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If $\vec{u},\vec{v},\vec{w}$, and $\vec{a}$ all lie in the same plane in $\mathbb{R_3}$, then $(\vec{u} \times \vec{v}) \times (\vec{w}\times\vec{a})=\vec{0}$.

I tried to multiply the whole thing out manually and ended up with a VERY messy expression that didn't simplify to the zero vector.

Any better ways of doing it?

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  • $\begingroup$ Are you permitted the geometric interpretation of the cross product? $\endgroup$ – Eric Stucky Nov 14 '15 at 7:25
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    $\begingroup$ When you take $ \vec{u} \times \vec{v}$, you get the direction vector of a normal vector to the plane. The same is true for $\vec{w} \times \vec{a}$. What happens when you take the cross product of two parallel vectors? $\endgroup$ – Nicholas Nov 14 '15 at 7:26
  • $\begingroup$ I know the cross product of two parallel vectors is the zero vector but why are $\vec{u} X \vec{v}$ and $\vec{w}X\vec{a}$ parallel just because they are on the same plane. @Nicholas $\endgroup$ – Rloc Nov 14 '15 at 7:32
  • $\begingroup$ Both $\vec{u} \times \vec{v}$ and $\vec{w} \times \vec{a}$ give the normal vector of the same plane. The normal vector to the same plane must be parallel. $\endgroup$ – Nicholas Nov 14 '15 at 7:35
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    $\begingroup$ Okay thanks! that makes a lot of sense now. $\endgroup$ – Rloc Nov 14 '15 at 7:36
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A somewhat nicer formula to prove is for any vectors $a,b,c\in \mathbb R^3$ is $a\times (b\times c) = \langle a,c\rangle b -\langle a, b\rangle c$

Then we have

$(u\times v)\times(w\times a)= \langle (u\times v),a\rangle w - \langle (u\times v),w\rangle a$

Notice that $(u\times v)$ will be perpendicular to the plane where $a$ is lying, hence their dot product is 0. Same goes for $\langle(u\times v),w\rangle$

Thus, $(u\times v)\times(w\times a)=0-0=0$

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