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?

Denote $\left[\begin{matrix}u&v&w\end{matrix}\right]=(u\times v).w$ as the scalar triple product.
And note that $\left[\begin{matrix}u&v&w\end{matrix}\right]=0$ if any two vectors are identical. This follows from property of determinants if one expands $(u\times v).w$ as a determinant.
Take dot product by $w$ on both sides to get
\begin{align}
&\qquad\quad(u\times v+v\times w+w\times u).w=0.w
\\&\implies (u\times v).w+(v\times w).w+(w\times u).w=0
\\&\implies \left[\begin{matrix}u&v&w\end{matrix}\right]+ \left[\begin{matrix}v&w&w\end{matrix}\right]+ \left[\begin{matrix}w&u&w\end{matrix}\right]=0
\\&\implies \left[\begin{matrix}u&v&w\end{matrix}\right]=0
\end{align}
That last implication is a necessary and sufficient condition for the vectors $u,v$ and $w$ to be coplanar. We know that $\left[\begin{matrix}u&v&w\end{matrix}\right]$ geometrically represents the volume of a cuboid formed by the vectors $u,v$ and $w$. If this volume vanishes, then the vectors must lie on a plane. Hence, as you say, they are dependent.
