Distance of a Point from a Plane

The distance of the origin from a plane already written in normal form $\vec r\cdot\hat n=p$ is simply $p$ — this is the very meaning of the normal form (Fig. 6.13). To find the distance of an arbitrary point $A(\vec a)$ from the plane $\vec r\cdot\hat n=p$, construct the plane through $A$ parallel to the given one: a parallel plane shares the same normal $\hat n$, so its equation is $\vec r\cdot\hat n=\vec a\cdot\hat n$. Since this new plane is itself already in normal form, the distance of the origin from it is, by the same normal-form fact, simply $\vec a\cdot\hat n$ (Fig. 6.14 pictures the two parallel planes together — the original at distance $p$ from $O$, and the one through $A$ at distance $\vec a\cdot\hat n$ from $O$). The distance from $A$ to the original plane is then the difference of these two origin-distances, made non-negative:

This formula only works once the plane's equation is genuinely in normal form (unit normal, non-negative $p$) — a plane given as $\vec r\cdot\vec n=d$ with $\vec n$ not a unit vector must first be divided through by $|\vec n|$ before this formula can be used.

Worked example. Find the distance of the point $4\hat i-3\hat j+2\hat k$ from the plane $\vec r\cdot(-2\hat i+\hat j-2\hat k)=6$.

Here $\vec a=4\hat i-3\hat j+2\hat k$ and $\vec n=-2\hat i+\hat j-2\hat k$, so $|\vec n|=\sqrt{4+1+4}=3$, giving $\hat n=\dfrac{-2\hat i+\hat j-2\hat k}{3}$.

The normal form of the given plane is $\vec r\cdot\hat n=\dfrac{6}{3}=2$, so $p=2$. …