\[ax+by+cz=d\]
.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
\[-6x-3y+3z=-3\]
.