Suppose you have a vector $\vec{r}_0$ that is rotating counter clockwise with constant angular velocity $\vec{\omega}$. Our task is to determine the tangential velocity $\vec{v}$.

We can do this in the following way. Consider the infinitesimal difference 
$$d\vec{r}_0 = R(dt, \vec{\omega})\vec{r}_0-\vec{r}_0 \tag{1}$$
where $R(dt, \vec{\omega})$ is an infinitesimal rotation around the $\vec{\omega}$ direction. We can rewrite $R(dt, \vec{\omega})$ as
$$R(dt, \vec{\omega}) = e^{dt A} = 1 +dt A+\mathcal{O}(dt^2)$$
where 
$$A \equiv\begin{pmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{pmatrix}. \tag{2}$$
Truncating the exponential series at first order, we obtain
$$R(dt, \vec{\omega}) = 1 + dt A.$$
Hence, equation (1) becomes
$$d\vec{r}_0 = dt A\cdot\vec{r}_0.$$
We remark that we can rewrite $A \cdot \vec{r}_0$ as $\vec{\omega} \times \vec{r}_0$ (see the [wikipedia page][1] for alternate ways to compute the cross product).

Finally, dividing by $dt$, we obtain
$$\vec{v} \equiv \frac{d\vec{r}_0}{dt} = \omega\times\vec{r}_0.$$


  [1]: https://en.wikipedia.org/wiki/Cross_product#Alternative_ways_to_compute