Science:Math Exam Resources/Courses/MATH152/April 2022/Question A09

MATH152 April 2022

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Question A09
Find a constant $c$ such that the plane defined by $2x+cz=2$ is parallel to the line $$L:{\begin{cases}x+2y+3z=3\\x+y+z=4\end{cases}}.$$

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
The line $L$ is given as an intersection of two planes. Can you use the equations of the two planes to find a vector that is parallel to the line?

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. The planes defining the line have normal vectors ${\vec {v}}=\left[{\begin{array}{c}1\\2\\3\end{array}}\right]{\text{ and }}{\vec {w}}=\left[{\begin{array}{c}1\\1\\1\end{array}}\right]$. Their cross-product yields a vector that is parallel to $L$: $${\vec {v}}\times {\vec {w}}=\left[{\begin{array}{c}-1\\2\\-1\end{array}}\right].$$ For this latter vector to be parallel to the plane given by $2x+cz=2$, it must be orthogonal to the vector that is normal to the plane, namely $\left[{\begin{array}{c}2\\0\\c\end{array}}\right].$ In other words, we have $$\left[{\begin{array}{c}-1\\2\\-1\end{array}}\right]\cdot \left[{\begin{array}{c}2\\0\\c\end{array}}\right]=-1\cdot 2+2\cdot 0+-1\cdot c=0,$$ for which the solution is $c=-2.$