Angular velocity vector from linear velocity vector Linear velocity vector is $\vec{v}$ and the distance vector is $\vec{r} $.
How is it that angular velocity vector is $\vec{ω}$ =  $\frac {\vec{r} × \vec{v}}{|\vec{r}|^2}$
The equation I am aware of is  $\vec{v}$ $=$ $\vec{r}×\vec{ω}$. I tried taking a dot product on both sides, cross product etc but I wasn't able to derive the above equation. How is it possible ?
 A: The triple vector-product identity gives
$$
{\bf r}\times {\bf v}= {\bf r}\times({\boldsymbol \omega}\times {\bf r} )=-({\boldsymbol \omega}\cdot {\bf r}){\bf r}+ |{\bf r}|^2 {\boldsymbol \omega}. $$
But $({\boldsymbol \omega}\cdot {\bf r})$ is not necessarily zero, and so in general
${\boldsymbol \omega}\ne ({\bf r}\times {\bf v})/|r|^2$. It's only true if ${\bf r}$ is perpendicular to ${\boldsymbol \omega}$
In my original answer I miscopied fom a piece of paper and had $-({\boldsymbol \omega}\cdot {\bf v}){\bf r}$ in the first term, and this is zero, but wrong!
A: The correct definition of linear velocity is $\vec{v} = \vec{\omega}\times\vec{r}$ and your relation for $\vec{\omega}$ is then a simple application of the vector triple product, see https://en.wikipedia.org/wiki/Triple_product.
(Or the other way around if you prefer that.)
Edit: As @mikestone correctly notes in his answer this is only valid for circular motion where the position vector is orthogonal to the angular velocity vector.
