Three vectors, $V_1,V_2,V_3$ are in the $\mathbb{R}^2$ plane where $V_1+V_2+V_3=\vec{0}$ and the magnitudes of these vectors are the same. Show that the angle between any two of these vectors is 120 degrees.
My try:
$V_1+V_2+V_3=\vec{0}$, so $V_1+V_2=-V_3$. This would mean that $(V_1+V_2)^2=V_3^2$ which is $|V_1|+|V_2| + 2V_1V_2 =|V_3|$. But since the magnitude is the same, we get $2|V_1|+2V_1V_2=|V_1|$ which is $2V_1V_2=-|V_1|$ so $V_1*V_2=-0.5|V_1|$.
The formula for the angle is $\cos(a)= \frac{V_1*V_2}{|V_1|*|V_2|} = -0.5*\frac{|V_1|}{|V_1|^2}=\frac{-0.5}{|V1|}$
I know that the solution would be a triangle and that $\cos(a)=-0.5$
But that would mean that my angle is dependent on $V_1$ which isn't the case. So what am I doing wrong?