I am doing some review on cross products, and I forgot how do a cross product similar to this:
$$vQ \times B_1 = vQ \times B_2$$
where $Q,B_1,B_2$ are vectors, $\times$ is cross product, and $v$ is a scalar.
I would be trying to prove that $B_1$ and $B_2$ are equal to each other, other than saying that it is indeed true, but I don't know if an "inverse" cross product exists, and can't seem to figure out how to prove it exists, if it does. Any help would be awesome!
I'll be using JMoravitz's interpretation of the problem.
Suppose for every nonzero scalar $v$ and nonzero vector $Q$ we have $vQ\times B_1 = v Q \times B_2$. Then $$ vQ\times B_1 - v Q \times B_2= \vec{0} $$ $$ v\left(Q\times B_1 - Q \times B_2\right)= \vec{0} $$ $$ v\left(Q\times \big( B_1 - B_2\big)\right)= \vec{0} $$Since $Q\neq \vec{0}$, there's only one vector that crosses with any vector to yield $\vec{0}$, namely $\vec{0}$. Then $B_1-B_2=\vec{0}$.