I am asked to elaborate on the following proof:
Given that $ \{ \vec{u}, \vec{v} \}$ are linearly independent prove that if $ \vec{w} \times \vec{u} = \vec{w} \times \vec{v} = \vec{0}$ then $\vec{w} = \vec{0}$. How would you interpret it geometrically?
For the first part (the proof) I am honestly lost. I am not sure you can make such a proof algebraically.
For the second part, I would say that $\vec{w}$ cannot be parallel to $\vec{u}$ and $\vec{v}$ given that they are not parallel to each other ( since $\{ \vec{u}, \vec{v} \}$ are linearly independent).
Is my reasoning for the second part correct? Is there a way to prove the first part?
Possible answer for the first part:
$$ \Vert \vec{w} \times \vec{u} \Vert = \Vert \vec{w} \Vert \Vert \vec{u} \Vert \sin{\theta_1}\\ \Vert \vec{w} \times \vec{v} \Vert = \Vert \vec{w} \Vert \Vert \vec{v} \Vert \sin{\theta_2} \\ \vec{u} \neq \vec{0} \text{ and } \vec{v} \neq \vec{0} \text{ since } \{ \vec{u}, \vec{v} \} \text{ are linearly independent so } \vec{w} = \vec{0} $$
$w\times u=w\times v=0$ implies that $w=au=bv, a,b\in R$ which is equivalent to $au-bv=0$ since $u,v$ linear independent $a=b=0$ and $ w=0$.