrow {\sin }^2\theta =\frac{6}{11}\Rightarrow \sin \theta =\sqrt{\frac{6}{11}}$.

Area of $△ABP=\frac{1}{2}\cdot |AB|\cdot |AP|\cdot \sin \theta =\frac{1}{2}\cdot \sqrt{11}\cdot \frac{\sqrt{5}}{2}\cdot \sqrt{\frac{6}{11}}=\frac{\sqrt{30}}{4}$.