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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 $\mathbf{x}=\alpha \mathbf{p}+[k-\alpha |\mathbf{p}{|}^2]\mathbf{q}/(\mathbf{p}\cdot \mathbf{q})+\beta (\mathbf{p}\times \mathbf{q})$.

Use $(0,1,1)$ as an anchor. Two directions in the required plane areTheir cross product is $(-2,8,2)$, so a normal vector is $\mathbf{n}=(-1,4,1)$. The Cartesian equation of a plane isDirect substitution verifies all three given points. The independent direction vectors ensure the plane is uniquely determined.

Two directions within the plane areTheir cross product is $(1,1,1)/4$, so a unit vector normal to the plane isIts negative is equally valid. An equation of a plane through three points in vector form isor, parametrically, $\mathbf{r}=\mathbf{e}+s(\mathbf{f}-\mathbf{e})+t(\mathbf{g}-\mathbf{e})$. Since the coordinate sum of $E$ is one, the Cartesian equation of a plane isThe distance from a point to a plane is the absolute normal displacement, giving

A plane is a two-dimensional affine subspace: a point $\mathbf{a}$ together with every linear combination of two linearly independent direction vectors $\mathbf{u},\mathbf{v}$,In three-dimensional Euclidean space, a Cartesian equation of a plane fixes one linear equation $\mathbf{n}\cdot \mathbf{x}=d$, with a nonzero normal vector $\mathbf{n}$. The two free directions are orthogonal to $\mathbf{n}$. A plane extends indefinitely; a rectangular patch in a drawing marks only part of it, and does not define an edge of the plane.

For example, the plane $2x-y+2z=3$ contains $\mathbf{a}=(1,1,1)$. The vectors $\mathbf{u}=(1,2,0)$ and $\mathbf{v}=(0,2,1)$ are linearly independent and satisfy $(2,-1,2)\cdot \mathbf{u}=(2,-1,2)\cdot \mathbf{v}=0$. Hence its points are exactly $(1+s,1+2s+2t,1+t)$: substitution gives $2x-y+2z=3$ for every real $s,t$, and every solution has $s=x-1$, $t=z-1$.

The description separates position from direction. A translation of the plane changes $\mathbf{a}$ and possibly $d$, while retaining its parallel directions. A plane through the origin is a linear subspace; the displayed example is not, since the zero vector fails its linear equation. This distinction matters when applying orthogonal projection: subtract an anchor point before projecting onto the direction linear subspace.

Historically, Euclid's Euclid's Elements by Euclid, Book I Definition 7, describes a plane surface through the straight lines lying on it. David E. Joyce's accompanying guide notes that this older use can include a bounded square or circular figure. That usage differs from the whole unbounded affine subspace used here.^[ref]