When I have vectors
$\vec r_1=(4,0,0)$, $\vec v_1=(-2,-1,-2)$,
$\vec r_2=(-2,0,3)$, and $\vec v_2=(-1,0,1)$, and unknown vectors $\vec r_3$ and $\vec v_3$ that satisfy
$m_1 \vec{r_1} \times \vec v_1 + m_2 \vec r_2 \times \vec v_2 + m_3 \vec r_3 \times \vec v_3=0$, for scalars $m_1=1$, $m_2=2$, and $m_3=3$,
How can I calculate $\vec r_3$ and $\vec v_3$ out of these ?
Thanks a lot.
We have $(4,0,0)\times(-2,-1,-2) + 2(-2,0,3)\times(-1,0,1) +3\vec r_3\times\vec v_3 = 0$. Thus,
$$\vec r_3 \times \vec v_3 = \frac{-1}{3}\left((0,8,-4) + (0,-2,0)\right) = \frac{1}{3}(0,-6,4).$$
Unfortunately, there is no way to determine $\vec r_3$ and $\vec v_3$ from this: in particular, if $(\vec r_3, \vec v_3)$ is a pair of vectors satisfying this equation, then so is $(\alpha\vec r_3, \frac{1}{\alpha}\vec v_3)$ for any scalar $\alpha \neq 0$. We cannot even determine $\vec r_3$ and $\vec v_3$ up to scaling, as also $(\vec r_3 + \vec v_3,\vec v_3)$ satisfies the same equation.
However, expanding out what that equation means, with $\vec r_3 = (a,b,c)$, and $\vec v_3 = (d,e,f)$, we obtain the following equations:
$bf - ce = 0$
$cd - af = -2$
$ae - bd = \frac{4}{3}$.
Now, when $a = 0$, a general solution to these is given in terms of $d\neq 0$ and $e$ by
$d \neq 0$, $c = \frac{-2}{d}, b = \frac{-4}{3d}$, $f = \frac{2e}{3}$.
When $a \neq 0$, we have $f = \frac{2-cd}{a}$, and $e = \frac{4 + 3bd}{3a}$. Substituting these into the first equation and simplifying, we have $6b-3bcd - 4c - 3bcd = 0$, so $c = \frac{3b}{2}$. Substituting that back through, we have $f = \frac{4-3bd}{2a}$, which gives our general solution in terms of $a\neq 0$, $b$, and $d$.