Question

Find the distance of the point $(1,2,-1)$ from the plane $x-2y+4z-10=0$.

Answer 1

Distance of the point $(x_1,y_1,z_1)$ to plane $ax+by+cz+d=0$ is $D=\left|\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

We have $x_1=1,y_1=2,z_1=-1,a=1,b=-2,c=4,d=-10$

$\therefore D=\left|\frac{1\left(1\right)+(-2)2+4(-1)-10}{\sqrt{1^2+(-2{)}^2+4^2}}\right|$

$=\left|\frac{1-4-4-10}{\sqrt{1+4+16}}\right|$

$=\left|\frac{-17}{\sqrt{21}}\right|$

$\therefore D=\frac{17}{\sqrt{21}}units$