Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 29

MATH152 April 2017

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Question A 29
Give a vector in $\mathbb {R} ^{3}$ that makes an angle of ${\frac {\pi }{4}}$ radians with the vector ${\mathbf {v}}=[3,1,5].$ Describe briefly the procedure you used to get your answer.

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
It is easy to find vectors in $\mathbb {R} ^{3}$ such that the angle between it and $[1,0,0]$ is ${\frac {\pi }{4}}$. Find an orthogonal matrix that maps $[1,0,0]$ to $[3,1,5]$.

Hint 2
(For alternative solution) Consider a vector orthogonal to ${\textbf {v}}$, call it ${\textbf {u}}$. How can that help you? (Draw a picture!)

Hint 3
Continuing hint 2: what angle does the diagonal of a square make with its side?

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. An orthonormal basis that contains a positive scalar multiple of $[3,1,5]$ is $[{\frac {3}{\sqrt {35}}},{\frac {1}{\sqrt {35}}},{\frac {5}{\sqrt {35}}}],[{\frac {-5}{\sqrt {34}}},0,{\frac {3}{\sqrt {34}}}],[{\frac {3}{\sqrt {1190}}},{\frac {-34}{\sqrt {1190}}},{\frac {5}{\sqrt {1190}}}]$ The corresponding orthogonal matrix is ${\begin{pmatrix}{\frac {3}{\sqrt {35}}}&{\frac {-5}{\sqrt {34}}}&{\frac {3}{\sqrt {1190}}}\\{\frac {1}{\sqrt {35}}}&0&{\frac {-34}{\sqrt {1190}}}\\{\frac {5}{\sqrt {35}}}&{\frac {3}{\sqrt {34}}}&{\frac {5}{\sqrt {1190}}}\end{pmatrix}}$ The angle between the vectors $[1,0,0]$ and $[1,1,0]$ is ${\frac {\pi }{4}}$, so the angle between $[3,1,5]$ and the following vector is ${\frac {\pi }{4}}$ ${\begin{pmatrix}{\frac {3}{\sqrt {35}}}&{\frac {-5}{\sqrt {34}}}&{\frac {3}{\sqrt {1190}}}\\{\frac {1}{\sqrt {35}}}&0&{\frac {-34}{\sqrt {1190}}}\\{\frac {5}{\sqrt {35}}}&{\frac {3}{\sqrt {34}}}&{\frac {5}{\sqrt {1190}}}\end{pmatrix}}{\begin{pmatrix}1\\1\\0\end{pmatrix}}={\begin{pmatrix}{\frac {3}{\sqrt {35}}}-{\frac {5}{\sqrt {34}}}\\{\frac {1}{\sqrt {35}}}\\{\frac {5}{\sqrt {35}}}+{\frac {3}{\sqrt {34}}}\end{pmatrix}}$ $\color {blue}Answer:{\begin{pmatrix}{\frac {3}{\sqrt {35}}}-{\frac {5}{\sqrt {34}}}\\{\frac {1}{\sqrt {35}}}\\{\frac {5}{\sqrt {35}}}+{\frac {3}{\sqrt {34}}}\end{pmatrix}}$

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. (Alternative solution) First, we find a vector ${\textbf {u}}$ orthogonal to ${\textbf {v}}$ with the same magnitude with ${\mathbf {v}}$ As a solution to the equation ${\textbf {u}}\cdot {\textbf {v}}=0$, we can find an orthogonal vector ${\mathbf {u}}=c[0,5,-1]$ to ${\mathbf {v}}$. To make the same magnitude with ${\mathbf {v}}$, we choose $c$ which solves $3^{2}+1^{2}+5^{2}=|{\mathbf {v}}|^{2}=|{\mathbf {u}}|^{2}=c^{2}(5^{2}+(-1)^{2})\iff 35=26c^{2}\iff c=\pm {\sqrt {\frac {35}{26}}}$. Now, we find the desired vector ${\mathbf {u}}={\sqrt {\frac {35}{26}}}\ [0,5,-1]$. Note that the addition of two orthogonal vectors with the same magnitude makes an angle of ${\frac {\pi }{4}}$ radians with both vectors. Therefore, ${\textbf {w}}={\textbf {u}}+{\textbf {v}}=\color {blue}\left[3,1+5{\sqrt {\frac {35}{26}}},5-{\sqrt {\frac {35}{26}}}\right]$ is a vector that we are looking for.