$\mathbf{a}\times\mathbf{b}=\mathbf{b}\times\mathbf{c}=\mathbf{c}\times\mathbf{a}$

Question

For any vectors $\mathbf{a},\mathbf{b},\mathbf{c}$, if $$\mathbf{a}\times\mathbf{b}=\mathbf{b}\times\mathbf{c}=\mathbf{c}\times\mathbf{a}$$ where $\mid{\mathbf{a}\times\mathbf{b}}\mid\neq0$

Show that $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$, i.e. a triangle

What I have done so far:

Subtract each term from another, change the sign of cross product when we switch elements s.t. $$\mathbf{a}\times\mathbf{b}+\mathbf{c}\times\mathbf{b}=0, \mathbf{a}\times\mathbf{b}+\mathbf{a}\times\mathbf{c}=0,\mathbf{b}\times\mathbf{c}+\mathbf{a}\times\mathbf{c}=0$$

Which gives $$\mathbf{a}\times\mathbf{(b+c)}=\mathbf{b}\times\mathbf{(a+c)}=\mathbf{c}\times\mathbf{(a+b)}=0$$

Any hints for what to do next? And I think there should be a geometric proof somewhere as well.

Answer 1

Use $$(a\times b)\times c+(b\times c)\times a+(c\times a)\times b=0.$$

We got $$(a\times b)\times(a+b+c)=0,$$ which says $$(a\times b)||(a+b+c).$$

Answer 2

An alternative argument: as $a\times b$ is nonzero, $a$ and $b$ are nonzero and non-parallel. Also as $(a+b+c)\times a=0$ then $a+b+c$ is a multiple of $a$. Similarly $a+b+c$ is a multiple of $b$. A vector can only be a multiple of $a$ and of $b$ if it is zero.