Angular velocity vector from linear velocity vector

Question

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 ?

mike stone · Accepted Answer

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!

NDewolf · Answer

The correct definition of linear velocity is $vec{v} = vec{omega}timesvec{r}$ 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.

Angular velocity vector from linear velocity vector

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