Rotation of a vector

Question

My linear algebra textbook only explained the rotation of a vector in a counterclockwise direction. I'm just wondering what happens if I rotate a vector in the clockwise direction? Do I solve such problems using the same method as for rotation in the anticlockwise direction? Thanks.

Answer 1

It's likely that the scheme you're using has a parameter $\theta$ where you rotate $\theta$ radians clockwise if $\theta$ is positive, and that negative radians will instead carry out the clockwise rotation.

Alternatively, if you wanted to rotate $\theta$ radians clockwise, then you could just rotate $2\pi-\theta$ radians counterclockwise to get the same effect.

Answer 2

Well, if you rotate by an angle $\phi$, the rotation matrix (around the $Z$-axis) becomes:

$$ M_z=\begin{pmatrix} \cos{\phi} & -\sin{\phi}& 0\\ \sin{\phi} & \cos{\phi} & 0 \\ 0 & 0 & 1 \end{pmatrix} $$

You can also rotate by an angle $-\phi$. The matrix then becomes: $$ M_z=\begin{pmatrix} \cos{\phi} & \sin{\phi}& 0\\ -\sin{\phi} & \cos{\phi} & 0 \\ 0 & 0 & 1 \end{pmatrix} $$

because $\cos{(-\phi)}=\cos{\phi}$ and $\sin{(-\phi)} = -\sin{\phi}$. So you just replace $\phi$ by $-\phi$.

Answer 3

$$\begin{pmatrix}\cos(\theta) & -\sin(\theta) \\\sin(\theta) & \cos(\theta)\end{pmatrix}$$

Above is a rotation matrix in the plane, notice what we get when we change $\theta$ to $-\theta$

$$\begin{pmatrix}\cos(-\theta) & -\sin(-\theta) \\\sin(-\theta) & \cos(-\theta)\end{pmatrix}$$

so that:

$$\begin{pmatrix}\cos(\theta) & \sin(\theta) \\-\sin(\theta) & \cos(\theta)\end{pmatrix}$$