Question

Find the distance of point $(2,3,-5)$ from the plane $x+2y-2z-9=0$

Answer 1

We have,

$x+2y-2z-9=0$

Distance from the given point $(2,3,-5)$

$=\frac{2\times 1+3\times 2-5\times -2-9}{\sqrt{1+4+4}}$

$=\frac{2+6+10-9}{\sqrt{9}}$

$=\frac{9}{3}$

$=3units$