A vector $\vec{v}$ is called a unit vector if $\|\vec{v}\| = 1$.
Let $\vec{a}$, $\vec{b}$, and $\vec{c}$ be unit vectors, such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$. Show that the angle between any two of these vectors is $120^\circ$.
I can't figure out where the degrees part comes in
2 Answers
$\vec{a} + \vec{b} = -\vec{c}$ and hence $(\vec{a}+\vec{b})\cdot (\vec{a}+\vec{b}) = (-\vec{c})\cdot(-\vec{c}) = 1$. Thus, $|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b} = 1$. This gives $\vec{a}\cdot\vec{b} = -\frac{1}{2}$ and consequently, angle between $\vec{a}$ and $\vec{b}$ is $120^\circ$.
Hint: $a \cdot (a+b+c)=0=a\cdot a+a\cdot b+a\cdot c=1+\cos(\theta_1)+\cos(\theta_2)$ where $\theta_1$ is the angle between $a$ and $b$ and $\theta_2$ is the angle between $b$ and $c$. (Both angles, by the definition of "between", are less than or equal to 180 degrees.) By dotting with $b$ and $c$ you can find two more equations which involve the angle between $a$ and $c$.
Can you solve these equations for the three cosines? If so then you can solve for the angles themselves.
