# Is there a way to solve $\vec{y}=\vec{y}\times\vec{a}+\vec{b}$ for arbitrary vectors?

I have a vector equation, given by:
$$\vec{y}=\vec{y}\times\vec{a}+\vec{b}$$
and I'm trying to solve it for $$\vec{y}$$. I've tried to take $$\times\vec{a}$$ of both sides and use vector identities to get me somewhere, but it's really not looking promising and it's got me wondering whether there actually is a way to solve this problem without prior information about $$\vec{y}$$, $$\vec{a}$$ and $$\vec{b}$$

Crossing both sides with $$\mathbf a$$ gives \begin{align*} \mathbf y \times \mathbf a &= (\mathbf y \times \mathbf a) \times \mathbf a + \mathbf b \times \mathbf a \\ \mathbf y - \mathbf b &= (\mathbf y \cdot \mathbf a)\mathbf a - (\mathbf a \cdot \mathbf a)\mathbf y + \mathbf b \times \mathbf a. \end{align*} Dotting both sides with $$\mathbf a$$ gives $$\mathbf y \cdot \mathbf a = \mathbf b \cdot \mathbf a.$$ Thus, \begin{align*} \mathbf y - \mathbf b &= (\mathbf b \cdot \mathbf a)\mathbf a - (\mathbf a \cdot \mathbf a)\mathbf y + \mathbf b \times \mathbf a \\ (1 + (\mathbf a \cdot \mathbf a))\mathbf y &= (\mathbf b \cdot \mathbf a)\mathbf a + \mathbf b +\mathbf b \times \mathbf a \\ \mathbf y &= \frac{(\mathbf b \cdot \mathbf a)\mathbf a + \mathbf b + \mathbf b \times \mathbf a}{1 + |\mathbf a|^2}. \end{align*}
• You could also dot with $\vec{a}$ to start. Doing the dot and cross products then can be viewed as looking at the parallel and perpendicular components of $\vec{y}$ with respect to $\vec{a}$ which will identify the solution. Commented Mar 11 at 15:32
I will drop the vector arrow. We can rewrite the cross product as $$y \times a=Ay$$ for some $$3 \times3$$ matrix $$A$$. Then, we can re-write the equation to $$y=y \times a + b \iff Iy=Ay+b$$ with $$I$$ being the $$3 \times 3$$ identity matrix. Then you can re-write this as a linear system $$(I-A)y=b$$ and check for its space of solutions.