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

MATH152 April 2015

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Question B 3 (b)

Line $l_{1}$ and plane $P_{1}$ shown in the figure are given in parametric form as $l_{1}:\mathbf {x} =t\mathbf {a} ,\quad P_{1}:\mathbf {x} =\mathbf {q} +s_{1}\mathbf {b} _{1}+s_{2}\mathbf {b} _{2},$ where $\mathbf {a} ={\begin{bmatrix}0\\1\\2\end{bmatrix}},\quad \mathbf {q} ={\begin{bmatrix}1\\1\\1\end{bmatrix}},\quad \mathbf {b} _{1}={\begin{bmatrix}0\\1\\0\end{bmatrix}},\quad \mathbf {b} _{2}={\begin{bmatrix}1\\2\\-1\end{bmatrix}}.$ (b) Write the parametric form for plane $P_{2}$, which contains the line $l_{1}$ and is perpendicular to the plane $P_{1}$.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The form given in the question is actually vector form of a plane: given any point $a_{0}$ in the plane, and two nonparallel vectors $v,w$ in the plane, then any other point $X$ can be written as $$X=a_{0}+t\cdot v+s\cdot w$$ . Here, $t,s$are parameters. We already know one direction in the plane, i.e., the direction of line $l$, the target is to find one point and another direction in the plane.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\color {blue}\mathbf {x} =s_{1}{\begin{bmatrix}0\\1\\2\end{bmatrix}}+s_{2}{\begin{bmatrix}-1\\0\\-1\end{bmatrix}}.$