 Signed distance from a point to a plane

Orient the plane $\mathbf{n}\cdot \mathbf{r}=d$ with a unit normal $\mathbf{n}$. A point $\mathbf{p}$ then has signed distance from a point to a planeIts absolute value is the ordinary distance from a point to a plane. The closest point on a plane is $\mathbf{p}-s\mathbf{n}$, so the displacement from the point to its orthogonal projection onto a plane is $-s\mathbf{n}$. Reversing the unit normal and $d$ reverses $s$ without changing the plane or the distance from a point to a plane. A sphere of radius $q$ centred at $\mathbf{p}$ has sphere-plane intersection radius $\sqrt{q^2-s^2}$ when $|s|<q$.

For a general, not necessarily unit, normal vector ${\mathbf{n}}_0$ in ${\mathbf{n}}_0\cdot \mathbf{x}=d_0$, the oriented distance is $({\mathbf{n}}_0\cdot \mathbf{p}-d_0)/\Vert {\mathbf{n}}_0\Vert$. In the worked example of distance from a point to a plane, orientation toward $(2,-1,2)$ assigns $+8/3$ to $(4,1,2)$; reversing that unit normal assigns $-8/3$. The closest point on a plane and the unsigned distance are unaffected.