Rotation between two point in 2D dimension

Question

Having the coordinates x,x',y and y', how can i calculate the angle Ɵ between this two point.

Thank you for you help

Answer 1

Here is another simple possibility. You know the coordinates $(x,y)$ and $(x',y')$ of the point before and after the rotation, so you could convert each pair to polar form $(r,\phi)$ and $(r', \phi')$, where $r = \sqrt{x^2+y^2}$, $\phi = \tan^{-1}(y/x)$, and $r' = \sqrt{x'^2+y'^2}$, $\phi' = \tan^{-1}(y'/x')$. The inverse coordinate transformations are $x = r\cos{\phi}$, $y = r\sin\phi$, and $x' = r'\cos{\phi'}$, $y' = r'\sin\phi'$ (the polar angles $\phi$ and $\phi'$ are positive counterclockwise from the $x-$axis).

The distance from the origin to either point, however, is unchanged by the rotation, $x^2+y^2 = x'^2+y'^2$, so $r'= r$, hence $x' = r\cos{\phi'}$ and $y' = r\sin\phi'$.

Now substitute for $(x,y)$ and $(x',y')$ in your rotational transformation equations to get $$\begin{align}r\cos{\phi'} &= r\cos{\phi}\cos{\theta} - r\sin{\phi}\sin{\theta} = r \cos{(\phi+\theta)},\\ r\sin\phi' &= r\cos\phi\sin\theta + r\sin\phi\cos\theta = r \sin{(\phi+\theta)}.\end{align}$$ Cancelling the common factor of $r$ leaves you with $\cos\phi' = \cos(\phi+\theta)$ and $\sin\phi' = \sin(\phi+\theta)$, from which it follows that $\phi' = \phi + \theta,$ or the rotation angle $\boxed{\theta = \phi' - \phi}$, just the difference in polar angles of the two points. (If the angle is negative, then the rotation was clockwise rather than counterclockwise.) This result is very clear from a consideration of the geometry of the situation.

Directly, in terms of the rectangular coordinates, this difference is $$\boxed{\theta = \tan^{-1}(y'/x') - \tan^{-1}(y/x)~~({\text{ or }} \arctan(y'/x') - \arctan(y/x)\,\,) .\,}$$

Answer 2

Given $r^2=x^2+y^2=x'^2+y'^2$: $$\frac{xx'+yy'}{r^2} = \cos\theta\\ \frac{xy'-yx'}{r^2} = \sin\theta $$