How can I solve this vector algebra problem? Suppose that
$$\vec{\omega} = \frac{\vec{s} \times \vec{v}}{\lVert \vec{s}\rVert^2}$$
How do I solve for $\vec{s}$?
 A: For the problem to have a solution, $\vec{\omega}$ has to be orthogonal to both $\vec{s}$ and $\vec{v}$, hence you can write $\vec{s}$ as a linear combination of $\vec{\omega}\times\vec{v}$, and $\vec{v}$, which are orthogonal to $\vec{\omega}$, and additionally orthogonal to each other (thus they form a basis of the plane orthogonal to $\vec{\omega}$).
That is:
$$\vec{s}=\alpha\vec{\omega}\times\vec{v}+\beta\vec{v}$$
But you need also to have
$$\frac{\vec{s}\times\vec v}{\lVert \vec{s}\rVert^2}=\frac{\alpha(\vec{\omega}\times\vec{v})\times\vec v}{\lVert \vec{s}\rVert^2}=\vec\omega$$
Or by expanding the "triple product" (see here):
$$\frac{\alpha}{\lVert \vec{s}\rVert^2}\left(\vec{v}(\vec{v}\cdot\vec{\omega})-\vec{\omega}(\vec{v}\cdot\vec{v})\right)=\vec\omega$$
That is,
$$\alpha=-\frac{\lVert \vec{s}\rVert^2}{\vec{v}\cdot\vec{v}}=-\frac{\lVert \vec{s}\rVert^2}{\lVert \vec{v}\rVert^2}$$
And $\beta$ remains arbitrary: when there is a solution, there are actually infinitely many solutions.
