Question

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

Answer 1

We have,

$\frac{x-2}{1}=\frac{y-3}{2}=\frac{3+5}{-2}$

$Q=\left(r+1,2r+3,-2r-5\right)$

$\Rightarrow r+2+2\left(2r+3\right)-2\left(-2r-5\right)=9$

$\Rightarrow r+2+4r+6+4r+10=9$

$\Rightarrow 9r=-9$

$r=\frac{-9}{9}=-1$

$=(1,1,-3)$

Then,

We get $PQ=\sqrt{1+4+4}=3$