JEE Main 2022 (Online) 29th July Evening ShiftQuestion 1
Let $Q$ be the foot of perpendicular drawn from the point $P(1,2,3)$ to the plane $x+2y+z=14$. If $R$ is a point on the plane such that $\angle PRQ=60^\circ$, then the area of $\Delta PQR$ is equal to :
3D Geometry
Solution
Normal $(1,2,1)$; $\frac{14-(1+4+3)}{6}=1$, so $Q=P+(1,2,1)=(2,4,4)$ and $PQ=\sqrt6$. $PQ\perp$ plane, so $\triangle PQR$ is right-angled at $Q$ with $\angle R=60^\circ$: $QR=\frac{PQ}{\tan60^\circ}=\frac{\sqrt6}{\sqrt3}=\sqrt2$. Area $=\frac12\sqrt6\cdot\sqrt2=\sqrt3$. Option (B).