# Vectors / Linearly independent question

an easy question that I completely understand, just not sure how to algebraically prove.
$u,v,w$ are vectors in $R^3$
given $u\times v +v\times w + w\times u=0$

I need to prove that {u,v,w} are linearly dependent.

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are those $x$'s cross product in $\mathbb{R}^3$? – countinghaus Apr 28 '12 at 18:00
maybe not since then the claim is false (although the negation of the claim is true - perhaps you meant linearly dependent?) – countinghaus Apr 28 '12 at 18:01
@countinghaus - if $x$ is indeed $\times$, then $u,v,w$ are dependent. – nbubis Apr 28 '12 at 18:01
@nbubis I agree - so then what could OP mean? what operation $x$ is there on vectors such that the cyclic sum of three vectors $x$'d pairwise against one another equalling zero implies INdependence? – countinghaus Apr 28 '12 at 18:03
i'm sorry guys. it is cross product. and i meant to prove that they ARE dependent – Guy Apr 28 '12 at 18:04

Use the fact that $v\times v=0$: $$u\times v +v\times w + w\times u=u\times(v-w)+v\times w-v\times v= (u-v)\times(v-w)=0$$ Meaning that: $$(u-v)=c(v-w)$$ Or that: $$u=c(v-w)+v$$
Lemma: a nonzero vector $x$ in $\mathbb{R^3}$ can't be perpendicular to three linearly independent vectors $u, v, w$. Proof: three linearly independent vectors make a basis, so we can write $x = a_1u + a_2v + a_3w$. Dotting both sides with $x$ tells you that $x$ is 0.
Main result: Dot both sides of the given equality with $u$. We have $u\cdot (v \times w) = 0$. Thus $v \times w$ is perpendicular to $u, v,$ and $w$.
It follows that either $u, v,$ and $w$ are linearly dependent or $v \times w = 0$, but in this latter case certainly $u, v, w$ are linearly dependent since $v, w$ are.