<s>[INST] $\overline{x} \times \overline{a} = \overline{b}$ has a solution when $ \langle\overline{a},\overline{b} \rangle =0$ I'm trying to solve this exercise:

Let $\overline{a} \neq \overline{0}$, $\overline{b}$ be two vectors of the Euclidean vector space $V_{3}$. Prove the equation $\overline{x} \times \overline{a} = \overline{b}$ has a solution when $ \langle\overline{a},\overline{b} \rangle =0$ and find if it's the only one.

It does not seem that hard, my approach is the following, but I seem to be missing something:
Let $B = \left \{ \overline{\epsilon_{1}}, \overline{\epsilon_{2}}, \overline{\epsilon_{3}} \right \}$ be a base of $V_{3} $
Then the vectors $\overline{a}$. $\overline{b}$ can be represented as: 
$\overline{a} = a_{1}\overline{\epsilon_{1}} + a_{2}\overline{\epsilon_{2}} + a_{3}\overline{\epsilon_{3}} = \sum_{i=1}^{3}a_{i}\overline{\epsilon_{i}}$
$\overline{b} = \sum_{j=1}^{3}b_{j}\overline{\epsilon_{j}}$
I also analyze the vector $\overline{x}$ as follows:
$\overline{x} = x_{1}\overline\epsilon_{1} + x_{2}\overline\epsilon_{2} +x_{3}\overline\epsilon_{3}$
$\overline{x} \times \overline{a} = \overline{b}$
$(x_{2}a_{3} - x_{3}a_{2})\overline\epsilon_{1} + (x_{3}a_{1} -x_{1}a_{3})\overline\epsilon_{2} + (x_{1}a_{2} - x_{2}a_{1})\overline\epsilon_{3} = \sum_{j=1}^{3}b_{j}\overline\epsilon_{j}$
And I get:
$x_{2}a_{3} - x_{3}a_{2} = b_{1}$
$x_{3}a_{1} - x_{3}a_{3} = b_{2}$
$x_{1}a_{2} - x_{3}a_{1} = b_{3} $
I am aware that $\langle\overline{a},\overline{b} \rangle =0$ means that the two vectors at vertical with each other, but how to I proceed from here?
 [/INST] We know that $\vec x \times \vec a = \vec b$ means that $\vec a \perp \vec b$ 
If the two vectors are not perpendicular then there are no solutions. The following system has no solutions:
$$\left\{ \begin{gathered}
  {a_3}{x_2} - {a_2}{x_3} = {b_1}  \\
  {a_1}{x_3} - {a_3}{x_1} = {b_2}  \\
  {a_2}{x_1} - {a_1}{x_3} = {b_3}  \\ 
\end{gathered}  \right.$$
We know that $\vec a\perp\vec b$ which means $a_1b_1+a_2b_2+a_3b_3=0$
we can solve for $b_3$ getting $b_3=-\dfrac{a_1b_1+a_2b_2}{a_3}$ and substitute into the previous system
$$\left\{ \begin{gathered}
  {a_3}{x_2} - {a_2}{x_3} = {b_1}  \\
  {a_1}{x_3} - {a_3}{x_1} = {b_2}  \\
  {a_2}{x_1} - {a_1}{x_3} = -\dfrac{a_1b_1+a_2b_2}{a_3}  \\ 
\end{gathered}  \right.$$
Solving the system we get:
$$\left\{ \begin{gathered}
  {x_1} = \frac{{{b_2}{b_3} - {a_1}{b_3}{x_3}}}{{{a_1}{b_1} + {a_2}{b_2}}}  \\
  {x_2} =  - \frac{{{b_1}{b_3} - {a_2}{b_3}{x_3}}}{{{a_1}{b_1} + {a_2}{b_2}}}  \\
  {x_3} = {x_3}  \\ 
\end{gathered}  \right.$$
Which shows that the solution is not unique
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