# Angle between three vectors in $\mathbb{R}$

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?

• With respect to the two answers provided, is either one useful to you? If not, please edit your query re (1) what is the background of the question - re what theorems or previously solved problems might be pertinent to you here? (2) please show your work, re what you have tried. Commented Oct 5, 2020 at 10:54

$$V_3=-V_1-V_2$$ implies that $$\|V_3\|^2=\langle V_1+V_2,V_1+V_2\rangle =\|V_1\|^2+\|V_2\|^2+2\langle V_1,V_2\rangle =\|V\|^2$$ where $$\|V_1\|=\|V_2\|=\|v_3\|=\|V\|$$

This implies that $$2\|V\|^2 cos(V_1,V_2)=-\|V\|^2$$.

You cannot "square" a vector: the statement "$$(V_1 + V_2)^2 = V_3^2$$" makes no sense.

I sense that you are trying to use the dot product. In that case, you will instead obtain:

$$|V_3|^2 = V_3 \cdot V_3 = (V_1 +V_2) \cdot (V_1 +V_2) = |V_1|^2 + 2 V_1 \cdot V_2 + |V_2|^2$$

Now we continue with your method. We have from $$|V_1|^2 = -2V_1\cdot V_2$$

$$\implies \cos (a) = \frac {V_1 \cdot V_2}{|V_1||V_2|} = \frac {V_1 \cdot V_2}{|V_1|^2} = -\frac12$$