JEE Main 2022 (Online) 29th July Morning ShiftQuestion 1
If the foot of the perpendicular from the point $A(-1,4,3)$ on the plane $P:2x+my+nz=4$, is $\left(-2,\frac72,\frac32\right)$, then the distance of the point $A$ from the plane $P$, measured parallel to a line with direction ratios $3,-1,-4$, is equal to :
3D Geometry
Solution
$\vec{AF}=(-1,-\tfrac12,-\tfrac32)\parallel(2,1,3)$ is the normal, so plane: $2x+y+3z=k$; the foot gives $k=-4+\tfrac72+\tfrac92=4$ (matches $=4$). Line from $A$ with direction $(3,-1,-4)$: $(-1+3t,\,4-t,\,3-4t)$. On plane: $-2+6t+4-t+9-12t=4\Rightarrow11-7t=4\Rightarrow t=1$. Distance $=\sqrt{9+1+16}=\sqrt{26}$. Option (B).