Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 15

MATH152 April 2011

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[hide] Question A 15

Consider the linear transformation $T$ which takes vectors in $\mathbb {R} ^{2}$, first rotates them by $\pi /6$ (30°) counter-clockwise and then projects the result onto the direction $\left[{\begin{array}{c}1\\2\end{array}}\right].$ Find $T\left(\left[{\begin{array}{c}1\\0\end{array}}\right]\right).$ Note: $\cos \pi /6={\sqrt {3}}/2$ and $\displaystyle \sin \pi /6=1/2$. Your final answer should not contain trigonometric functions.

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!

[show] Hint
How can we use right-triangle trig techniques to get the components of a rotated vector? What is the formula for projection?

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[show] 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. We could start by finding both of the separate transformation matrices and then multiplying them together which would give us the general transformation matrix T. However since we only need T([1,0]), the transformation of the unit vector in the x-direction, it will suffice to just consider how that single vector is transformed and not deal with matrices at all. We first rotate the vector [1,0] counter-clockwise by $\pi /6$ to get a new vector $\mathbf {v} _{1}$. After doing so, the new vector forms a right triangle with a vector in the x-direction and a vector in the y-direction. We can get its components by using trigonometry. The new vector will have length 1 (since it is a rotation of [1,0] which also has length 1) so its components are $\mathbf {v} _{1}=\left[\cos \left({\frac {\pi }{6}}\right),\sin \left({\frac {\pi }{6}}\right)\right]=\left[{\frac {\sqrt {3}}{2}},{\frac {1}{2}}\right].$ We then must project this vector onto u=[1,2]. Recall that a projection of a vector v onto a vector u is ${\textrm {proj}}_{\mathbf {u} }\mathbf {v} ={\frac {\mathbf {v} \cdot \mathbf {u} }{|\mathbf {u} |^{2}}}\mathbf {u} ={\frac {\mathbf {v} \cdot \mathbf {u} }{\mathbf {u} \cdot \mathbf {u} }}\mathbf {u} .$ Using our vector $\mathbf {v} _{1}$ and projecting it onto u which has length $|\mathbf {u} |={\sqrt {5}}$, we have that our new vector, $\mathbf {v} _{2}$, is ${\begin{aligned}\mathbf {v_{2}} &={\textrm {proj}}_{\mathbf {u} }\mathbf {v} _{1}={\frac {\left[{\frac {\sqrt {3}}{2}},{\frac {1}{2}}\right]\cdot \left[1,2\right]}{5}}[1,2]\\&={\frac {{\frac {\sqrt {3}}{2}}(1)+({\frac {1}{2}})2}{5}}[1,2]={\frac {{\sqrt {3}}+2}{10}}[1,2].\end{aligned}}$ Therefore, the transformation on the vector [1,0] is $T\left({\begin{bmatrix}1\\0\end{bmatrix}}\right)={\frac {{\sqrt {3}}+2}{10}}{\begin{bmatrix}1\\2\end{bmatrix}}.$

[show] 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. As practice for doing linear transformations, we will create the matrices anyway and show that they lead to the same conclusion as going step by step in Solution 1. First recall that a rotation matrix rotating counter-clockwise with an angle $\theta$ has the form $A={\begin{bmatrix}\cos(\theta )&-\sin(\theta )\\\sin(\theta )&\cos(\theta )\end{bmatrix}}.$ If you are having trouble remembering how to form a rotation matrix, see the solution to Question A28 on this exam for a more detailed explanation of these entries (they are components of the vectors generated by rotating the unit vectors). Since in this example the angle rotating is $\pi /6$ then we get $A={\frac {1}{2}}{\begin{bmatrix}{\sqrt {3}}&-1\\1&{\sqrt {3}}\end{bmatrix}}.$ Next for the projection matrix, to get the entries in column k, we have to do a projection of the kth unit vector onto our desired vector u=[1,2]. We get that the first column is ${\textrm {proj}}_{\mathbf {u} }[1,0]={\frac {[1,0]\cdot [1,2]}{5}}[1,2]={\frac {1}{5}}[1,2].$ The second column is ${\textrm {proj}}_{\mathbf {u} }[0,1]={\frac {[0,1]\cdot [1,2]}{5}}[1,2]={\frac {2}{5}}[1,2].$ Therefore we get that the projection matrix is $B={\frac {1}{5}}{\begin{bmatrix}1&2\\2&4\end{bmatrix}}.$ The total transformation then is (remember that the rotation is applied first), ${\begin{aligned}T&=BA={\frac {1}{10}}{\begin{bmatrix}1&2\\2&4\end{bmatrix}}{\begin{bmatrix}{\sqrt {3}}&-1\\1&{\sqrt {3}}\end{bmatrix}}\\&={\frac {1}{10}}{\begin{bmatrix}{\sqrt {3}}+2&2{\sqrt {3}}-1\\2{\sqrt {3}}+4&4{\sqrt {3}}-2\end{bmatrix}}\end{aligned}}$. Therefore, ${\begin{aligned}T\left({\begin{bmatrix}1\\0\end{bmatrix}}\right)&={\frac {1}{10}}{\begin{bmatrix}{\sqrt {3}}+2&2{\sqrt {3}}-1\\2{\sqrt {3}}+4&4{\sqrt {3}}-2\end{bmatrix}}{\begin{bmatrix}1\\0\end{bmatrix}}\\&={\frac {1}{10}}{\begin{bmatrix}{\sqrt {3}}+2\\2{\sqrt {3}}+4\end{bmatrix}}={\frac {{\sqrt {3}}+2}{10}}{\begin{bmatrix}1\\2\end{bmatrix}}\end{aligned}}$ like we recovered in the first solution.