Let $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ be unit vectors, with $\mathbf{a+b+c=0}$. The angle between any two of these vectors is $120^\circ$. Let $\mathbf{a}$,  $\mathbf{b}$, and $\mathbf{c}$ be unit vectors, such that  $\mathbf{a}+\mathbf{b}+\mathbf{c} = \mathbf{0}$. Show that the angle between any two of these vectors is $120^\circ$.
Hi, 
I have been having some trouble with this problem.  
I have tried to assign variables to the vectors, and creating various equations with them.  However, I can't figure out how to use the $\mathbf{a+b+c=0}$ information.  
Thanks!

Geometrically, $\mathbf a + \mathbf b + \mathbf c = \mathbf 0$ means we can form a triangle $\triangle XYZ$ whose oriented sides $\overrightarrow{XY}, \overrightarrow{YZ}, \overrightarrow{ZX}$ are parallel copies of $\mathbf a$, $\mathbf b$, $\mathbf c$.
Since $\mathbf a$, $\mathbf b$, $\mathbf c$ are unit vectors, $\triangle XYZ$ is equilateral, so $\angle X = \angle Y = \angle Z = 60^\circ$.
But $\angle X$ is the angle between $\overrightarrow{XY}$ and $\overrightarrow{XZ}$, or between $\mathbf a$ and $-\mathbf c$. If the angle between $\mathbf a$ and $-\mathbf c$ is $60^\circ$, then the angle between $\mathbf a$ and $\mathbf c$ is its supplementary angle, $120^\circ$.
Similarly, we conclude from $\angle Y = \angle Z = 60^\circ$ that the angle between $\mathbf a$ and $\mathbf b$, and the angle between $\mathbf b$ and $\mathbf c$, are both $120^\circ$.
