Determining an unknown vector from its cross and dot product with known vector Let $\vec{k}$, $\vec{v}$, and $\vec{u}$ be vectors, where $\vec{u}$ is unknown and $\vec{k}$ and $\vec{v}$ are known vectors. Given:  


*

*$\vec{u}\cdot\vec{k}=c$ 

*$\vec{u} \times \vec{k}= \vec{v}$


From this relations, how can i determine the vector $u$? I tried to construct orthogonal coordinate system from $(\vec{k}, \vec{v},\vec{k}\times \vec{v})$ but i couldn't proceed from there. Any idea?
 A: From
$$
\vec k\times(\vec u\times\vec k)=\vec u(\vec k\cdot\vec k)-\vec k(\vec k\cdot\vec u)
$$
you get
$$
\vec u=\frac{\vec k\times(\vec u\times\vec k)+\vec k(\vec k\cdot\vec u)}{\vec k\cdot\vec k}=\frac{\vec k\times\vec v+c\vec k}{\vec k\cdot\vec k}\;.
$$
(Of course you need $\vec k\ne\vec0$.)
A: In geometric algebra you can divide by non-zero vectors. So to calculate $u$ given $uk = u\cdot k + u\wedge k$, where in 3D $u\wedge k = (u\times k)I$, you just multiply on the right by $k^{-1} = k/k^2$.  This gives $$u = ukk^{-1}= \frac{(u\cdot k)k + (u\wedge k)k}{k^2} = \frac{(u\cdot k)k-(u\times k)\times k}{k^2}$$
where I also used $(AI)\cdot v = (A\wedge v)I$, $I^{-1} = -I$, $v\wedge v = 0$ and $(a\wedge b)c = (a\wedge b)\cdot c + a\wedge b\wedge c$.
If being able to divide by vectors sounds like something you'd find useful, you should look at geometric algebra.  (Actually, you should look at geometric algebra regardless. Note, other than the translation to the cross product, everything above can be done in any dimension, though the sign of $I^{-1}$ varies with dimension.)
A: You can get to the same answer as joriki more along your original lines of thought. To do this we will utilize a new orthogonal coordinate system by embedding $\vec{u}$ in the plane formed by $\vec{k}$ and $\vec{k} \times \vec{v}$.
First to find the $\vec{k}$ component, take the projection of $\vec{u}$ onto $\vec{k}$.
\begin{equation}
\text{proj}_\vec{k}(\vec{u}) %
= \frac{\vec{k} \cdot \vec{u}}{\left|\vec{k}\right|^2}\:\vec{k} %
= \frac{c\vec{k}}{\left|\vec{k}\right|^2}
\end{equation}
Now we will find the $\vec{k} \times \vec{v}$ component.
$\vec{u} \times \dfrac{\vec{k}}{\left|\vec{k}\right|}$ will have the magnitude of the desired vector, but will be in the $\vec{v}$ direction. Since crossing anything in the $\vec{k}$ direction with anything in the $\vec{v}$ direction will naturally be in the $\vec{k} \times \vec{v}$ direction, we will pre-cross our expression with a unit vector (as not to alter the magnitude) in the $\vec{k}$ direction.
\begin{equation}
\text{proj}_{\vec{k} \times \vec{v}}(\vec{u}) %
= \frac{\vec{k}}{\left|\vec{k}\right|} \times \left(\vec{u} \times \frac{\vec{k}}{\left|\vec{k}\right|}\right) %
= \frac{\vec{k}\times \vec{v}}{\left|\vec{k}\right|^2}
\end{equation}
The final solution is just a matter of adding these two componants,
\begin{equation}
\vec{u} = \frac{\vec{k}\times \vec{v} + c\vec{k}}{\left|\vec{k}\right|^2}
\end{equation}
The choice to use $\:\vec{k}\cdot\vec{k}$ or $\:\left|\vec{k}\right|^2$ is largly dependent on setting and application as these two are equal.
