Equation of a Plane

Given a normal vector to the plane $n$ , and $r$ , $r_{0}$ the position vectors of two points in the plane, then we have the vector form equation of a plane:

$\begin{eqnarray*} n \cdot (\vec{r} - \vec{r_{0}}) &=& 0 \end{eqnarray*}$

This can be expanded using commutativity:

$\begin{eqnarray*} n \cdot (\vec{r} - \vec{r_{0}}) &=& n\cdot\vec{r} - n\cdot\vec{r_{0}} \end{eqnarray*}$

in order to obtain:

$\begin{eqnarray*} n\cdot\vec{r} &=& n\cdot\vec{r_{0}} \end{eqnarray*}$

Since each vector in the plane most be orthogonal to the normal vector $n$ , then it means that the vector $r-r_{0}$ is a vector in the plane.

With some simplification and substituting $n=<a,b,c>$ , $r=<x,y,z>$ and $r_{0}=<x_{0}, y_{0}, z_{0}>$ we obtain a parametric equation:

$\begin{eqnarray*} a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) &=& 0 \end{eqnarray*}$

Find the Equation of a Plane Defined by Three Points

Since a plane is defined by three points in geometry, we can use the parametric equation form to determine the equation of the plane that contains all three points - there is only one such plane but can be described by several equations.

Suppose that the given points are:

$\begin{eqnarray*} R &=& (r_{x}, r_{y}, r_{z}) \\ S &=& (s_{x}, s_{y}, s_{z}) \\ T &=& (t_{x}, t_{y}, t_{z}) \end{eqnarray*}$

We compute the normal vector by computing the cross-product of the vectors between any two points - for example, the vector $\vec{a}$ from $R$ to $S$ and the vector $\vec{b}$ from $S$ to $T$ :

$\begin{eqnarray*} \vec{a} &=& <s_{x}-r_{x}, s_{y}-r_{y}, s_{z}-r_{z}> \\ \vec{b} &=& <t_{x}-s_{x}, t_{y}-s_{y}, t_{z}-s_{z}> \end{eqnarray*}$

leadning to:

$\begin{eqnarray*} \vec{n} &=& \vec{a} \times \vec{b} \\ &=& \left| \begin{matrix} i & j & k \\ s_{x}-r_{x} & s_{y}-r_{y} & s_{z}-r_{z} \\ t_{x}-s_{x} & t_{y}-s_{y} & t_{z}-s_{z} \end{matrix} \right| \end{eqnarray*}$

and expanding the determinant by the cofactors of $i$ , $j$ and $k$ :

$\begin{eqnarray*} \vec{n} &=& <n_{i}, n_{j}, n_{k}> \end{eqnarray*}$

where:

$\begin{eqnarray*} n_{i} &=& (s_{y}-r_{y})*(t_{z}-s_{z})-(s_{z}-r_{z})*(t_{y}-s_{y}) \\ n_{j} &=& (s_{x}-r_{x})*(t_{z}-s_{z})-(s_{z}-r_{z})*(t_{x}-s_{x}) \\ n_{k} &=& (s_{x}-r_{x})*(t_{y}-s_{y})-(s_{y}-r_{y})*(t_{x}-s_{x}) \end{eqnarray*}$

Next, in addition to the normal vector, we need a point to describe the plane by and we can pick any of the three points chosen. Any of the resulting equation will be valid. Let's pick the first one $R$ which will lead to the plane equation:

$\begin{eqnarray*} n_{i}(x-r_{x}) + n_{j}(y-r_{y}) + n_{k}(z-r_{z}) = 0 \end{eqnarray*}$

conversely, if we picked point $S$ , we would have obtained:

$\begin{eqnarray*} n_{i}(x-s_{x}) + n_{j}(y-s_{y}) + n_{k}(z-s_{z}) = 0 \end{eqnarray*}$

and, for point $T$ :

$\begin{eqnarray*} n_{i}(x-t_{x}) + n_{j}(y-t_{y}) + n_{k}(z-t_{z}) = 0 \end{eqnarray*}$

Any of these three equations describe a plane that contains all of the points $R$ , $S$ , $T$ .

We can transform the equations back into vector form by knowing the normal vector $n$ and making pairs of $R$ , $S$ and $T$ :

$\begin{eqnarray*} n \cdot (S-R) &=& 0 \\ n \cdot (T-S) &=& 0 \\ n \cdot (T-R) &=& 0 \\ \cdots \end{eqnarray*}$