How to rotate two vectors (2d), where their angle is larger than 180.

Question

The rotation matrix $$\begin{bmatrix} \cos\theta & -\sin \theta\\ \sin\theta & \cos\theta \end{bmatrix}$$ cannot process the case that the angle between two vectors is larger than $180$ degrees. (counter-clockwise rotation).

Answer 1

The matrix $M(\theta)$, where $$M(\theta) = \left[ \begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\right]$$

Can take any value of $\theta$ whatsoever. For example, let $\theta = 270^{\circ}$ then we have $\sin\theta = -1$ and $\cos\theta =0$ giving $$M(270^{\circ}) = \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right]$$ The vector $[1,0]^{\top}$ gets sent to $[0,-1]^{\top}$ while the vector $[0,1]^{\top}$ to $[1,0]^{\top}$. Notice, moreover, that $\det(M(\theta)) = 1$ for all $\theta$. This is exactly an anti-clockwise rotation of $270^{\circ}$.

Answer 2

Suppose you have a point $\,P=(x,y)\in\Bbb R^2\,$ in the first quadrant, and you want to get its reflection through the origin, i.e.: the point on the line through $\,P\,$ and the origin at the same distance from the latter as $\,P\,$ : then you simply take the point $\,P'=(-x,-y)\,$ , right?

Well, now just do the same for your case...but using a little trigonometry: since

$$\cos(\pi+\alpha)=-\cos\alpha\;\;,\;\;\;\sin(\pi+\alpha)=-\sin\alpha$$

then if $\;\,\theta=\pi+\alpha\;$ we get

$$\begin{pmatrix}\cos\theta&\!\!-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}=\begin{pmatrix}-\cos\alpha&\;\;\;\sin\alpha\\-\sin\alpha&-\cos\alpha\end{pmatrix}=-\begin{pmatrix}\cos\alpha&\;\!\!-\sin\alpha\\\sin\alpha&\cos\alpha\end{pmatrix}$$

and you get the additive inverse of a usual rotation matrix...(not that the original option was bad, though)