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!
 A: The angle $\alpha$ betwwen unit vectors $\mathbf x,\mathbf y$ is defined by the property $\cos\alpha=\mathbf x\cdot\mathbf y$. So here we want to show that $\mathbf a\cdot\mathbf b=\mathbf a\cdot\mathbf c=\mathbf b\cdot\mathbf c=-\frac12$. By multiplying the given equation with $\mathbf a$, you get 
$$1+\mathbf a\cdot\mathbf b+\mathbf a\cdot \mathbf c=0 $$
If you do the same with mutltiplicatoin by $\mathbf b$ and multiplication by $\mathbf c$, you obtain three equations in the three unknowns $\mathbf a\cdot \mathbf b$, $\mathbf b\cdot \mathbf c$, $\mathbf a\cdot \mathbf c$ that you can solve.
A: Let $\theta$ be the angle between $a$ and $b$. Then $a+b+c=0$ gives $a+b=-c$, hence $\|a+b\|=\|c\|=1$. Since the cosine theorem gives:
$$1=\|a+b\|^2 = \|a\|^2+\|b\|^2+2\|a\|\|b\|\cos\theta =2+2\cos\theta$$
we have $\cos\theta=-\frac{1}{2}$, hence $\theta\in 2\pi\mathbb{Z}\pm\frac{2\pi}{3}$, and the same holds if we exchange two vectors among $a,b,c$.
A: taking the dot product with $a,b$ and $c$ of $a+b+c = 0$ and using the fact that all vectors are unit, we get 
$$\begin{align}a.b+ a.c &= -1\\a.b+b.c &= -1\\a.c + b.c &= -1\end{align}$$ adding the three up we have $$a.b + b.c + a.c=-\frac32$$ and this implies $$a.b = b.c = c.a = -\frac 12 $$ therefore the angle between any two vectors is $120^\circ.$
A: 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 $120^\circ$.
