Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (a) ii

MATH200 December 2013

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Question 01 (a) ii
The line $\displaystyle L$ has vector parametric equation $\displaystyle {\textbf {r}}(t)=(2+3t){\textbf {i}}+4t{\textbf {j}}-{\textbf {k}}$ ii. Let $\displaystyle \alpha$ be the angle between the line $\displaystyle L$ and the plane given by the equation $\displaystyle x-y+2z=0$. Find $\displaystyle \alpha$.

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
Recall the definition of the dot product: ${\overrightarrow {A}}\cdot {\overrightarrow {B}}=||{\overrightarrow {A}}||\ ||{\overrightarrow {B}}||cos(\theta )$ What does it tell you about two intersecting vectors?

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Solution

The line $L$ has a directional vector of ${\overrightarrow {v}}=<3,4,0>$ while the plane has a normal vector of ${\overrightarrow {n}}=<1,-1,2>$ By the definition of the dot product, $|{\overrightarrow {v}}||{\overrightarrow {n}}|cos(\theta )={\overrightarrow {v}}\cdot {\overrightarrow {n}}$ (where $\theta$ is the angle between ${\overrightarrow {v}}$ and ${\overrightarrow {n}}$) $\rightarrow ({\sqrt {3^{2}+4^{2}}}{\sqrt {1^{2}+(-1)^{2}+2^{2}}})cos(\theta )=3(1)+4(-1)+0(2)$ $\rightarrow 5{\sqrt {6}}cos(\theta )=-1$ $\rightarrow cos(\theta )={\frac {-1}{5{\sqrt {6}}}}$ $\rightarrow \theta =cos^{-1}({\frac {-1}{5{\sqrt {6}}}})$ $\theta$ however, is the angle between ${\overrightarrow {v}}$ and ${\overrightarrow {n}}$ as shown below. $\alpha$, the angle we're interested in is simply $\alpha ={\frac {\pi }{2}}-\theta =\color {blue}{\frac {\pi }{2}}-cos^{-1}({\frac {-1}{5{\sqrt {6}}}})$