# How to Find the Distance Between a Point and a Plane

The distance from a point $P$ to a plane $\alpha$ is the straight line from the point down onto the plane. To find this distance, you can use the equation for the distance between a point and a plane.

Formula

### DistanceBetweenPointandPlane

The distance between a point $P=\phantom{\rule{-0.17em}{0ex}}\left({x}_{1},{y}_{1},{z}_{1}\right)$ and a plane $\alpha :$ $ax+by+cz+d=0$ is

 $D=\frac{\phantom{\rule{-0.17em}{0ex}}|a{x}_{1}+b{y}_{1}+c{z}_{1}+d|}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}.$

Example 1

You are given the point $\phantom{\rule{-0.17em}{0ex}}\left(2,1,3\right)$ and the plane $2x-y+2z=9$. In that case, $a=2$, $b=-1$, $c=2$, and $d=-9$. Put it into the formula and find the distance: $\begin{array}{llll}\hfill D& =\frac{\phantom{\rule{-0.17em}{0ex}}|2\cdot 2-1\cdot 1+2\cdot 3-9|}{\sqrt{{2}^{2}+\phantom{\rule{-0.17em}{0ex}}{\left(-1\right)}^{2}+{2}^{2}}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\phantom{\rule{-0.17em}{0ex}}|4-1+6-9|}{\sqrt{4+1+4}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\phantom{\rule{-0.17em}{0ex}}|0|}{\sqrt{9}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The point is actually in the plane! You can also find this out by inserting the coordinates of the point into the equation of the plane and see that both sides become $9$.