Any vector $\mathbf{v}$ is a unit vector if $\|\mathbf{v}\| = 1$.
Let $\mathbf{x}$, $\mathbf{y}$, and $\mathbf{z}$ be unit vectors, such that $\mathbf{x} + \mathbf{y} + \mathbf{z} = \mathbf{0}$. Show that the angle between any two of these vectors is $120^\circ$.
I know how to prove this using geometry, but the problem instructs me not to. How should I start to/prove this?
Thanks in advance!
Consider the dot product between $\vec x$ and $\vec y$. Then $\vec x\cdot \vec y=\cos (\alpha)$, where $\alpha$ is the angle between the vectors $\vec x$ and $\vec y$. But $\vec y=-(\vec x+\vec z)$ and so $\vec x\cdot (-\vec x-\vec z)=\cos(\alpha).$ This implies that $$-|x|^2-\vec x\cdot \vec z=\cos(\alpha)$$ $$\Rightarrow -1-\cos (\beta)=\cos (\alpha)$$ $$\Rightarrow \cos(\alpha)+\cos(\beta)=-1$$ Use a similar argument by considering $\vec y \cdot \vec z$ and $\vec z\cdot \vec x$ and you will get $$\cos(\beta)+\cos(\gamma)=-1$$ And $$\cos(\gamma)+\cos(\alpha)=-1$$ Add all these three equations to get $2(\cos(\alpha)+\cos(\beta)+\cos(\gamma))=-3$ or $$\cos(\alpha)+\cos(\beta)+\cos(\gamma)=\frac{-3}{2}.$$ Now we know that $$\frac{-3}{2}\leq \cos(\alpha)+\cos(\beta)+\cos(\gamma)\leq 3$$ so clearly $\alpha, \beta \text{ and }\gamma$ are at their minimum values, which is $$\alpha= \beta= \gamma=\frac{2\pi}{3}.$$ For a rigorous justification of this claimm, check out this link.
We have that $$(\mathbf{x} + \mathbf {y} + \mathbf {z})^2 =0$$ $$\Rightarrow \mathbf {x}^2 + \mathbf {y}^2 + \mathbf {z}^2 +2 (\mathbf {x}\cdot \mathbf {y} + \mathbf {y}\cdot \mathbf {z} + \mathbf {z}\cdot \mathbf {x}) =0$$ $$\Rightarrow (\mathbf {x}\cdot \mathbf {y} + \mathbf {y}\cdot \mathbf {z} + \mathbf {z}\cdot \mathbf {x}) = -\frac {3}{2} $$ $$(|\mathbf {x}||\mathbf {y}|\cos \theta_{xy} + |\mathbf {y}||\mathbf {z}|\cos \theta_{yz} + |\mathbf {z}||\mathbf {x}|\cos \theta_{zx}) = -\frac {3}{2} $$ $$=\cos \theta_{xy} + \cos \theta_{yz} + \cos \theta_{zx} = -\frac {3}{2}$$
Can you conclude using this? Hope it helps.
