Question

Perpendicular are drawn form points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x+y+z=3$. The feet of perpendicular lie on the line.

Answer 1

The direction of the line on which feets of perpendicular line will be same as that of line $L:\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$

Plane $:x+y+z-3=0$

$\therefore$ directions of normal are $1,1,1$

Let feet of ${\perp }^r$ be $r-2,r-1,r$

$\therefore r-2+r-1+r-3=0$

$\therefore r=2$

Feet of ${\perp }^r$ is$(0,1,2)$ for $(-2,-1,0)$ on line $L$.

$\therefore$ equation of required line is

$\frac{(x-0)}{2}=\frac{(y-1)}{-1}=\frac{(z-2)}{3}$