Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 6 (c)

MATH152 April 2015

  •  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 2(c)  •  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 5(d)  •  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 6 (c)
The matrix $A={\begin{bmatrix}1/2&-{\sqrt {2}}/2&1/2\\{\sqrt {2}}/2&0&-{\sqrt {2}}/2\\1/2&{\sqrt {2}}/2&1/2\end{bmatrix}}$ represents a rotation in 3D relative to some axis. (c) Find the angle of rotation around the axis in (b). Hint: rotate a vector that is perpendicular to the rotation axis.

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
To find the angle of a rotation, once the axis of the rotation is known, select a vector $\mathbf {v}$ perpendicular to the axis. Then the angle of the rotation is the angle between $\mathbf {v}$ and $\mathbf {A} \mathbf {v} .$

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. From part (b), we know that the direction vector of the axis of the rotation is ${\begin{bmatrix}1\\0\\1\end{bmatrix}}$, then a vector perpendicular to this direction vector is $\mathbf {v} ={\begin{bmatrix}0\\1\\0\end{bmatrix}}$. Now following the hint, we first calculate $\mathbf {A} \mathbf {v} ={\begin{bmatrix}1/2&-{\sqrt {2}}/2&1/2\\{\sqrt {2}}/2&0&-{\sqrt {2}}/2\\1/2&{\sqrt {2}}/2&1/2\end{bmatrix}}{\begin{bmatrix}0\\1\\0\end{bmatrix}}={\begin{bmatrix}-{\sqrt {2}}/2\\0\\{\sqrt {2}}/2\end{bmatrix}}$. The angle $\mathbf {\theta }$ between the two vectors $\mathbf {v} ,\mathbf {A} \mathbf {v}$ is calculated by using the Cosine Formula $\cos(\theta )={\frac {v\cdot Av}{\|v\|\|Av\|}}={\frac {{\begin{bmatrix}0\\1\\0\end{bmatrix}}\cdot {\begin{bmatrix}-{\sqrt {2}}/2\\0\\{\sqrt {2}}/2\end{bmatrix}}}{\|{\begin{bmatrix}0\\1\\0\end{bmatrix}}\|\|{\begin{bmatrix}-{\sqrt {2}}/2\\0\\{\sqrt {2}}/2\end{bmatrix}}\|}}=0.$ Thus the angle of rotation is $\color {blue}\mathbf {\theta } =\cos ^{-1}(0)={\frac {\pi }{2}}.$