## Cartesian Form of a Plane From the Vector Form

A common form for the equation of a plane is the Cartesian form
$ax+by+cz=d$
.
We can deduce the Cartesian form of a plane from the vector form
$\vec{r}(s,t)=\vec{r}_0 + s \vec{u}+t \vec{v}$
where
$\vec{s}, \: \vec{t}$
are twop vectors in the plane.
$\vec{u} \times \vec {v}$
(the vector or cross product) is a vector perpendicular to both vectors, and the components of theis vector will form the coefficients
$a, \: b, \: c$
of the Cartesian form. To find the contstant
$d$
substitute values of
$s, \: t$
into the vector form of the plane to get values of
$x, \: y, \: z$
and thence
$d$
.
This works because the Cartesian equation of the plane is taken from
$\vec{n} \cdot (\vec{r}- \vec{r}_0)=\begin{pmatrix}n_1\\n_2\\n_3\end{pmatrix} \cdot (\begin{pmatrix}x\\y\\z\end{pmatrix} - \begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix})=0 \rightarrow n_1x+n_2y+n_3z=d$
.
where
$d=n_1x_0+n_2y_0+n_3z_0$
.
Example: Let
$\vec{r}(s,t)= \begin{pmatrix}1\\2\\3\end{pmatrix}+ s \begin{pmatrix}1\\0\\2\end{pmatrix}+ t \begin{pmatrix}-1\\3\\1\end{pmatrix}$
.
$\vec{u} = \begin{pmatrix}1\\0\\2\end{pmatrix}, \: \vec{v} = \begin{pmatrix}-1\\3\\1\end{pmatrix}$
.
$\vec{u} \times \vec{v} = \begin{pmatrix}1\\0\\2\end{pmatrix} \times \begin{pmatrix}-1\\3\\1\end{pmatrix}= \begin{pmatrix}0 \times 1 - 2 \times 3\\ -1 \times 2 -1 \times 1\\ 1 \times 3 -(-1) \times0\end{pmatrix}= \begin{pmatrix}-6\\-3\\3\end{pmatrix}$
.
The equation of the plane at the moment is
$-6x-3y+3z=d$
. (1)
Let
$s=t=0$
then
$x=1, \: y=2, \: z=3$
.
Substitute these into (1) then
$-6 \times 1 -3 \times 2+3 \times 3=-3=d$
.
The equation of the plane is
$-6x-3y+3z=-3$
.