Equation on cross product I have to solve the equation $\vec x\times\vec a=\vec b-\vec x$  ,   where $\vec x, \vec a,\vec b$ are vectors and the last two are known.
I have proven that $\vec a\cdot \vec x=\vec a\cdot\vec b$, but that is all.
Thank you in advance !
<< Symbol : $\vec a \vec b = \vec a \cdot \vec b$ >>
 A: Given:
$$\vec{x} = \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}, \vec{a} = \begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix} ~\text{and}~ \vec{b} = \begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix},$$
the equation
$$\vec x\times\vec a=\vec b-\vec x$$
corresponds to:
$$\begin{cases}
a_3 x_2 - a_2 x_3 +x_1 - b_1 = 0\\
a_1 x_3 - a_3 x_1 +x_2 - b_2 = 0\\
a_2 x_1 - a_1 x_2 +x_3 - b_3 = 0
\end{cases}.$$
The solution is:
$$x_1 = \frac{a_1(\vec{a} \cdot \vec{b}) + b_1 + a_2 b_3 - a_3 b_2}{|\vec{a}|^2 + 1},$$
$$x_2 = \frac{a_2(\vec{a} \cdot \vec{b}) + b_2 + a_3 b_1 - a_1 b_3}{|\vec{a}|^2 + 1},$$
$$x_2 = \frac{a_3(\vec{a} \cdot \vec{b}) + b_3 + a_1 b_2 - a_2 b_1}{|\vec{a}|^2 + 1}.$$
Note that $\vec{a} \cdot \vec{b}$ is the scalar product, while $|\vec{a}|^2 = \vec{a} \cdot \vec{a}$ is the squared norm of the vector $\vec{a}$.
A: If $\vec{x}=\vec{y},\,\vec{x}=\vec{y}+\vec{z}$ both work, $\vec{z}\times\vec{a}=-\vec{z}$, so $\vec{z}$ is self-orthogonal and vanishes, and the solution $\vec{x}$ is unique. Now try the Ansatz $\vec{x}=A\vec{a}+B\vec{b}+C\vec{a}\times\vec{b}$. As I noted in a comment on @thecandyman's answer, you should find$$\vec{x}=\frac{(\vec{a}\cdot\vec{b})\vec{a}+\vec{b}+\vec{a}\times\vec{b}}{a^2+1}.$$
