Determining an unknown vector from its cross and dot product with known vector

Question

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:

1. $\vec{u}\cdot\vec{k}=c$
2. $\vec{u} \times \vec{k}= \vec{v}$

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?

Answer 1

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 $\vec k\ne\vec0$.)

Answer 2

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.)