= Active planar rotation
{title2=$v'=R_\theta v$}
For column <coordinate vectors> in the fixed <Cartesian basis>, an active <planar rotation> sends $(x,y)^T$ to $(x\cos\theta-y\sin\theta,\,x\sin\theta+y\cos\theta)^T$. For $\theta=\pi/3$ <radians>,
$$
R_{\pi/3}=\begin{pmatrix}1/2&-\sqrt3/2\\\sqrt3/2&1/2\end{pmatrix},\qquad
R_{\pi/3}\begin{pmatrix}2\\0\end{pmatrix}=\begin{pmatrix}1\\\sqrt3\end{pmatrix}.
$$
The endpoint travels counterclockwise through $60^\circ$ along a <circle> of radius $2$. Its <Euclidean norm> stays $\sqrt{1^2+(\sqrt3)^2}=2$. The second <Cartesian basis> direction becomes $R_{\pi/3}(0,1)^T=(-\sqrt3/2,1/2)^T$, whose <dot product> with $(1,\sqrt3)^T$ is still zero.
\Image[/rotation-counterclockwise-column-vector.png]
{title=An active counterclockwise <planar rotation> by $\pi/3$: the <vector> moves while the <Cartesian basis> stays fixed}
A quick sign check is to look at the first column: $R_\theta(1,0)^T=(\cos\theta,\sin\theta)^T$ must point above the horizontal axis for $0<\theta<\pi$. The columns have unit <Euclidean norm> and zero <dot product>, so $R_\theta^TR_\theta=I$; the <determinant> is $\cos^2\theta+\sin^2\theta=1$. Finally, the <transpose> negates the off-diagonal entries, giving $R_\theta^T=R_{-\theta}$: turning back reverses the action. This is the same <matrix> used for <passive rotation of coordinates>, though that interpretation keeps the <vector> fixed.
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