Plane ($P$)

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]

Two planes $n_1\cdot r=d_1$, $n_2\cdot r=d_2$ with independent normal vectors meet in a straight line with direction $n_1\times n_2$. Any difference of two points on both planes is perpendicular to both normal vectors, which proves the direction statement. A particular point follows by solving the two linear equations; adding a real multiple of the direction gives the parametric equation of a straight line. For parallel normal vectors, the planes are either coincident or disjoint instead.

Suppose $\mathbf{a}\ne 0$ and $\mathbf{a}\cdot \mathbf{b}=0$. The line $\mathbf{r}\times \mathbf{a}=\mathbf{b}$ has parametrization $\mathbf{r}={\mathbf{r}}_0+t\mathbf{a}$, where ${\mathbf{r}}_0=(\mathbf{a}\times \mathbf{b})/|\mathbf{a}{|}^2$. For a point $\mathbf{y}$ outside the line, a normal vector of the unique plane containing both is $\mathbf{N}=\mathbf{a}\times (\mathbf{y}-{\mathbf{r}}_0)=\mathbf{b}+\mathbf{a}\times \mathbf{y}$. Thus its equation is the one in the title. The vector triple product verifies ${\mathbf{r}}_0\times \mathbf{a}=\mathbf{b}$, and cyclic invariance of the scalar triple product verifies that every point of the line satisfies the equation. The excluded case $\mathbf{N}=0$ is exactly $\mathbf{y}$ lying on the line, when the plane is not unique.

The acute angle between two unoriented planes is the acute angle between their nonzero normal vectors, giving the displayed formula. For an oriented convex polyhedron, the interior dihedral angle of a convex polyhedron is instead $\pi$ minus the angle between outward face normals. The distinction matters when a face normal is reversed: its unoriented plane is unchanged, but its vector angle is replaced by its supplement.

For smooth level surfaces $f=0$, $g=0$ meeting at a point where both gradients are nonzero, their acute angle is the angle between two planes tangent to the surfaces there. Each gradient is a normal vector to its tangent plane, proving the displayed formula. Vanishing gradients require further analysis because the level-set equation then need not specify a unique tangent plane.

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

Planes with common unit normal $n$ have distance $|(a_1-a_2)\cdot n|$. Every joining displacement has this normal component, giving a lower bound by its absolute value; a displacement parallel to the normal attains the bound.

For a common nonzero normal vector $\mathbf{n}$, not necessarily of unit Euclidean norm, the planes $\mathbf{n}\cdot \mathbf{x}=d_1$ and $\mathbf{n}\cdot \mathbf{x}=d_2$ have distance $|d_2-d_1|/\Vert \mathbf{n}\Vert$. Thus $2x-y+2z=3$ and $2x-y+2z=9$ are distance $6/3=2$ apart. First put both plane equations into the same normal-vector scale: subtracting offsets from independently rescaled equations need not give a geometric distance.

A line $r=b+sm$ intersects a plane $(r-a)\cdot n=0$ exactly when the scalar equation $s(m\cdot n)=(a-b)\cdot n$ can be solved. If $m\cdot n\ne 0$, there is one point. If that product vanishes, the line is either contained in the plane or disjoint from it. Direction and normal vectors must be nonzero.

For three non-collinear points with position vectors $\mathbf{a},\mathbf{b},\mathbf{c}$, a nonzero normal vector isThe plane is the set of $\mathbf{r}$ satisfying $(\mathbf{r}-\mathbf{a})\cdot \mathbf{N}=0$. Expanding the cross product and using the scalar triple product givesNon-collinearity is essential: otherwise $\mathbf{N}=0$ and the equation fails to select a unique plane.

The Euclidean plane is ${\mathbb{R}}^2$ with its standard Euclidean norm and the resulting distance, angles, and Riemannian metric.

A connected open subset of the Euclidean plane. For a bounded polygonal example, the Dirichlet Laplacian is the self-adjoint operator associated with the Dirichlet energy on $H_0^1(\Omega )$; its eigenfunctions are smooth in the interior but need not be smooth at reentrant boundary corners.