If a and b are vectors perpendicular to each other Then is it okay to say that the general solution to $\vec{r} \times \vec{a} = \vec{b}$ is $$\vec{r}= \frac{\vec{r}\cdot \vec{a}}{\vec{a} \cdot \vec{a}}\vec{a}+\frac{1}{\vec{a} \cdot \vec{a}}(\vec{a} \times \vec{b})\,?$$
I did cross product with a vector to the given vector equation to get the above result will we really say the solution is that only as such $x$ has dependance on $r$ again isn't ?
Starting with $\vec{r} \times \vec{a} = \vec{b} $
where $\vec{a}, \vec{b} $ are given, we have from the properties of cross product that $\vec{r}$ must be perpendicular to $\vec{b}$. Clearly we also must have $\vec{b} $ perpendicular to $\vec{a}$. Therefore, our solution $\vec{r}$ lies in the plane spanned by $\vec{a}$ and $\vec{a} \times \vec{b} $, i.e.
$ \vec{r} = t \vec{a} + s \left(\vec{a} \times \vec{b} \right) $
Plug this expression into the original equation
$ \left( t \vec{a} + s \left(\vec{a} \times \vec{b} \right)\right) \times \vec{a} = \vec{b}$
Using the fact that $\vec{a} \times \vec{a} = 0 $ and
$\left( \vec{a} \times \vec{b} \right) \times \vec{a} = - \vec{a} (\vec{a} \cdot \vec{b} ) + \vec{b} (\vec{a} \cdot \vec{a} ) $
and using the fact that $ \vec{a} \cdot \vec{b} = 0 $ because they are pendicular, then
$s = \dfrac{1}{\vec{a} \cdot \vec{a} } $
Therefore, the general solution is
$\vec{r} = t \vec{a} + \dfrac{ \vec{a} \times \vec{b} } { \vec{a} \cdot \vec{a}} $
where $t \in \mathbb{R}$.