Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 21

MATH152 April 2015

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Question A 21
Let $L:{\begin{bmatrix}1\\2\\-1\end{bmatrix}}+t{\begin{bmatrix}1\\3\\a\end{bmatrix}}$ be a line in $\mathbb {R} ^{3}$ that is parallel to $2x+2y+z=8$. Find $a$.

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
You want the line $L$ to be parallel to the plane $2x+2y+z=8$. What does this imply about the direction vector of $L$ and a normal vector for this 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. A normal vector for the plane $2x+2y+z=8$ is $(2,2,1).$ In order for the line $L$ to be parallel to the plane $2x+2y+z=8$, we need the direction vector of $L$ to be perpendicular to any normal vector for the plane. This means that the dot product of the vectors $(1,3,a)$ and $(2,2,1)$ must be zero: ${\begin{aligned}{\begin{bmatrix}1\\3\\a\end{bmatrix}}\cdot {\begin{bmatrix}2\\2\\1\end{bmatrix}}&=0\\\Longleftrightarrow 2+6+a&=0.\end{aligned}}$ Hence $a=\color {blue}-8$.