Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 3 (b)/Solution 1

According to definition of parametric form of plane, we will write $P_{2}$ as

${\mathbf {x}}={\mathbf {c}}+s_{1}{\mathbf {v_{1}}}+s_{2}{\mathbf {v_{2}}}$

Our purpose is to find $c,{\mathbf {v_{1}}},{\mathbf {v_{2}}}$. As hinted, since $P_{2}$ contains $l_{1}$, we can set ${\mathbf {v_{1}}}={\mathbf {a}}$. Now let's look for ${\mathbf {v_{2}}}$, in fact, we know $P_{2}$ is perpendicular to $P_{1}$, it not hard to see that the normal vector of $P_{1}$ is on the plane $P_{1}$, this normal vector is going to be ${\mathbf {v_{2}}}$.

We are already given the parametric form of $P_{1}$, by the standard method of finding normal vector, we need to apply cross product to two nonparallel vectors ${\mathbf {b}}_{1},{\mathbf {b}}_{2}$ in the plane $P_{1}$.

The cross product ${\mathbf {b}}_{1}\times {\mathbf {b}}_{2}$ is given by the formal determinant

$\det {\begin{bmatrix}{\hat {\mathbf {i} }}&{\hat {\mathbf {j} }}&{\hat {\mathbf {k} }}\\0&1&0\\1&2&-1\end{bmatrix}}=\det {\begin{bmatrix}1&0\\2&-1\end{bmatrix}}{\hat {\mathbf {i} }}\ -\ \det {\begin{bmatrix}0&0\\1&-1\end{bmatrix}}{\hat {\mathbf {j} }}\ +\ \det {\begin{bmatrix}0&1\\1&2\end{bmatrix}}{\hat {\mathbf {k} }}=-{\hat {\mathbf {i} }}-{\hat {\mathbf {k} }}={\begin{bmatrix}-1\\0\\-1\end{bmatrix}},$

so we set

${\mathbf {v_{2}}}={\begin{bmatrix}-1\\0\\-1\end{bmatrix}}.$

It remains to compute ${\mathbf {c}}$. Notice that the origin is in $l_{1}$, so it is in $P_{2}$ since $P_{2}$ contains $l_{1}$. Hence we can take ${\mathbf {c}}=0$. It follows that a parametric form of the equation for $P_{2}$ is

$\color {blue}\mathbf {x} =s_{1}{\begin{bmatrix}0\\1\\2\end{bmatrix}}+s_{2}{\begin{bmatrix}-1\\0\\-1\end{bmatrix}}.$