# Fin the Rotation $\alpha$ Centered in $(2,1) \in \mathbb{R^2}$.

Let $\alpha \in [0, 2\pi]$. Write the rotation $\alpha$ centered in $(2,1) \in \mathbb{R^2}$ under the form Q(x) + v

attempt at finding the solution

Let x =$\left[ \begin{array}{ccc} x_1 \\ x_2 \\ \end{array} \right]$

The rotation matrix is $\left[ \begin{array}{ccc} cos\alpha \ -sin\alpha \\ sin\alpha \ cos\alpha \\ \end{array} \right]$

Therefore the new coordinates of a point $x'$ after rotation is:

$x' =\left[ \begin{array}{ccc} x'_1 \\ x'_2 \\ \end{array} \right]$ =$\left[ \begin{array}{ccc} cos\alpha \ -sin\alpha \\ sin\alpha \ cos\alpha \\ \end{array} \right]\left[ \begin{array}{ccc} x_1 \\ x_2 \\ \end{array} \right]$ =$\left[ \begin{array}{ccc} x_1cos\alpha - x_2sin\alpha \\ x_1sin\alpha + x_2cos\alpha \\ \end{array} \right]$

But the rotation is centered in $[0, 2π)$ Therefore the rotation is $\left[ \begin{array}{ccc} x_1cos\alpha - x_2sin\alpha \\ x_1sin\alpha + x_2cos\alpha \\ \end{array} \right] +\left[ \begin{array}{ccc} 2 \\ 1 \\ \end{array} \right] =\left[ \begin{array}{ccc} cos\alpha \ -sin\alpha \\ sin\alpha \ cos\alpha \\ \end{array} \right]\left[ \begin{array}{ccc} x_1 \\ x_2 \\ \end{array} \right] + \left[ \begin{array}{ccc} 2 \\ 1 \\ \end{array} \right]$

end of attempt

Is that the right approach to the question?

You are wrong because the rotation matrix $$R= \begin{bmatrix} \cos \alpha& -\sin \alpha\\ \sin \alpha& \cos \alpha \end{bmatrix}$$ represents a rotation centered at $(0,0)$, so the first step is to translate the origin to the center of rotation $[2,1]^T$ . This is done by the translation: $$\vec x=\begin{bmatrix} x\\y\end{bmatrix} \rightarrow \vec x'=\begin{bmatrix} x-2\\y-1\end{bmatrix}=\begin{bmatrix} x\\y\end{bmatrix}-\begin{bmatrix} 2\\1\end{bmatrix}=\vec x - \vec t$$ Where $\vec t$ is the center of rotation.
Now we can do the rotation: $$R \vec x'= \begin{bmatrix} \cos \alpha& -\sin \alpha\\ \sin \alpha& \cos \alpha \end{bmatrix}\begin{bmatrix} x-2\\y-1\end{bmatrix}= \begin{bmatrix} (x-2)\cos \alpha-(y-1)\sin \alpha\\(x-2)\sin \alpha+(y-1)\cos \alpha\end{bmatrix}$$ Finally we have to return to the old coordinate system with a translation $\vec t$, so the final result is: $$R^*\vec x =R\vec x' +\vec t$$
that you can write as: $$R^*\vec x=R(\vec x-\vec t)+\vec t=R\vec x +(I-R)\vec t$$
• Wait why is it $\left[ \begin{array}{ccc} x \\ y \\ \end{array} \right] - \left[ \begin{array}{ccc} 2 \\ 1 \\ \end{array} \right]$ and not $\left[ \begin{array}{ccc} x \\ y \\ \end{array} \right] + \left[ \begin{array}{ccc} 2 \\ 1 \\ \end{array} \right]$? – philooooooo Mar 13 '16 at 15:09
• Hint: the point $(2,1)$ have to becomes the origin $(0,0)$ after the translation. – Emilio Novati Mar 13 '16 at 15:11