Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 03

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MATH152 April 2011

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Question A 03
Given the vectors ${\begin{aligned}\mathbf {x} &=[0,-3,2]\\\mathbf {y} &=[1,1,1]\\\end{aligned}}$ Compute the projection of x onto the direction of y.

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
Consider drawing a line segment to form a right triangle with the two vectors. If you knew the angle between the vectors, how would you get the lengths of the perpendicular sides? How could you determine the angle between the vectors?

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Solution
Consider connecting the two vectors into a right triangle similar the picture to the right . To get the component of $\mathbf {x}$ onto $\mathbf {y}$ , we follow like in the picture and take, ${\textrm {proj}}_{\mathbf {y} }\mathbf {x} =\|\mathbf {x} \|\cos(\theta ){\hat {\mathbf {y} }}$ where $\theta$ is the angle between the vectors $\mathbf {x}$ and $\mathbf {y}$ and ${\hat {\mathbf {y} }}$ is a unit vector in the direction of $\mathbf {y}$ . We can rewrite $\cos(\theta )$ in terms of the dot (scalar) product as, $\cos(\theta )={\frac {\mathbf {x} \cdot \mathbf {y} }{\|\mathbf {x} \|\|\mathbf {y} \|}}.$ Therefore, the projection of $\mathbf {x}$ onto $\mathbf {y}$ is ${\textrm {proj}}_{\mathbf {y} }\mathbf {x} ={\frac {\mathbf {x} \cdot \mathbf {y} }{\|\mathbf {y} \|}}{\hat {\mathbf {y} }}={\frac {\mathbf {x} \cdot \mathbf {y} }{\|\mathbf {y} \|^{2}}}\mathbf {y}$ where we have recognized that ${\hat {\mathbf {y} }}={\frac {\mathbf {y} }{\|\mathbf {y} \|}}.$ Computing the dot product we get, $\mathbf {x} \cdot \mathbf {y} =0-3+2=-1$ and for the magnitude of $\mathbf {y}$ , $\|\mathbf {y} \|={\sqrt {1^{2}+1^{2}+1^{2}}}={\sqrt {3}}.$ Therefore we get that the projection of $\mathbf {x}$ onto $\mathbf {y}$ is, ${\textrm {proj}}_{\mathbf {y} }\mathbf {x} ={\frac {\mathbf {x} \cdot \mathbf {y} }{\|\mathbf {y} \|^{2}}}\mathbf {y} =-{\frac {1}{3}}\mathbf {y} =\left[-{\frac {1}{3}},-{\frac {1}{3}},-{\frac {1}{3}}\right].$