= Passive rotation of coordinates
{title2=$[v]_{\rm new}=R_\theta^T[v]_{\rm old}$}
Rotate an <orthonormal basis> counterclockwise by $\theta$ while keeping the <vector> $v$ fixed. The new <basis vectors>, written in the old <basis>, are the columns of $P=R_\theta$. The <change-of-basis matrix> convention therefore gives
$$
[v]_{\rm old}=R_\theta[v]_{\rm new},\qquad
[v]_{\rm new}=R_\theta^{-1}[v]_{\rm old}=R_\theta^T[v]_{\rm old}=R_{-\theta}[v]_{\rm old}.
$$
For example, the fixed <vector> with old <coordinate vector> $(2,0)^T$ has new <coordinate vector> $(1,-\sqrt3)^T$ after the <basis> turns by $\pi/3$ <radians>. Its negative second coordinate means that the arrow lies below the new first axis. By contrast, an <active planar rotation> through that same <angle> moves the arrow to old coordinates $(1,\sqrt3)^T$. The sign difference follows from which object moves. The passive component formulas are $x'=x\cos\theta+y\sin\theta$, $y'=-x\sin\theta+y\cos\theta$.<The Feynman Lectures on Physics, Volume I by Richard Feynman and others/Chapter 20: Rotation in space>[§20–1, equation (20.3) and the preceding paragraph: the angle relates coordinate axes, not rotating objects]{ref}
If <coordinate vectors> are instead written as rows, transposing the active equation gives $(v')^T=v^TR_\theta^T$. Switching to rows changes the side and the <transpose>; it does not by itself turn an active <planar rotation> into <passive rotation of coordinates>.
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