Find $\vec r$ if $\vec r \times \vec a = \vec b$ In $\mathbf R^3$, let $\vec r$ be a unit vector satisfying the condition
$$\vec r \times \vec a = \vec b\tag {i}$$
Here, $\vec a$ and $\vec b$ are vectors whose magnitudes are
$$\begin{alignat}{2}
|\vec a|&=&\sqrt 3\tag{ii}\\
|\vec b|&=&\sqrt 2\tag{iii}
\end{alignat}$$
Then what is the expression of $\vec r$ in terms of $\vec a$ and $\vec b$?
 A: There is a well know procedure used in mechanics for this.
First of all note that if $\mathbf{a}=\mathbf{0}$ then also $\mathbf{b}$ must be zero, so there is no interest in this case, and we suppose $\mathbf{a}\neq\mathbf{0}.$
Next, if you perform a scalar product by $\mathbf{a}$ you obtain
$$
\mathbf{r}\times\mathbf{a}\cdot\mathbf{a}=\mathbf{a}\cdot\mathbf{b}
$$
and given that the RHS vanished, then a necessary condition for the existence of solutions is $\mathbf{a}\cdot\mathbf{b}=0$
Now try a solution of type $\mathbf{r}=\lambda\,\mathbf{a}\times\mathbf{b}$ with $\lambda$ to be determined. By substituting in the equation we have
\begin{align}
&(\lambda\,\mathbf{a}\times\mathbf{b})\times\mathbf{a}=\mathbf{b}\\
&\lambda\,|\mathbf{a}|^2\,\mathbf{b}-\lambda(\mathbf{a}\cdot\mathbf{b})\mathbf{a}=\mathbf{b}\\
&\lambda\,|\mathbf{a}|^2\,\mathbf{b}=\mathbf{b}\\
&\lambda=\frac{1}{|\mathbf{a}|^2}
\end{align}
so a solution is 
$$
\mathbf{r}=\frac{\mathbf{a}\times\mathbf{b}}{|\mathbf{a}|^2}
$$
But this is not the only solution, because adding $\mu\,\mathbf{a}$ to this is still a solution, so the set of all solutions is given by
$$
\mathbf{r}=\frac{\mathbf{a}\times\mathbf{b}}{|\mathbf{a}|^2}+\mu\,\mathbf{a},\qquad\mu\in\mathbb{R}
$$
(all solutions because it is easy to see that the difference of two solution must be parallel to $\mathbf{a}$.)
Should be added that only knowing the magnitudes of $\mathbf{a}$ and $\mathbf{b}$ cannot be enough to obtain a solution.
Edit
The requirement that $|\mathbf{r}|=1$ gives
$$
|\mathbf{r}|^2=\frac{|\mathbf{a}|^2|\mathbf{b}|^2}{|\mathbf{a}|^4}+\mu^2|\mathbf{a}|^2=1\implies\mu^2=\frac{|\mathbf{a}|^2-|\mathbf{b}|^2}{|\mathbf{a}|^4}
$$
so there are two vector satisfying the condition.
A: $||r\wedge a||^2=\underbrace{||r||^2}_{=1}||a||^2\sin(\theta)^2=||b||^2$ 
$(r\cdot a)^2=\underbrace{||r||^2}_{=1}\,||a||^2\cos(\theta)^2=||a||^2-||a||^2\sin(\theta)^2=||a||^2-||b||^2$
From the triple vector product formula we get $(r\wedge a)\wedge a=(r\cdot a)a-(a\cdot a)r=b\wedge a$
$r=\frac 1{||a||^2}\bigg(a\wedge b\pm\sqrt{||a||^2-||b||^2}\ a\bigg)$
A: 
Clearly, $\vec b=(\vec r \times \vec a) \perp \vec r,\,\vec a$.
From the figure, it is clear that $\vec a \times \vec b$ lies in the plane of $\vec r$ and $\vec a$. Also, since $\vec a \perp \vec b$, we have
$ |\vec a \times \vec b| = |\vec a||\vec b| = \sqrt6$. Hence we may write $\vec r = \lambda \vec a + \mu (\vec a \times \vec b)$ for some real $\lambda ,\, \mu$. Since $\vec a \times \vec b \perp \vec a$, we have
$$\begin{eqnarray}
|\vec r|&=&|\lambda \vec a + \mu (\vec a \times \vec b)|\\
        &=& \sqrt{\lambda^2|\vec a|^2 + \mu^2|\vec a \times \vec b|^2}\\
        &=& \sqrt{3\lambda^2 + 6\mu^2}
\end{eqnarray}$$
But, $\vec r$ is a unit vector. So, $3\lambda^2 + 6\mu^2=1$. Hence,
$$ \boxed{\vec r = \lambda \vec a \pm \sqrt{\frac{1-3\lambda^2}{6}}(\vec a \times \vec b)} \quad \forall \quad \lambda \in \mathbf R - \Bigg\{-\frac1{\sqrt3},\,\frac1{\sqrt3}\Bigg\}$$
Here $\lambda \neq \pm \frac1{\sqrt3}$ because the coefficient of $(\vec a \times \vec b)$ cannot be zero.
