If $\vec{a}\times\vec{b}=\vec{c}\times\vec{d}$ and $\vec{a}\times\vec{c}=\vec{b}\times\vec{d}$ then $\vec{a}-\vec{d}$ is parallel to $\vec{b}-\vec{c}$

Question

I need some help with cross product:

Prove that if $\vec{a}\times\vec{b} = \vec{c} \times \vec{d}$ and $\vec{a} \times \vec{c} = \vec{b} \times \vec{d}$ then $\vec{a} - \vec{d}$ is parallel to $\vec{b} - \vec{c}$.

Honestly, I don't know where to start.

At least I know that if $\vec{a} - \vec{d}$ is parallel to $\vec{b} - \vec{c}$ then $(\vec{a} - \vec{d})\times(\vec{b} - \vec{c}) = 0$

Answer 1

$\begin{align} (\vec{a}-\vec{d})\times (\vec{b}-\vec{c})&=(\vec{a}\times\vec{b})-(\vec{a}\times\vec{c})-(\vec{d}\times\vec{b})+(\vec{d}\times\vec{c}) \tag{1}\\ &=(\vec{a}\times\vec{b})-(\vec{a}\times\vec{c})+(\vec{b}\times\vec{d})-(\vec{c}\times\vec{d})\tag{2}\\ &=\left((\vec{a}\times\vec{b})-(\vec{c}\times\vec{d})\right)+\left((\vec{b}\times\vec{d})-(\vec{a}\times\vec{c})\right)\\ &=\vec{0}+\vec{0}\\ &=\vec{0} \end{align}$ $(1)$: By distribution.

$(2)$: Since $\vec{x}\times\vec{y}=-\vec{y}\times\vec{x}$

So, according to at least what you know $\vec{a}-\vec{d}$ and $\vec{b}-\vec{c}$ are parallel.