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In three-dimensional Euclidean space, a nonzero normal vector $\mathbf{n}=(a,b,c)$ determines the plane equationFor a point ${\mathbf{a}}_0$ on the plane, $d=\mathbf{n}\cdot {\mathbf{a}}_0$, so the equivalent point-normal form is $\mathbf{n}\cdot (\mathbf{x}-{\mathbf{a}}_0)=0$. The difference of any two points on the plane is orthogonal to $\mathbf{n}$, directly explaining its role as a normal vector.

Multiplying $a,b,c,d$ by the same nonzero scalar preserves the set of solutions. Thus $2x-y+2z=3$ and $-6x+3y-6z=-9$ describe the same plane. Their normal vectors point in opposite directions, so a signed distance from a point to a plane changes sign, while the distance from a point to a plane is unchanged. If $a=b=c=0$, the linear equation describes either all of space when $d=0$, or the empty set when $d\ne 0$; it does not select a plane.

The chosen unit normal points toward the plane from the origin, so the plane equation isTo locate the circle centre, project $\mathbf{p}$ onto this plane. PutThen $\mathbf{n}\cdot \mathbf{h}=d$, so $\mathbf{h}$ lies in the plane. For any other point $\mathbf{r}$ of the plane, $\mathbf{r}-\mathbf{h}$ is perpendicular to $\mathbf{n}$. The Pythagorean theorem therefore givesRestricting the sphere equation to the plane consequently gives a circle centred at $\mathbf{h}$. Thus $\mathbf{m}=\mathbf{h}$, proving that the displacement of its centre is purely normal:The same Pythagorean theorem gives $q^2={\rho }^2+{\gamma }^2$, and henceThe sign of $\gamma$ is fixed by which side of the plane contains $\mathbf{p}$; $q$ and $\rho$ alone determine only its magnitude. Substituting the signed offset yields the sphere-plane intersection radiusA genuine circle requires $|d-\mathbf{n}\cdot \mathbf{p}|<q$; equality gives tangency and $\rho =0$, while a larger distance gives no intersection. The signed distance from a point to a plane in the direction $\mathbf{n}$ is $\mathbf{n}\cdot \mathbf{p}-d=-\gamma$; the ordinary distance from a point to a plane is its absolute value.

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.