We denote:
- $R(a,\alpha)$ :"rotation about point $a$ by angle $\alpha$.
- $T_a$ : "translation" (e.g: $T_a(x)=x+a$).
I'm trying to simplify the composition of two rotations $R(a,\alpha)$ and $R(b,-\alpha)$.
But, I would like to clarify if one can move a center of a rotation.
More precisely, can $R(b,-\alpha)$ be inverted to a composition of rotation and translation? (i.e. decompose $R(b,-\alpha)$ with $R(a,\alpha)^{-1}$ and some translation.) So far, I have something like $R(b,\alpha)^{-1}=(T_{b-a}R(a,\alpha)T_{a-b})^{-1}$, but is there a simpler decomposition?
I'm hoping to get $R(a,\alpha)R(b,-\alpha)=T$ with some translation $T$ in terms of $a, b$
P.S. Every map I'm talking about here are defined over a plane ($\mathbb{R}^2$)
Let $\mathcal A(E)$ be an affine plane over the vector space $E\;$ (therefore $\dim(E)=2).$
An affine rotation around the point $c$ and with angle $\theta$ is given, for all $p\in \mathcal A(E)$, by
$$ R(c, \theta)(p) = c + R_\theta(p-c) $$
where $R_\theta$ is the vector rotation with angle $\theta$ in $E$.
If $a$ and $b$ are any two points of $\mathcal A(E)$ and $\alpha$ and $\beta$ are any two angles, therefore, we have
\begin{align} R(b,\beta)\circ R(a,\alpha)(p) &= R(b,\beta)(a+R_\alpha(p-a)) =\\[1ex] &= b+R_\beta(a+R_\alpha(p-a)-b) =\\[1ex] &= b+R_\beta(a-b)+R_\beta\circ R_\alpha(p-a). \end{align}
If $\;\beta=-\alpha$, we then obtain
\begin{align} R(b,-\alpha)\circ R(a,\alpha)(p) &= b+R_{-\alpha}(a-b)+R_{-\alpha}\circ R_\alpha(p-a) =\\[1ex] &= b+R_{-\alpha}(a-b)+p-a =\\[1ex] &= p+(b-a)+R_{-\alpha}(a-b) \end{align}
i.e. the translation with $\;b-a+R_{-\alpha}(a-b)\;$ as translation vector.