JEE Main 2022 (Online) 28th July Morning ShiftQuestion 1
The foot of the perpendicular from a point on the circle $x^2+y^2=1$, $z=0$ to the plane $2x+3y+z=6$ lies on which one of the following curves ?
3D Geometry
Solution
Point on circle $(\cos\theta,\sin\theta,0)$; foot $=(\cos\theta,\sin\theta,0)+q(2,3,1)$ with $q=\frac{6-2\cos\theta-3\sin\theta}{14}$. So $x=\cos\theta+2q$, $y=\sin\theta+3q$, $z=6-2x-3y$ (on plane). Eliminating $\theta$ gives a relation $\cos^2\theta+\sin^2\theta=1$ in terms of $x,y$; solving $\cos\theta,\sin\theta$ as linear functions of $x,y$ and substituting yields $(5x+6y-12)^2+4(3x+5y-9)^2=1$. Numerical check at several $\theta$ values: only option (B) evaluates to exactly 1. Option (B).