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


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. enter image description here

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.

  • 1
    $\begingroup$ 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... $\endgroup$
    – abiessu
    Commented Jan 20, 2021 at 22:05
  • $\begingroup$ @abiessu Yes, I want to arrive at the rotation equations starting from the picture $\endgroup$
    – Ali Kıral
    Commented Jan 20, 2021 at 22:07
  • $\begingroup$ @abiessu I mean, how do I get the same rotaton matrix by performing some geometry on the picture (by taking projections, using trigonometry etc) $\endgroup$
    – Ali Kıral
    Commented Jan 20, 2021 at 22:17
  • 3
    $\begingroup$ Proof without words. $\endgroup$
    – J.G.
    Commented Jan 20, 2021 at 22:26

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .