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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?

Salihcyilmaz
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  • I suppose $\nu$ is known as well? – Calvin Khor Mar 05 '16 at 10:15
  • yes, editing know. Thanks for pointing... – Salihcyilmaz Mar 05 '16 at 10:16
  • @Salihcyilmaz Are they vectors in $R^3$ – user41736 Mar 05 '16 at 10:21
  • @user41736 i was thinking in $R^3$ since i was dealing with electromagnetics. But it is interesting question anyways, which lead me to think about it. At first glance cross product seemed meaningless to me at other dimensions. Then I found this Q/A, check it out: http://math.stackexchange.com/questions/720813/do-four-dimensional-vectors-have-a-cross-product-property – Salihcyilmaz Mar 05 '16 at 10:27

3 Answers3

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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 $\vec k\ne\vec0$.)

joriki
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  • Brilliant solution, i salute you... But from where did you get the idea? – Salihcyilmaz Mar 05 '16 at 10:23
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    @Salihcyilmaz: The operation $\vec k\times$ turns the component perpendicular to $\vec k$ by $\pi/2$, so I figured you need to apply it again to get something useful. From there it was just looking up the BAC CAB rule and doing the arithmetic. – joriki Mar 05 '16 at 10:33
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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.)

Derek Elkins left SE
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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.

DonkoXI
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