Plane

Topics: Geometry - 3D Space

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Let $P=(P_1,P_2,P_3)$ be a point in ${\mathbb{R}}^3$ and $u=(u_1,u_2,u_3),v=(v_1,v_2,v_3)$ be linearly independent vectors in ${\mathbb{R}}^3$.

For relationships between planes and points, refer to Planes and Points.

Representations §

Vectorial Representation §

In its vectorial representation, the plane that passes through $P$ and is generated by $u$ and $v$ is the set:

$$\Pi :=\{P+tu+sv:t,s\in \mathbb{R}\}$$

Parametric Representation §

In its parametric representation, the plane that passes through $P$ and is generated by $u$ and $v$ is:

$$\Pi =\{\begin{array}{l}x=P_1+s{u}_1+t{v}_1\\ y=P_2+s{u}_2+t{v}_2\\ z=P_3+s{u}_3+t{v}_3\end{array}$$

Normal Representation §

In its normal representation, the plane that passes through $p$ and is generated by $u$ and $v$ is:

$$\Pi =\{q\in {\mathbb{R}}^3\mid n\cdot (q-p)=0\}$$

…where $n=u\times v$ is the plane’s normal vector, and $q$ is any point such that the equation is satisfied.

Equation §

All planes have an associated linear equation with three variables:

$$Ax+By+Cz+D=0$$

Example

The following plane in its parametric representation:

$$\Pi =\{\begin{array}{l}x=P_1+s{u}_1+t{v}_1\\ y=P_2+s{u}_2+t{v}_2\\ z=P_3+s{u}_3+t{v}_3\end{array}$$

…has the normal representation:

$$n\cdot (\hat{x}-p)=\left|\begin{array}{ccc}x& x-5& x\\ 5& 2& -2\\ 0& -2& -2\end{array}\right|=0$$

Upon solving this determinant, we get the normal equation for this plane:

$$-8x+10y-10z-50=0$$

…and from there, we can see that the normal vector is $n=(-8,10,-10)$.