JEE Main 2021 (Online) 26th August Morning ShiftQuestion 1
Let the line $L$ be the projection of the line $\frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}$ in the plane $x-2y-z=3$. If $d$ is the distance of the point $(0,0,6)$ from $L$, then $d^{2}$ is equal to $\_\_\_\_\_\_$ .
3D Geometry
Answer$26$
Solution
Line meets the plane at $t=-6$: $(-11,-3,-8)$. Direction of projection $=d-\frac{d\cdot n}{|n|^2}n=(2,1,2)+\frac13(1,-2,-1)\parallel(7,1,5)$. $d^2=\frac{|(11,3,14)\times(7,1,5)|^2}{75}=\frac{1950}{75}=26$.