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

$\mathbf{𝐱}=\mathbf{𝐜}+s_1{\mathbf{v}}_{\mathbf{𝟏}}+s_2{\mathbf{v}}_{\mathbf{𝟐}}$

Our purpose is to find $c,{\mathbf{v}}_{\mathbf{𝟏}},{\mathbf{v}}_{\mathbf{𝟐}}$. As hinted, since $P_2$ contains $l_1$, we can set ${\mathbf{v}}_{\mathbf{𝟏}}=\mathbf{𝐚}$. Now let's look for ${\mathbf{v}}_{\mathbf{𝟐}}$, 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 ${\mathbf{v}}_{\mathbf{𝟐}}$.

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

$\det \left[\begin{array}{ccc}\widehat{\mathbf{𝐢}}& \widehat{\mathbf{𝐣}}& \widehat{\mathbf{𝐤}}\\ 0& 1& 0\\ 1& 2& -1\end{array}\right]=\det \left[\begin{array}{cc}1& 0\\ 2& -1\end{array}\right]\widehat{\mathbf{𝐢}}-\det \left[\begin{array}{cc}0& 0\\ 1& -1\end{array}\right]\widehat{\mathbf{𝐣}}+\det \left[\begin{array}{cc}0& 1\\ 1& 2\end{array}\right]\widehat{\mathbf{𝐤}}=-\widehat{\mathbf{𝐢}}-\widehat{\mathbf{𝐤}}=\left[\begin{array}{c}-1\\ 0\\ -1\end{array}\right],$

so we set

${\mathbf{v}}_{\mathbf{𝟐}}=\left[\begin{array}{c}-1\\ 0\\ -1\end{array}\right].$

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

$\mathbf{𝐱}=s_1\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]+s_2\left[\begin{array}{c}-1\\ 0\\ -1\end{array}\right].$