John Abbott 201-SN4-RE tutoring

Distance to a plane: find the closest point

Find the closest point on a plane, calculate the distance, and verify both answers. You need dot products and the length of a vector. Coordinates use the same length unit on three perpendicular axes.

Before you start: review dot products and projection in the matrix and vector guide.

The shortest segment is perpendicular to the plane

Let $\Pi$ be the plane $ax+by+cz=k$, where $a,b,c,k$ are real constants and $\mathbf n=(a,b,c)\ne\mathbf0$. The normal vector $\mathbf n$ is perpendicular to the plane. For a point $P=(p_x,p_y,p_z)$, its perpendicular foot $H$ is the point on the plane reached by starting at the given point and travelling in the normal direction. If $P$ is already on the plane, $H=P$ and the distance is zero.

Build the formula from two conditions

The foot must lie on the plane and on the line through $P$ perpendicular to the plane. Write $H=P-t\mathbf n$, where $t$ is a real scalar. Substitute this into the plane equation: $$\begin{aligned}\mathbf n\cdot H&=k,\\\mathbf n\cdot P-t\|\mathbf n\|^2&=k.\end{aligned}$$

Define the signed numerator $s=\mathbf n\cdot P-k$. Since $\mathbf n\ne\mathbf0$, its squared length is positive, so $$\begin{aligned}t&=\frac{s}{\|\mathbf n\|^2},\\H&=P-\frac{s}{\|\mathbf n\|^2}\mathbf n,\\d&=\|t\mathbf n\|=|t|\,\|\mathbf n\|\\&=\frac{|s|}{\|\mathbf n\|}.\end{aligned}$$ Here $d$ is the distance, in the coordinate length unit. Use the squared norm for the foot and the norm for the distance.

Worked example

Find the closest point to $P=(4,1,0)$ on $\Pi:2x-y+2z=-2$, and find the distance.

Why this point really is closest

For any point $Q$ on the plane, $H-Q$ has zero dot product with $P-H$. Pythagoras gives $$\begin{aligned}\|P-Q\|^2&=\|P-H\|^2\\&\quad+\|H-Q\|^2.\end{aligned}$$ The second term is nonnegative and is zero only when $Q=H$. So $H$ is the unique closest point.

Try two independent checks