How to solve $\vec x \times \vec a + \vec y \times \vec b = \vec r \times \vec v$?

Question

Given the equation:

$\vec x \times \vec a + \vec y \times \vec b = \vec r \times \vec v$

and knowing $\vec x , \vec y , \vec r$ and $\vec v$,

is is it possible to find $\vec a$ and $\vec b$?

example:

$\vec x$ = [-10, 0, 0]
$\vec y$ = [10,0,0]
$\vec r$ = [4, 20, -3]
$\vec v$ = [-3,1,-5]

Answer 1

Set $a=(a_1,a_2,a_3), b=(b_1,b_2,b_3)$ and try. You get:

$$x_2a_3-x_3a_2+y_2b_3-y_3b_2=r_2v_3-r_3v_2$$ $$x_3a_1-x_1a_3+y_3b_1-y_1b_3=r_3v_1-r_1v_3$$ $$x_1a_2-x_2a_1+y_1b_2-y_2b_1=r_1v_2-r_2v_1$$

Your example boils down to: $$0=-97$$ $$10(a_3-b_3)=-11$$ $$10(b_2-a_2)=64$$ Obviously the first never holds, so this has no solutions, however I suspect that is the work of the $0$ co-ordinates in $x$ and $y$. There will be solutions for others, but don't expect them to be unique.

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WRT Comment:

You get:

$$a_3-a_2+b_3-b_2=-80\tag1$$ $$a_1+10a_3+b_1-10b_3=22\tag2$$ $$-10a_2-a_1+10b_2-b_1=71\tag3$$ We can trivially satisfy $(2)$ with $a_1=b_1=a_3=1, b_3=-1$

cancelling these out reduces $(1)$ and $(3)$ to

$$a_2+b_2=80\tag4$$ $$10b_2-10a_2=73\tag5$$

which yield $a_2=\frac{727}{20}, b_2=\frac{873}{20}$, so

$$\vec{a}=(1,\frac{727}{20},1), \vec{b}=(1,\frac{873}{20},-1)$$ satisfy this.

Answer 2

The equation can be simplified to be written as $$\left(\begin{matrix}10\\0\\0\end{matrix}\right)\times(\underline{b}-\underline{a})=\left(\begin{matrix}-97\\29\\64\end{matrix}\right)$$

This leads to $0=-97$ so there is no vector $\underline{b}-\underline{a}$ satisfying this. Therefore there are no solutions.