# Equations of rotation of a vector

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.

• You mean you want to work out the intermediate steps between your picture and the completed rotation equations above? You have the rotation matrix, just create your initial vector and multiply it by the matrix... Commented Jan 20, 2021 at 22:05
• @abiessu Yes, I want to arrive at the rotation equations starting from the picture Commented Jan 20, 2021 at 22:07
• @abiessu I mean, how do I get the same rotaton matrix by performing some geometry on the picture (by taking projections, using trigonometry etc) Commented Jan 20, 2021 at 22:17
• – J.G.
Commented Jan 20, 2021 at 22:26

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 $$\begin{pmatrix}\cos\theta&-\sin\theta&0\\\sin\theta&\cos\theta&0\\0&0&1\end{pmatrix}$$