Why do we use the cross product in relative motion? The equation of motion for $V_{b/a}$ is:
$V_{b/a} = \dot{r}_{b/a} = \omega \times r_{b/a}$
Why do we use the cross product? For some reason I am unable to gather the intuition for its use. It's the same with other dynamic equations, the difference between two motions is usually the cross product. For example:
$\alpha = \frac{d\omega}{dt}\vert_{xyz} + \Omega \times \omega$
Many resources just say "we use the cross product" but that does not give meaningful intuition. My question is why? It's ubiquitous in my dynamics course and rotation and I don't understand why. 
 A: These are vector equations. E.g. if we consider a particle spinning with frequency $f$ on a circle with radius $r$ it's velocity is given by $|v| = (2\pi f) \cdot r$. However, this is only the absolut value of the velocity. 
The velocity is actually a 3 dimensional vector,
$\vec v = (v_x, v_y, v_z)^T$. Thus, in order to obtain the vector components of the velocity we need to consider the vector $\vec \omega = 2\pi \vec f$ and $\vec r$. Now, the cross product of two vectors $\vec a$ and $\vec b$ is perpendicular to both vectors,
\begin{align}
\vec a \, \times\, \vec b &\perp \vec a\\
\vec a \, \times\, \vec b &\perp \vec b
\end{align}
Now, imagine the particle counter-clockwise spins in the $(x,y)$-plane. The angular velocity for this particle is given by $\vec \omega = \omega \cdot \vec{e}_z$, while the distance to the centre-point of the circle is given by $\vec r = r\cdot \vec{e}_{\rho}$, where $\vec{e}_{\rho}$ is the radial unit vector in polar coordinates. Now, the velocity in this example is given by $\vec{v} = v \cdot \vec{e}_{\phi}$, where $\vec{e}_{\phi}$ is the azimuthal unit vector in polar coordinates. Hence, we need to find a mathematical formula to obtain this direction from $\vec{e}_z$ and $\vec{e}_{\rho}$. If you think about this, you realise that 
$$
\vec v = \vec \omega \times \vec r
$$
is correct.
Finally, consider add a constant $z$-motion to the rotating particle. This constant change of the $z$ position does not change the azimuthal velocity part. However, it shows up in $\vec r$. So we need a way to get rid of the parallel component of the two vectors 
$\vec \omega$ and $\vec r$.  This is handled by the cross product as well, 
$$
|\vec \omega \, \times\, \vec r| 
= |\vec \omega| |\vec r| \cdot \sin{(\alpha_{ab})}
= \omega \cdot \rho
$$
where $\rho$ is the distance of the particle to the $z$-axis.
