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 $$d\vec{r}_0 = R(dt, \vec{\omega})\vec{r}_0-\vec{r}_0 \tag{1}$$$$d\vec{r} = R(dt, \vec{\omega})\cdot \vec{r}-\vec{r} \tag{1}$$ where $R(dt, \vec{\omega})$ is an infinitesimal rotation matrix 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 $$d\vec{r}_0 = dt A\cdot\vec{r}_0.$$$$d\vec{r} = dt A\cdot\vec{r}.$$ We remark that we can rewrite $A \cdot \vec{r}_0$$A \cdot \vec{r}$ as $\vec{\omega} \times \vec{r}_0$$\vec{\omega} \times \vec{r}$ (see the wikipedia page for alternate ways to compute the cross product).
Finally, dividing by $dt$, we obtain $$\vec{v} \equiv \frac{d\vec{r}_0}{dt} = \omega\times\vec{r}_0.$$$$\vec{v} \equiv \frac{d\vec{r}}{dt} = \omega\times\vec{r}_0.$$