According to definition of parametric form of plane, we will write
as
![{\displaystyle {\mathbf {x}}={\mathbf {c}}+s_{1}{\mathbf {v_{1}}}+s_{2}{\mathbf {v_{2}}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/65eb47a0fa2506b128af7858194e51b6f219f3ad)
Our purpose is to find
. As hinted, since
contains
, we can set
. Now let's look for
, in fact, we know
is perpendicular to
, it not hard to see that the normal vector of
is on the plane
, this normal vector is going to be
.
We are already given the parametric form of
, by the standard method of finding normal vector, we need to apply cross product to two nonparallel vectors
in the plane
.
The cross product
is given by the formal determinant
![{\displaystyle \det {\begin{bmatrix}{\hat {\mathbf {i} }}&{\hat {\mathbf {j} }}&{\hat {\mathbf {k} }}\\0&1&0\\1&2&-1\end{bmatrix}}=\det {\begin{bmatrix}1&0\\2&-1\end{bmatrix}}{\hat {\mathbf {i} }}\ -\ \det {\begin{bmatrix}0&0\\1&-1\end{bmatrix}}{\hat {\mathbf {j} }}\ +\ \det {\begin{bmatrix}0&1\\1&2\end{bmatrix}}{\hat {\mathbf {k} }}=-{\hat {\mathbf {i} }}-{\hat {\mathbf {k} }}={\begin{bmatrix}-1\\0\\-1\end{bmatrix}},}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/52fd1e6bce4135d121ba07edd649d476b69548f6)
so we set
![{\displaystyle {\mathbf {v_{2}}}={\begin{bmatrix}-1\\0\\-1\end{bmatrix}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d3e35d51f95207e10db3f7645ce90b1cf8f5d080)
It remains to compute
. Notice that the origin is in
, so it is in
since
contains
. Hence we can take
. It follows that a parametric form of the equation for
is
![{\displaystyle \color {blue}\mathbf {x} =s_{1}{\begin{bmatrix}0\\1\\2\end{bmatrix}}+s_{2}{\begin{bmatrix}-1\\0\\-1\end{bmatrix}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/1f9dfc580836f16c658718f5c8ed253048eaaa83)