Question

Find the distance of the point $P(1,0,-3)$ from the plane $\omega :x-y-z=9$, measured parallel to the line $L$ given by $\frac{x-2}{2}=\frac{y+2}{3}=\frac{z-6}{-6}$

Answer 1

Perpendicular distance of $P(1,0,-3)$ from plane is $\frac{\mid 1-0+3-9\mid }{\sqrt{1+1+1}}=\frac{5}{\sqrt{3}}$

angle between the plane and line is ${\text{sin}}^{-1}(\frac{2\left(1\right)+3(-1)+(-6)(-1)}{\sqrt{4+9+36}\sqrt{1+1+1}})={\text{sin}}^{-1}(\frac{5}{7\sqrt{3}})$

$⟹$ required distance $=\frac{5}{\sqrt{3}}\times \frac{7\sqrt{3}}{5}=7$