If $\vec x \times \vec y = \vec x \times \vec z = \vec z \times \vec y \neq0 $ then $\vec z = $?

Question

If$$\vec x \times \vec y = \vec x \times \vec z = \vec z \times \vec y \neq0 ,$$ then it must be: $$\vec z = 1\vec x + 1\vec y+0\vec x \times \vec y\ ? $$ How can I come to this conclusion?

Answer 1

From $x\times y=x\times z$ and $x\ne0$ we can conclude that $z-y=\lambda x$ for some $\lambda\in{\mathbb R}$, or $z=y+\lambda x$. Plugging this into the equation $x\times z=z\times y$ gives $$x\times(y+\lambda x)=(y+\lambda x)\times y\ ,$$ which allows to conclude that $\lambda=1$, since $x\times y\ne0$. It follows that necessarily $z=x+y$. On the other hand it is easily verfied that this $z$ satisfies all requirements.