The perpendicular distance of the origin from the plane $x-3 y+4 z-6=0$ is
The perpendicular distance of the origin from the plane $x-3 y+4 z-6=0$ is
$6$
$\frac{6}{\sqrt{26}}$
$\frac{1}{\sqrt{26}}$
$\frac{3}{\sqrt{26}}$
Solution
Length of perpendicular from point $\mathrm{O}(0,0,0)$ to plane $x-3 y+4 z-6=0$ is given by
$\begin{aligned}
d & =\left|\frac{0(1)+0(-3)+0(4)-6}{\sqrt{(1)^2+(-3)^2+(4)^2}}\right| \\
& =\left|\frac{-6}{\sqrt{1+9+16}}\right|=\frac{6}{\sqrt{26}}
\end{aligned}$