It can be much simpler and much less error prone (user errors) to transform forces and moments between two points by using a cross product instead of using transformation matrices like:

$\begin{bmatrix} 1&0&0\\ 0&\cos{\theta} & \sin{\theta}\\ 0&-\sin{\theta}& \cos{\theta} \end{bmatrix}$

The simple trick is to recognize that a cross product can be expressed as a matrix multiply as follows:

$\begin{matrix} a\times{b} & = & skew(a) \cdot{b}\\ \begin{bmatrix} a_x &a_y &a_z \end{bmatrix}\times{b}&=&\left[ \begin{matrix}0&-a_z&a_y\\a_z&0&-a_x\\-a_y&a_x&0 \end{matrix}\right]\cdot{b} \end{matrix}$

So to show how we would use the cross product to our advantage, see the the following example.  We want to transform the forces and moments applied to a rigid body at point A to point B:

$\begin{matrix} F_b&=&T_{Fa2b}\cdot{F_a} \\ \begin{bmatrix} F_{bx}\\ F_{by}\\ F_{bz}\\ F_{brx}\\ F_{bry}\\ F_{brz} \end{bmatrix} & = & \begin{bmatrix} I&0\\skew(ptb-pta)&I\end{bmatrix}\\ \end{matrix}\cdot{\begin{bmatrix} F_{ax}\\ F_{ay}\\ F_{az}\\ F_{arx}\\ F_{ary}\\ F_{arz} \end{bmatrix}}$

It’s quite easy to see that the forces along the cartesian coordinates remain the same between the two points, but the moments change due to the moment arm (ptb-pta).

It is also useful that displacements can be easily transformed using the transpose of the transformation matrix above:

$\begin{matrix} X_b&=&T_{Fa2b}'\cdot{X_a} \\ \begin{bmatrix} X_{bx}\\ X_{by}\\ X_{bz}\\ X_{brx}\\ X_{bry}\\ X_{brz} \end{bmatrix} & = & \begin{bmatrix} I&skew(ptb-pta)\\0&I\end{bmatrix}\\ \end{matrix}\cdot{\begin{bmatrix} X_{ax}\\ X_{ay}\\ X_{az}\\ X_{arx}\\ X_{ary}\\ X_{arz} \end{bmatrix}}$