Given $\vec{v_1} \times \vec{v_2}$ and $\vec{v_1} \times \vec{v_3}$, find $\vec{v_2} \times \vec{v_3}$

Question

Is it possible to get the cross product of two vectors $\vec{v_2}$ and $\vec{v_3}$, given the result of each cross product with a common vector $\vec{v_1}$?

e.g. $$ \vec{v_i} \in \mathbf{R}^2_+ \\ \lVert\vec{v_1} \times \vec{v_2}\lVert = 0.10007117 \\ \lVert\vec{v_1} \times \vec{v_3}\lVert = 0.05423194 \\ $$ find $\lVert\vec{v_2} \times \vec{v_3}\lVert$ . which is $0.01196676$, since i randomly generated 3 vectors: $$ v_1 = (0.97986514, 0.40872297) \\ v_2 = (0.69637269, 0.39259962) \\ v_3 = (0.2602131 , 0.16388685) $$

Is there any way to get to the answer without knowing the explicit vectors?

Answer 1

No. For example, $\vec{v1} = (0,1)$

The following to set of $\vec{v2}, \vec{v3}$ give the same $\vec{v1} \times \vec{v2}$ and $\vec{v1} \times \vec{v3}$, yet their $\vec{v2}\times\vec{v3}$ are different.

• $\vec{v2} = (1,1), \vec{v3} = (2,1)$
• $\vec{v2} = (1,10), \vec{v3}=(2,10)$

Answer 2

Here is a counter-example $\begin{cases} \vec{v_1}=(a,b)\\ \vec{v_2}=(\frac {\alpha}{b},\frac {2\alpha}{a})\\ \vec{v_3}=(\frac {\alpha\beta+ab\gamma}{b\alpha},\frac {2\alpha\beta+ab\gamma}{a\alpha})\end{cases}$ then $\begin{cases} \lVert\vec{v_1}\times\vec{v_2}\lVert=\alpha\\ \lVert\vec{v_1}\times\vec{v_3}\lVert=\beta\\ \lVert\vec{v_2}\times\vec{v_3}\lVert=\gamma\end{cases}$

with $\ a,b,\alpha,\beta, \gamma >0$

Since all three numbers $\alpha,\beta,\gamma$ can be chosen independently, it shows that the problem is not solvable in the general case.

Answer 3

Let $x=v_1\times v_2,\,y=v_1\times v_3$ so $(v_2+av_1)\times(v_3+bv_1)=v_2+v_3+ay-bx$ isn't uniquely determined.