If $l(a, b, c) = l(a', b', c')$, then $(a, b, c) = (a ', b', c')k$ for some $k \in F$?

Question

Let $F$ be a division ring. Define $l(a, b, c) = \{(x, y, z) \in F^3 : xa + yb + cz = 0\}$.

Question: If $l(a, b, c) = l(a', b', c')$ is it true that $(a, b, c) = (a', b', c')k$ for some $ k \in F$?

I'm guessing the answer is yes, but I haven't been able to prove it.

Answer 1

If $a',b',c'\ne0$ the following are equivalent:

1. $\exists k\in F$: $\,(a,b,c)=(a',b',c')k$
2. $(a')^{-1}a=(b')^{-1}b=(c')^{-1}c$
3. $b'(a')^{-1}=ba^{-1}$ and $c'(b')^{-1}=cb^{-1}$

Thus $l(a,b,c)=l(a',b',c')\Rightarrow (a,b,c)=(a',b',c')k$ follows if $(ba^{-1},cb^{-1})$ is an invariant of the locus set $l(a,b,c)$, since if it is an invariant then $l(a,b,c)=l(a',b',c')$ yields $(3)$ above.

Clearly $xa+yb+zc=0$ admits solutions of the form $(x,y,0)$ and $(0,u,v)$ with $y,v\ne0$. If we have any such solutions, we get $ba^{-1}=-y^{-1}x$ and $cb^{-1}=-v^{-1}u$, so the pair $(ba^{-1},cb^{-1})$ is in fact an invariant of the locus set $l(a,b,c)$ as it can be computed directly from its elements.

Consider the cases of one or more of $a',b',c'$ being zero separately to conclude the proof.