Let $v, w$ be linearly independent vector in $\mathbb{R}^3$. Given length of $v$ is 1.
If we take $w_1 = w- <v,w>v$.
Then $w_1$ is in $v^{\perp}$ or {av: a$\in\mathbb{R}\}^{\perp}$.
Now $v^{\perp}$ is 2 dimensional. There exist only 2 vector of length $w_1$ and perpendiular to $w_1$(can be easily proved existence of exactly and only of 2 here), then these vectors are $v\times w$ or $-v\times w$.
We want to reach that this vector will be same as $v\times w$ but without proving it in way that $v\times w$ satisfy mentioned condition
We can assume $v=(1,0,0)$ and $w=(a,b,0)$. Then $w_1=w-<v,w>v=(0,b,0)$. So a vector perpendicular to $v$ and $w_1$ must have the form $(0,0,c)$. From your length criterion this vector must be $(0,0,\pm b)=\pm v\times w$.