Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 15/Solution 1

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

${\mathbf{𝐯}}_1=[\cos \left(\frac{\pi }{6}\right),\sin \left(\frac{\pi }{6}\right)]=[\frac{\sqrt{3}}{2},\frac{1}{2}].$

We then must project this vector onto u=[1,2]. Recall that a projection of a vector v onto a vector u is

${\text{proj}}_{\mathbf{𝐮}}\mathbf{𝐯}=\frac{\mathbf{𝐯}\cdot \mathbf{𝐮}}{|\mathbf{𝐮}{|}^2}\mathbf{𝐮}=\frac{\mathbf{𝐯}\cdot \mathbf{𝐮}}{\mathbf{𝐮}\cdot \mathbf{𝐮}}\mathbf{𝐮}.$

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

$\begin{array}{rl}{\mathbf{v}}_{\mathbf{𝟐}}& ={\text{proj}}_{\mathbf{𝐮}}{\mathbf{𝐯}}_1=\frac{[\frac{\sqrt{3}}{2},\frac{1}{2}]\cdot [1,2]}{5}[1,2]\\ & =\frac{\frac{\sqrt{3}}{2}(1)+(\frac{1}{2})2}{5}[1,2]=\frac{\sqrt{3}+2}{10}[1,2].\end{array}$

Therefore, the transformation on the vector [1,0] is

$T\left(\left[\begin{array}{c}1\\ 0\end{array}\right]\right)=\frac{\sqrt{3}+2}{10}\left[\begin{array}{c}1\\ 2\end{array}\right].$