Finding the General Solution of a Linear System in Compact Form Consider the following two vector equations
\begin{align*}
\mathbf{a} \times (\mathbf{x}-\mathbf{y}) &= \mathbf{b}_1 \tag{1}\\
\mathbf{x} + \mathbf{y} \phantom{)} &= \mathbf{b}_2 
\tag{2}
\end{align*}
where $\mathbf{a}$, $\mathbf{b}_1$ and $\mathbf{b}_2$ are known vectors in $\Bbb{R}^3$ and $\mathbf{x}$ and $\mathbf{y}$ are unknowns which we wish to find. Here are my questions

  
*
  
*Does this system have a unique solution?
  
*Can we obtain a compact formula for $\mathbf{x}$ and $\mathbf{y}$?
  

My Thought
I think that due to the presence of  cross product, having a unique solution is too much to expect! For finding a compact formula, the difficulty lies in finding all of the solutions of $(1)$ by thinking of $\mathbf{x}-\mathbf{y}$ as a single unknown and thinking of $(1)$ as a linear systems of equations. After finding $\mathbf{x-y}$ from the first equation, it will be straight forward to find $\mathbf{x}$ and $\mathbf{y}$ by simply adding and subtracting the equations. However, I don't know how to present the general solution of $(1)$ in a compact form. Any hint or help is appreciated.
 A: In general, if $\vec a=(a_x,a_ya_z)^T$ and $\vec b=(b_x,b_yb_z)^T$ we can represents the cross product in matrix form as:
$$
\vec a\times \vec b=
\begin{pmatrix}
0&-a_z&a_y\\
a_z&0&-a_x\\
-a_y&a_x&0
\end{pmatrix}
\begin{pmatrix}
b_x\\b_y\\b_z
\end{pmatrix}
$$
but note that the matrix is singular, so  the existence of a  solution of $\vec a \times \vec b = \vec c$ is not guaranties and, in fact, exists only if $\vec c$ is orthogonal to $\vec a$ and $\vec b$, as noted in the comment of Winther.
A: Solving the second equation for ${\bf y}$ and substituting into the first one gives us
$${\bf a}\times{\bf x} = {\bf c}~~~\text{where}~~~{\bf c} = \frac{{\bf b_1} + {\bf a}\times {\bf b_2}}{2}$$
Now since ${\bf a}\cdot ({\bf a}\times {\bf x}) = {\bf x}\cdot ({\bf a}\times {\bf x}) = 0$ we have ${\bf a}\cdot {\bf c} = {\bf x}\cdot {\bf c} = 0$ so ${\bf c}$ has to be orthogonal to both ${\bf a}$ and ${\bf x}$. This is satisfied only when  ${\bf a}\cdot {\bf b_1} = 0$. By assuming that ${\bf a} \cdot {\bf x}=0$ we can get
$${\bf a}\times{\bf c} ={\bf a}\times ({\bf a}\times {\bf x}) = ({\bf a} \cdot {\bf x}) {\bf a} - ({\bf a} \cdot {\bf a}) {\bf x} = -{\bf x}\|{\bf a}\|^2$$ 
and hence we find
$${\bf x} = - \frac{{\bf a} \times {\bf c}}{\|{\bf a}\|^2}$$
Finally if ${\bf x}$ is a solution then so is ${\bf x} + k{\bf a}$ for any $k\in\mathbb{R}$ so the most general solution is 
$${\bf y} = {\bf b_2} - {\bf x}~~~\text{with}~~~{\bf x} = k{\bf a} - \frac{{\bf a} \times {\bf c}}{\|{\bf a}\|^2}~~~\text{when}~~~{\bf a}\cdot {\bf b_1} = 0$$
