Distance from a point to a plane

The distance from a point to a plane is the length of the “shortest” line segment from the point to the plane.

The problem

We have a plane given by the Cartesian equation r: 3x + 1y − 2z + 4 = 0 and a point X(5, 7, 2). We want to know the distance from the point X to the plane r. We could draw it like this:

In the figure we have some plane ABCD and a point X. The length of the line segment XF gives the distance from the point X to the plane. Again this line segment is perpendicular to the plane, so it is the shortest distance from the point to the plane. (The figure does not show the plane r: 3x + 1y − 2z + 4 = 0 or the point X(5, 7, 2), only the distance from a point to a plane in general.)

Using the formula

We can use practically the same formula as for the distance from a point to a line; we just add one more dimension. So the distance between a point X(x₁, x₂, x₃) and a plane r: ax + by + cz + d = 0 is

$$d(X,r)=\frac{|a\cdot x_1+b\cdot x_2+c\cdot x_3+d|}{\sqrt{a^2+b^2+c^2}}$$

For the plane r and the point X(5, 7, 2) from the problem, we substitute a = 3, b = 1, c = −2, d = 4 and x₁ = 5, x₂ = 7, x₃ = 2:

$$\begin{array}{rcl}d(X,r)& =& \frac{|3\cdot 5+1\cdot 7-2\cdot 2+4|}{\sqrt{3^2+1^2+(-2{)}^2}}\\ & =& \frac{22}{\sqrt{14}}\approx 5.880\end{array}$$