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

MATH152 April 2015

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Question B 3 (c)

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 $\mathbf {a} ={\begin{bmatrix}0\\1\\2\end{bmatrix}},\quad \mathbf {q} ={\begin{bmatrix}1\\1\\1\end{bmatrix}},\quad \mathbf {b} _{1}={\begin{bmatrix}0\\1\\0\end{bmatrix}},\quad \mathbf {b} _{2}={\begin{bmatrix}1\\2\\-1\end{bmatrix}}.$ (c) Planes $P_{1}$ and $P_{2}$ intersect along the line $l_{2}$. Write the equation form for the line $l_{2}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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
Like in part (a), convert this problem into a system of linear equations, then use the solution(s) to parametrize the intersection.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. We have $P_{1}:\ \ \ {\mathbf {x}}={\begin{pmatrix}1\\1\\1\end{pmatrix}}+s_{1}{\begin{pmatrix}0\\1\\0\end{pmatrix}}+s_{2}{\begin{pmatrix}1\\2\\-1\end{pmatrix}}$ and $P_{2}:\ \ \ {\mathbf {x}}=t_{1}{\begin{pmatrix}-1\\0\\-1\end{pmatrix}}+t_{2}{\begin{pmatrix}0\\1\\2\end{pmatrix}}$ for some $s_{1}$, $s_{2}$, $t_{1}$ and $t_{2}$ from the question and part (b). It is easy to see that the line $l_{2}$ is just the intersection of plane $P_{1},P_{2}$, i. e., any point ${\mathbf {x}}$ on the line has to satisfy both equations. Thus we have following equation for line $l_{2}$ . ${\mathbf {x}}={\begin{pmatrix}1\\1\\1\end{pmatrix}}+s_{1}{\begin{pmatrix}0\\1\\0\end{pmatrix}}+s_{2}{\begin{pmatrix}1\\2\\-1\end{pmatrix}}=t_{1}{\begin{pmatrix}-1\\0\\-1\end{pmatrix}}+t_{2}{\begin{pmatrix}0\\1\\2\end{pmatrix}}$. Now let's find out the relation among parameters $s_{1},s_{2},t_{1},t_{2}$ . The first coordinate shows $1+s_{2}=-t_{1}$, while the third shows $1-s_{2}=-t_{1}+2t_{2}$. Summing these equations gives $2=-2t_{1}+2t_{2}$, hence $t_{2}=1+t_{1}$. Substituting it back gives: for any point ${\mathbf {x}}$ on line it satisfies $l_{2}:{\mathbf {x}}=t_{1}{\begin{pmatrix}-1\\0\\-1\end{pmatrix}}+(1+t_{1}){\begin{pmatrix}0\\1\\2\end{pmatrix}}=t_{1}{\begin{pmatrix}-1\\1\\1\end{pmatrix}}+{\begin{pmatrix}0\\1\\2\end{pmatrix}},$ which has is equivalent equation form; for any point ${\mathbf {x}}=(x_{1},x_{2},x_{3})^{T}$ on the line $l_{2}$ satisfies $\color {blue}{\begin{aligned}{\begin{cases}x_{1}+x_{2}=1\\x_{1}+x_{3}=2\end{cases}}\end{aligned}}$