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The formula for the velocity of a point $\vec {r}$ with angular velocity $\vec{\omega}$ is given by $\vec{v}=\vec{\omega} \times \vec{r}$.

But, how can we derive the formula $\vec{\omega}=\frac{\vec{r} \times \vec{v}}{|\vec{r}|^2}$ from the above formula, which is known?

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$$\begin{align} \vec v&=\vec \omega \times \vec r\\ \vec r \times \vec v&=\vec r \times (\vec \omega \times \vec r)\\ &=\vec \omega (\vec r \cdot \vec r)-\vec r(\vec r\cdot \vec \omega)^*\\ &=\vec \omega |\vec r|^2~^{**}-0^{***}\\ \frac{\vec r \times \vec v}{|\vec r|^2}&=\vec \omega\\ \end{align}$$

$^*$ The tripple rule: $$A\times (B\times C)=B(A\cdot C)-C(A\cdot B)$$

$^{**}$ Dot product with itself:

$$A\cdot A=|A|^2$$

$^{***}$ Direction of $\vec \omega$ is out of (or into) the circle plane and thus always perpendicular to any vector drawn on the circle plane. So $\vec r \perp \vec \omega$. Dot product of perpendicular vectors:

$$A \cdot B=0\qquad \text{for} \quad A \perp B$$

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Use the identity (see this) $$ {\bf v} = \left({\bf v} \bullet \hat{\bf r}\right) \hat{\bf r} + \left(\hat{\bf r} \times {\bf v}\right) \times \hat{\bf r} $$ and the fact that ${\bf v} \bullet \hat{\bf r} = \left(\omega \times {\bf r}\right) \bullet \hat{\bf r} = 0$.

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