Equations of rotation of a vector

Question

Let there be two completely coincident right-handed orthogonal frames and then one of them is rotated CCW by $\theta$ degrees around the z axis. Let the unit vectors of the rotated frame be denoted as $ x^{'} y^{'} z^{'} $ (those of stationary frame are $x y z$) The equations which express the rotated frame with respect to the stationary frame are

$x^{'}=x\cos\theta-y\sin\theta$
$y^{'}=x\sin\theta+y\cos\theta$
$z^{'}=z$

The rotation matrix is $$ \begin{bmatrix} \cos \theta & -\sin\theta & 0 \\ \sin\theta & \cos \theta & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$ We know that this matrix is also a rotational operator and the operation above is equivalent to rotating a vector CCW around the z axis by the same amount.

I want to derive the rotation equations by geometrical means considering the picture above. The components of the vector before and after rotation are as depicted.

Answer 1

Use the angle sum formulas for sine and cosine. If $$ \begin{pmatrix}x\\y\\0\end{pmatrix}=\begin{pmatrix}r\cos\alpha\\r\sin\alpha\\0\end{pmatrix} $$ Then $$ \begin{pmatrix}x'\\y'\\0\end{pmatrix}=\begin{pmatrix}r\cos(\alpha+\theta)\\r\sin(\alpha+\theta)\\0\end{pmatrix}. $$ Now, $$ \cos(\alpha+\theta)=\cos\alpha\cos\theta-\sin\alpha\sin\theta $$ and $$ \sin(\alpha+\theta)=\sin\alpha\cos\theta+\sin\theta\cos\alpha $$ so $$ \begin{pmatrix}x'\\y'\\0\end{pmatrix}=r\cos\alpha\begin{pmatrix}\cos\theta\\\sin\theta\\0\end{pmatrix}+r\sin\alpha\begin{pmatrix}-\sin\theta\\\cos\theta\\0\end{pmatrix}. $$ This formula is equivalent to multiplying $\begin{pmatrix}x\\y\\0\end{pmatrix}$ by $\begin{pmatrix}\cos\theta&-\sin\theta&0\\\sin\theta&\cos\theta&0\\0&0&1\end{pmatrix}$