# 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 ${\displaystyle \pi /6}$ to get a new vector ${\displaystyle \mathbf {v} _{1}}$. 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

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

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

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

Using our vector ${\displaystyle \mathbf {v} _{1}}$ and projecting it onto u which has length ${\displaystyle |\mathbf {u} |={\sqrt {5}}}$, we have that our new vector, ${\displaystyle \mathbf {v} _{2}}$, is

{\displaystyle {\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

${\displaystyle T\left({\begin{bmatrix}1\\0\end{bmatrix}}\right)={\frac {{\sqrt {3}}+2}{10}}{\begin{bmatrix}1\\2\end{bmatrix}}.}$