Quick question on solving some vectors.

Question

So, I have the following equation, $\mathbf{a}_s = \mathbf{a}_r + 2(\omega\times\mathbf{v}_r) + \omega\times(\omega\times\mathbf{r})$. And now I will let $\mathbf{a}_s = \mathbf{a}_r$ so we get, \begin{equation} 0 = 2(\omega\times\mathbf{v}_r) + \omega\times(\omega\times\mathbf{r}) \end{equation} Now, the goal is to solve for $\mathbf{v}_r$. I thought about using the identity, $a\times(b\times c) = (a\cdot c)b - (a\cdot b)c$ (and everything present are vectors) to simplify the right hand side, then act on everything a $\omega\times$ to get $\mathbf{v}_r$ by itself, but it doesn't work. So, I am stuck. Any suggestions are helpful, thanks.

Answer 1

Use associativity: $a\times b+a\times c=a\times(b+c)$. So you get $$\omega\times(2{\bf v_r}+\omega\times {\bf r})=0$$ The cross product of two vectors is zero if the two vectors are are collinear:$$2{\bf v_r}+\omega\times {\bf r}=2c\omega$$or $${\bf v_r}=c\omega-\frac12\omega\times {\bf r}$$ with $c\in\mathbb R$