Show that $u\times v+v\times w+w\times u=\textbf{0}$ implies that $u,v,w$ are linearly dependent I have what I think might be a solution, but I'm not sure it's formal enough. I begin by noting that $u,v,w$ are linearly dependent iff. they lie on the same plane. Then I construct the following chain of equivalences:
\begin{align*}
u\times v+v\times w+w\times u=\textbf{0}&\Longleftrightarrow u\times v+v\times w=-w\times u\\&\Longleftrightarrow v\times (w-u)=u\times w
\end{align*}
Then my reasoning is that since $w-u$ is obviously on the plane that $u,w$ spans, then for $v\times (w-u)$ to form the same normal vector as $u\times w$, v must lie on the same plane. I think the reasoning is correct, but how do  I formalize this last part using mathematical notation? Is it necessary to do so?

Note that $$v \times w + w \times u + u \times v =v \times w - u \times w + u \times v $$
$$=(v-u) \times  (w-v)$$
Thus $$  v \times w + w \times u + u \times v = 0$$
$$\implies (v-u) \times (w-v)=0$$
  Which implies $u$,$v$,and $w$ are linearly dependent.
