Given a coordinate frame \(\{B\}\) which is at an angle of \(\theta\) with respect to \(\{A\}\) and there is no translation between \(\{B\}\) and \(\{A\}\), we can represent the points in \(\{B\}\) with respect to \(\{A\}\) by multiplying them with the rotation matrix \(^{A}_{B}{R}\). Here \(^{A}_{B}{R}\) is a \(3\times 3\) matrix that maps a vector from coordinate frame \(B\) to coordinate frame \(A\).
We can get the rotation matrix by projecting the unit vectors of the principle axes of \(\{B\}\) to the unit vectors of the principle axes of \(\{A\}\) and stacking them horizontally in the form of a matrix as \(\begin{bmatrix}^{A}\hat{X}_{B}, ^{A}\hat{Y}_{B}, ^{A}\hat{Z}_{B}\end{bmatrix}\).
We can expand \(\begin{bmatrix}^{A}\hat{X}_{B}, ^{A}\hat{Y}_{B}, ^{A}\hat{Z}_{B}\end{bmatrix}\) as
\[\begin{align*} ^{A}_{B}{R} &= \begin{bmatrix}^{A}\hat{X}_{B}, ^{A}\hat{Y}_{B}, ^{A}\hat{Z}_{B}\end{bmatrix} \\ &= \begin{bmatrix} ^{B}\hat{X}^{A}\hat{X} && ^{B}\hat{Y}^{A}\hat{X} &&^{B}\hat{Z}^{A}\hat{X} \\ ^{B}\hat{X}^{A}\hat{Y} && ^{B}\hat{Y}^{A}\hat{Y} && ^{B}\hat{Z}^{A}\hat{Y} \\ ^{B}\hat{X}^{A}\hat{Z} && ^{B}\hat{Y}^{A}\hat{Z} && ^{B}\hat{Z}^{A}\hat{Z} \end{bmatrix} \end{align*}\]If we are rotating along the \(z\) axis, then only the \(^{B}\hat{X}\) and \(^{B}\hat{Y}\) axes rotate by an angle \(\theta\) where as the angle between \(^{B}\hat{Z}\) and \(^{A}\hat{Z}\) will be \(0\). In this case we represent the rotation matrix as \(^{A}_{B}{R}_z\) as
\[\begin{align*} ^{A}_{B}{R}_z &= \begin{bmatrix}^{A}\hat{X}_{B}, ^{A}\hat{Y}_{B}, ^{A}\hat{Z}_{B}\end{bmatrix} \\ &= \begin{bmatrix} ^{B}\hat{X}^{A}\hat{X} && ^{B}\hat{Y}^{A}\hat{X} &&^{B}\hat{Z}^{A}\hat{X} \\ ^{B}\hat{X}^{A}\hat{Y} && ^{B}\hat{Y}^{A}\hat{Y} && ^{B}\hat{Z}^{A}\hat{Y} \\ ^{B}\hat{X}^{A}\hat{Z} && ^{B}\hat{Y}^{A}\hat{Z} && ^{B}\hat{Z}^{A}\hat{Z} \end{bmatrix} \\ &= \begin{bmatrix} \mid ^{B}\hat{X}\mid \mid^{A}\hat{X}\mid \cos \theta && \mid ^{B}\hat{Y}\mid \mid^{A}\hat{X}\mid \cos (90 + \theta) && \mid^{B}\hat{Z}\mid \mid^{A}\hat{X}\mid \cos 90 \\ \mid ^{B}\hat{X}\mid \mid^{A}\hat{Y}\mid \cos (90- \theta) && \mid ^{B}\hat{Y}\mid \mid ^{A}\hat{Y}\mid \cos \theta && \mid ^{B}\hat{Z}\mid \mid ^{A}\hat{Y}\mid \cos 90 \\ \mid ^{B}\hat{X}\mid \mid^{A}\hat{Z}\mid \cos 90 && \mid ^{B}\hat{Y}\mid \mid^{A}\hat{Z}\mid \cos 90 && \mid ^{B}\hat{Z}\mid \mid ^{A}\hat{Z}\mid \cos 0 \end{bmatrix} \end{align*}\]Now we know that \(^{B}\hat{X}\), \(^{B}\hat{Y}\), \(^{B}\hat{Z}\), and \(^{A}\hat{X}\), \(^{A}\hat{Y}\), \(^{A}\hat{Z}\) are unit basis vectors. Therefore \(\mid ^{B}\hat{X}\mid\), \(\mid ^{B}\hat{Y}\mid\), \(\mid ^{B}\hat{Z}\mid\) and \(\mid ^{A}\hat{X}\mid\), \(\mid ^{A}\hat{Y}\mid\), \(\mid ^{A}\hat{Z}\mid\) will be equal to \(1\).
Therefore we can write
\[\begin{align*} ^{A}_{B}{R}_z &= \begin{bmatrix} \cos \theta && -\sin \theta && 0\\ \sin \theta && \cos \theta && 0\\ 0 && 0 && 1 \end{bmatrix} \end{align*}\]Similarly we find \(^{A}_{B}R_y = \begin{bmatrix} \cos \theta && 0 && \sin \theta \\ 0 && 1 && 0 \\ -sin \theta && 0 && \cos \theta \end{bmatrix}\) if we are rotating along the \(y\) axes.
If we are rotating along the \(x\) axes, \(^{A}_{B}R_x = \begin{bmatrix} 1 && 0 && 0 \\ 0 && \cos \theta && -\sin \theta \\ 0 && sin \theta && \cos \theta \end{bmatrix}\).
Rotation matrices are invertible and the inverse of a rotation matrix \(R\) is \(R^T\).
\[\begin{align*} ^{B}_{A}{R} &= \begin{bmatrix}^{B}\hat{X}_{A}, ^{B}\hat{Y}_{A}, ^{B}\hat{Z}_{A}\end{bmatrix} \\ &= \begin{bmatrix} ^{A}\hat{X}^{B}\hat{X} && ^{A}\hat{Y}^{B}\hat{X} && ^{A}\hat{Z}^{B}\hat{X} \\ ^{A}\hat{X}^{B}\hat{Y} && ^{A}\hat{Y}^{B}\hat{Y} && ^{A}\hat{Z}^{B}\hat{Y} \\ ^{A}\hat{X}^{B}\hat{Z} && ^{A}\hat{Y}^{B}\hat{Z} && ^{A}\hat{Z}^{B}\hat{Z} \end{bmatrix}\\ &= ^{A}_{B}{R}^T \end{align*}\]Thus if we use \(R\) to represent points in \(B\) in terms of \(A\) then we can use \(R^{T}\) to represent points in \(A\) in terms of \(B\).
They are also used to describe the [Orientation] of a point in space.