Cross product: Suppose $a \times b = a' \times b$ for all $b$. Is it true that $a=a'$?

Question

Cross product: Suppose $a \times b = a' \times b$ for all $b$. Is it true that $a=a'$?

Here's what I got so far: I can manipulate the equation to get $b \times (a'-a)=0$. I have a feeling this can help me, but I'm not sure how.

Answer 1

Your equation $b\times(a-a')=0$ tells you that $a-a'$ is colinear with $b$, that is $a-a'=\gamma\, b$ for some scalar $\gamma$. If you choose $b$ so that it is perpendicular to $a-a'$, then $$ 0=\|b\times(a-a')\|=\|b\|\,\|a-a'\|, $$ and it follows that $a'=a$, since $\|a-a'\|=0$.

Answer 2

You know $b \times (a' - a) = 0$ for all $b$. Now you could make a clever choice of $b$, say, let $b$ be a nonzero vector perpendicular to $a' - a$. Then consider the length $|b \times (a' - a)| = |b| \cdot |a' - a|$.

Answer 3

In fact, you don't even need for all values, just for $b=e_1,e_2,e_3$. Since Since $a$ is determined by its coordinates, it suffices to show that we can recover the coordinates of $a$ from the various cross products. Let us consider the identity $(a \times e_i) \cdot e_j = (e_i \times e_j) \cdot a$. Then since $e_1=e_2 \times e_3$, we have $a_1=e_1 \cdot a = (a \times e_2) \cdot e_3$, and so we can read off the first coordinate of $a$ by looking at $a\times e_2$. Things are similar for determining the other coordinates.

More generally, and more geometrically, if $a\times b =0$, then $a$ and $b$ are parallel, so if $a\times b = a'\times b$ for all $b$, then $(a-a')\times b = 0$ for all $b$. Now, pick two non-parallel vectors, $v$ and $w$. If $(a-a')\times v = (a-a') \times w = 0$, then $a-a'$ is parallel to both $v$ and $w$, but since $v$ and $w$ are not parallel, the only way for this to happen is if $a-a'=0$. Thus, we don't have to test against all vectors, just two of them.