Determining value of unit vector that satisfies cross product equation

Question

Consider the equation given by: $\langle 0,-1, 0 \rangle$ = $-qV \times \langle -1,0,0 \rangle$, where $\times$ denotes the cross product. I have to find a suitable standard basis vector $V$ that satisfies this equation. Where I'm getting stuck is since $\langle 0,-1, 0 \rangle$ is orthogonal to $-qV$ and $\langle -1,0,0 \rangle$, then I can express $-qV$ as $\langle 0,-1,0 \rangle \times \langle -1,0,0 \rangle$, which gives $-qV$ = $\langle 0,0,-1 \rangle$ and hence $V = \langle 0,0,1/q \rangle$. However, the answer to this problem is $V = \langle 0,0,-1/q \rangle$. I'm not sure where I am going wrong, but it seems I am off by a factor of negative $1$.

I am utilizing the result found in this link Opposite of a cross product

If anyone could find my algebra mistake, I would greatly appreciate it.

Answer 1

You missed the negative sign, $\langle 0 ,-1, 0\rangle \times \langle -1, 0, 0 \rangle = \langle0, 0, 1\rangle$ i.e $-j \times -i = k$. So $v = \frac {-1}{q} k$

Answer 2

More generally, consider $ c = a \times b$. Then if $c$ is orthogonal to $a$ and $b$, the solutions are given by $ b = (c \times a)/(a\, . a) + t a$ for arbitrary scalars $t$, and $a = (c \times -b)/(b\, . b) + t b$ for arbitrary scalars $t$. In the problem above, we can take $c = <0,-1,0>, a = -qV, b = <-1,0,0>.$ Solving for $a$ gives $a = <0,-1,0> x$ $- <-1,0,0>$, hence I dropped the negative here.

The answer posted on the hyperlink above was incomplete.