Cartesian Form of a Plane From the Vector Form

A common form for the equation of a plane is the Cartesian form  
We can deduce the Cartesian form of a plane from the vector form  
\[\vec{r}(s,t)=\vec{r}_0 + s \vec{u}+t \vec{v}\]
\[\vec{s}, \: \vec{t}\]
  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  
  substitute values of  
\[s, \: t\]
  into the vector form of the plane to get values of  
\[x, \: y, \: z\]
  and thence  
This works because the Cartesian equation of the plane is taken from  
\[\vec{n} \cdot (\vec{r}- \vec{r}_0)=\begin{pmatrix}n_1\\n_2\\n_3\end{pmatrix} \cdot (\begin{pmatrix}x\\y\\z\end{pmatrix} - \begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix})=0 \rightarrow n_1x+n_2y+n_3z=d\]
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  
. (1)
\[x=1, \: y=2, \: z=3\]
Substitute these into (1) then  
\[-6 \times 1 -3 \times 2+3 \times 3=-3=d\]
The equation of the plane is  

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