$\vec{x}\times\vec{y}=\vec{a}\times\vec{b}$ and $\vec{x}\times\vec{a}=\vec{y}\times\vec{b}$, prove that $\vec{x}-\vec{b}\parallel\vec{y}-\vec{a}$.

Question

How to prove that $\vec{x}-\vec{b}\parallel\vec{y}-\vec{a}$, knowing $\vec{x}\times\vec{y}=\vec{a}\times\vec{b}$ and $\vec{x}\times\vec{a}=\vec{y}\times\vec{b}$?

Answer 1

Hint: Compute \begin{align} (\vec{x}-\vec{b})\times (\vec{y}-\vec{a}) \end{align} directly and show it equals zero.

Answer 2

Their cross-product is the zero vector: $$(\vec{x}-\vec{b})\times(\vec{y}-\vec{a})=\vec{x}\times \vec{y}-\vec{x} \times \vec{a} - \vec{b} \times {y} + \vec{b} \times \vec{a}=(\vec{x} \times \vec{y} - \vec{a} \times \vec{b})+(\vec{x} \times \vec{a}-\vec{y} \times \vec{b})=\vec{0}$$ Where I used that $\vec{a} \times \vec{b} + \vec{b} \times \vec{a} = \vec{0}$