I have this equation:$$ -\nu(v)\mathbf{F}(v)+\Omega(\mathbf{b\times F}(v))= -\mathbf{Q}(v) $$ I want to solve this equation to find $\mathbf{F}(v)$, but I don't really know where to start. The tip in the book is to take $\mathbf{b\times}$ the former equation and "solve the system of equations to find the components of $\mathbf{F}$ perpendicular to $\mathbf{b}$.
Does anyone have any insights?
Further to some comments, I have tried the tip, and got stuck here
$$-\nu \mathbf{b \times F}+\Omega\mathbf{b \times (b \times F)} = -\mathbf{b \times Q} \\ \nu(\mathbf{b \times F}) + \Omega[\mathbf{(b \cdot F)b - (b \cdot b)F}]=-\mathbf{b\times Q}$$
But I don't know where to go from here, to find the next steps.
You can consider operation of $\mathbf b\times \mathbf F$ as a multiplication by dual matrix:
$$ \mathbf b\times\mathbf F = \begin{pmatrix}0&-b_3&b_2\\b_3&0&-b_1\\-b_2&b_1&0\end{pmatrix}\mathbf F= \mathbf B\mathbf F $$
At the same time you can consider $-\nu\mathbf F=-\nu\mathbf I\mathbf F$, where $\mathbf I$ is identity matrix.
Thus, your equation can be rewritten as: $$ (\Omega\mathbf B - \nu\mathbf I)\mathbf F = \mathbf Q,\\ \mathbf F=(\Omega\mathbf B - \nu\mathbf I)^{-1}\mathbf Q $$