Rotation Transformation

When going from one Reference Frame to another, and the relation between them is just a rotation.

This relation can be written as

$${\mathbf{r}}^b=R_o^b(\alpha ){\mathbf{r}}^o$$

Where $R_{from}^{to}$ is a Rotation Matrix

---

Example

Given the following frames We want to find a vector

$$\vec{r}=e{\vec{i}}_o+f{\vec{j}}_o+g{\vec{k}}_o=e^{\prime }{\vec{i}}_b+f^{\prime }{\vec{j}}_b+g^{\prime }{\vec{k}}_b$$

This can be done by finding ${\vec{i}}_o,\dots$ as a function of ${\vec{i}}_b,\dots$

$$\begin{array}{rl}{\vec{k}}_o& ={\vec{k}}_b\\ {\vec{i}}_0& =\cos \alpha {\vec{i}}_b-\sin \alpha {\vec{j}}_b\\ {\vec{j}}_o& =\sin \alpha {\vec{i}}_b+\cos \alpha {\vec{j}}_b\end{array}$$

Giving us the new value of $\vec{r}$

$$\begin{array}{rl}\vec{r}& =e({\vec{i}}_b\cos \alpha -\vec{j}\sin \alpha )+f({\vec{i}}_b\sin \alpha +{\vec{j}}_b\cos \alpha )+g{\vec{k}}_b\end{array}$$

Where

$$\begin{array}{rl}{e}^{\prime }& =e\cos \alpha +f\sin \alpha \\ f^{\prime }& =-e\sin \alpha +f\cos \alpha \\ g^{\prime }& =g\end{array}$$

Written more formally, ${\mathbf{r}}^b=R_o^b(\alpha ){\mathbf{r}}^o$

$$\begin{array}{rl}{\mathbf{r}}^b& =\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0\\ -\sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}e\\ f\\ g\end{array}\right]=\left[\begin{array}{c}e\cos \alpha +f\sin \alpha \\ -e\sin \alpha +f\cos \alpha \\ g\end{array}\right]\end{array}$$

Which is a Principal Rotation Matrix about the z-axis.