Question

A distance between the line $\frac{x-2}{1}=\frac{y+2}{1}=\frac{z-3}{4}$ and the plane x +5 y +s = 5 is.

Answer 1

$\frac{x-2}{1}=\frac{y+2}{1}=\frac{z-3}{4}=\lambda$

$x+5y+z=5$

Any point on line call be

$(\lambda +2,\lambda -2,4\lambda +3)$

Lets find point of intersection of line and plane

$\Rightarrow \lambda +2+5\lambda -10+4\lambda +3=5$

$\Rightarrow 10\lambda -5=5$

$\Rightarrow \lambda =1$

$\therefore$ line intersects plane at $(1+2,1-2,4+3)$

$=(3,-1,7)$

$\therefore$ Distance between line & plane is $0$.