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

MATH152 April 2011

  •  QA 1  •  QA 2  •  QA 3  •  QA 4  •  QA 5  •  QA 6  •  QA 7  •  QA 8  •  QA 9  •  QA 10  •  QA 11  •  QA 12  •  QA 13  •  QA 14  •  QA 15  •  QA 16  •  QA 17  •  QA 18  •  QA 19  •  QA 20  •  QA 21  •  QA 22  •  QA 23  •  QA 24  •  QA 25  •  QA 26  •  QA 27  •  QA 28  •  QA 29  •  QA 30  •  QB 1(a)  •  QB 1(b)  •  QB 1(c)  •  QB 2(a)  •  QB 2(b)  •  QB 3(a)  •  QB 3(b)  •  QB 3(c)  •  QB 4(a)  •  QB 4(b)  •  QB 4(c)  •  QB 4(d)  •  QB 5(a)  •  QB 5(b)  •  QB 5(c)  •  QB 6(a)  •  QB 6(b)  •  QB 6(c)  •  QB 6(d)  •  QB 6(e)  •  

   Other MATH152 Exams

  •  April 2016  •  April 2015  •  April 2014  •  April 2022  •  April 2011  •  April 2010  •  April 2012  •  April 2013  •  April 2017  •  

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?

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. 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, ${\textrm {proj}}_{\mathbf {y} }\mathbf {x} ={\frac {\mathbf {x} \cdot \mathbf {y} }{|\mathbf {y} |^{2}}}\mathbf {y} =-{\frac {1}{3}}\mathbf {y} =\left[-{\frac {1}{3}},-{\frac {1}{3}},-{\frac {1}{3}}\right].$