Science:Math Exam Resources/Courses/MATH152/April 2013/Question B 01 (b)

MATH152 April 2013

  •  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 1(d)  •  QB 2(a)  •  QB 2(b)  •  QB 2(c)  •  QB 2(d)  •  QB 3(a)  •  QB 3(b)  •  QB 3(c)  •  QB 4(a)  •  QB 4(b)  •  QB 4(c)  •  QB 5(a)  •  QB 5(b)  •  QB 5(c)  •  QB 6(a)  •  QB 6(b)  •  QB 6(c)  •  

   Other MATH152 Exams

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

Question B 01 (b)
Let $L$ be the line $3x=4y$. Find the matrix A which represents ${\text{Ref}}_{L}$, the reflection across the line L.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
To reflect a vector about a line that goes through the origin, let $v=(v_{x},v_{y})$ be a vector in the direction of the line. The matrix which represents ${\text{Ref}}_{L}$ is given by $\mathbf {A} ={\frac {1}{\lVert v\rVert ^{2}}}{\begin{bmatrix}v_{x}^{2}-v_{y}^{2}&2v_{x}v_{y}\\2v_{x}v_{y}&v_{y}^{2}-v_{x}^{2}\end{bmatrix}}$

Hint 2
For an alternative solution, we can recover the rotation matrix A if we know how it acts on two linearly independent vectors. Consider a vector on the given line, and a vector orthogonal to that line.

Checking a solution serves two purposes: helping you if, after having used all the hints, 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 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. To reflect a vector about a line that goes through the origin, let $v=(v_{x},v_{y})$ be a vector in the direction of the line. The matrix which represents ${\text{Ref}}_{L}$ is given by $\mathbf {A} ={\frac {1}{\lVert v\rVert ^{2}}}{\begin{bmatrix}v_{x}^{2}-v_{y}^{2}&2v_{x}v_{y}\\2v_{x}v_{y}&v_{y}^{2}-v_{x}^{2}\end{bmatrix}}$ Since $v=(4,3)$ is on the line, we have $\mathbf {A} ={\frac {1}{4^{2}+3^{2}}}{\begin{bmatrix}4^{2}-3^{2}&2(4)(3)\\2(4)(3)&3^{2}-4^{2}\end{bmatrix}}={\frac {1}{25}}{\begin{bmatrix}7&24\\24&-7\end{bmatrix}}={\begin{bmatrix}{\frac {7}{25}}&{\frac {24}{25}}\\{\frac {24}{25}}&{\frac {-7}{25}}\end{bmatrix}}$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We can alternatively consider what a reflection does to vectors. Any vector pointing along the line 3x=4y, <4,3>, for example, will be reflected onto itself. Any vector perpendicular to <4,3> will point in the completely opposite direction after reflection. For example, <-3,4> will become <3,-4>. Note that these two vectors are linearly independent. Thus if A denotes the reflection, $A{\begin{bmatrix}4\\3\end{bmatrix}}={\begin{bmatrix}4\\3\end{bmatrix}}$ and $A{\begin{bmatrix}-3\\4\end{bmatrix}}={\begin{bmatrix}3\\-4\end{bmatrix}}$, which we can write as a matrix equation $A{\begin{bmatrix}4&-3\\3&4\end{bmatrix}}={\begin{bmatrix}4&3\\3&-4\end{bmatrix}}$. Thus $A={\begin{bmatrix}4&3\\3&-4\end{bmatrix}}{\begin{bmatrix}4&-3\\3&4\end{bmatrix}}^{-1}={\begin{bmatrix}7/25&24/25\\24/25&-7/25\end{bmatrix}}.$

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER QGT flag, MER Tag Linear transformation definition, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag