Point-plane distance

Property — Point–plane distance

The distance of a point $P$ from the plane $\pi\colon ax+by+cz+d=0$ is $d(P,\pi) = \frac{|ax_P+by_P+cz_P+d|}{\sqrt{a^2+b^2+c^2}}.$

Example

Let $P(0;1;6)$ and $\pi: 4x-6y+2z-6=0$. The magnitude of the normal is $|\vec n|=\sqrt{16+36+4}=\sqrt{56}$, and $d=\frac{|0-6+12-6|}{\sqrt{56}} = 0:$ the point lies on the plane. Changing $P$ to $(0;1;4)$ one obtains $d = \frac{|0-6+8-6|}{\sqrt{56}}=\frac{|-4|}{\sqrt{56}}\approx 0{,}53.$

Links

Topics: Analytic geometry in space
Concepts: Point-plane distance · Plane · Cartesian plane
Skills: Analytic geometry · Using formulae