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If $a$, $b$, $c$, and $d$ lie in the same plane, then $(a \times b) \times (c \times d) = 0$

I can't really see a way to prove this other than

if $(a \times b)$ is an orthogonal vector $v$ in $\mathbb{R}^n$ and $(c \times d)$ is an orthogonal vector $u$ in $\mathbb{R}^n$ then cross product of $u\times v$ has to lie in the original plane as the cross product is perpendicular to both planes

As you can see this isn't very general or formal for that matter.

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Let $n =a \times b$. $n$ is perpendicular to the plane. Let $m = c\times d$. $m$ is perpendicular to the plane as well. Then $n$ is parallel to $m$. Then...?

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The way I remember it is that $(\vec a \times \vec b)$ will produce a vector $\vec v$ which will be orthogonal to the vectors $\vec a$ and $\vec b$, which in this case will be perpendicular to the plane.

However in this case you also have $(\vec c \times \vec d) = \vec u$ which would also be perpendicular to the plane.

Therefore $(\vec u \times \vec v) = \lVert u \rVert \lVert v \rVert \sin(\theta)$, since $\vec u$ and $\vec v$ are perpendicular to the plane, you'll either have $\theta = 0$ or $\theta = \pi$, showing that $\vec u$ and $\vec v$ are either parallel or in opposite directions. So in the end you ll finally have $(\vec a \times \vec b) \times (\vec c \times \vec d) = \vec u \times \vec v = \lVert u \rVert \lVert v \rVert \sin(\theta) = 0$

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  • $\begingroup$ They are either parallel or they have an 180 angle between them. Since the perpendicular vector can go "down" the plane. No? $\endgroup$ – Sorfosh Oct 20 '15 at 6:23
  • $\begingroup$ Yes, but it would still yield to zero since you'll have $\theta = \pi$. I should have added that to my answer sorry. $\endgroup$ – Aldon Oct 20 '15 at 6:24
  • $\begingroup$ No problem, just want to make sure the answer is complete. +1 $\endgroup$ – Sorfosh Oct 20 '15 at 6:25

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