 Closest point on a plane

For a nonzero normal vector $\mathbf{n}$ and the plane equation $\mathbf{n}\cdot \mathbf{x}=d$, the closest point on a plane to $\mathbf{p}$ isIt subtracts only the normal displacement. Substituting gives $\mathbf{n}\cdot \mathbf{q}=d$, so $\mathbf{q}$ lies on the plane; the difference $\mathbf{p}-\mathbf{q}$ is a scalar multiple of its normal vector. These two checks identify the orthogonal projection onto a plane without solving a three-variable minimization problem.

For $\mathbf{n}=(2,-1,2)$, $d=3$, and $\mathbf{p}=(4,1,2)$, the residual $8$ givesIndeed, $(40-17+4)/9=3$, and $\Vert \mathbf{p}-\mathbf{q}\Vert =(8/9)\cdot 3=8/3$. Choosing the plane point $(1,1,1)$ instead gives distance $\sqrt{10}>8/3$: being on the plane does not make a point the nearest one.

As a function of $\mathbf{p}$, this projection isIt is an affine map, becoming a linear map when $d=0$. Projecting a second time changes nothing. It preserves displacements along the plane and removes the normal displacement, so it is a concrete example of an idempotent map.