Cartesian equation of a plane ($\mathbf{n}\cdot \mathbf{x}=d$)

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.

Let $\mathbf{p}\ne 0$ and choose an auxiliary vector $\mathbf{q}$ with $\mathbf{p}\times \mathbf{q}\ne 0$. Its component perpendicular to $\mathbf{p}$ is ${\mathbf{q}}_{\perp }=\mathbf{q}-(\mathbf{p}\cdot \mathbf{q})\mathbf{p}/|\mathbf{p}{|}^2$. Then all solutions of the Cartesian equation of a plane $\mathbf{x}\cdot \mathbf{p}=k$ areThe two free directions are nonzero, perpendicular to each other and perpendicular to $\mathbf{p}$, so they span its two-dimensional orthogonal complement. This proves completeness as well as direct substitution. If $\mathbf{p}\cdot \mathbf{q}=0$, then ${\mathbf{q}}_{\perp }=\mathbf{q}$. If the dot product is nonzero, another complete form is $\mathbf{x}=\alpha \mathbf{p}+[k-\alpha |\mathbf{p}{|}^2]\mathbf{q}/(\mathbf{p}\cdot \mathbf{q})+\beta (\mathbf{p}\times \mathbf{q})$.