Maximum and Minimum Angles between Vectors I have a question about the angles between vectors.
Problem:
Given three vectors $a, b, c \in \mathbb{R}^n$, where $\theta_1 = \theta_2 =\pi/3$, where $\theta_1$ is the angle between $a, b$ and $\theta_2$ is the angle between $b,c$, what are the maximum and minimum values of $\theta_3$, the angle between vectors $a, c$. 
I know the answer is $(2 \pi /3, 0)$, and I figured this out by just imagining three vectors in $\mathbb{R}^3$. I was wondering how would you actually prove this in $\mathbb{R}^n$ (or $\mathbb{R}^3$ for that matter). I am kind of rusty on geometry. I know that we have that
$\langle a, b \rangle = |a| |b| \cos \theta_1$, and the same for the other 2 pairs of vectors, but I'm not sure how to use this to find a bound on $\theta_3$, or if we need to use anything else.
 A: Write $a = a_\parallel + a_\perp$ and $c = c_\parallel + c_\perp$, where $a_\parallel, c_\parallel$ are the projections onto $b$ and $a_\perp, c_\perp$ are perpendicular to $b$. 
We are given $\langle a,b \rangle = |a||b|\cos\theta_1 = \frac12|a||b|$, but we also have $\langle a, b \rangle = |a_\parallel||b|$, so we know that $|a_\parallel| = \frac12 |a|$. Similarly, $|c_\parallel| = \frac12|c|$.
Because $|a|^2 = |a_\parallel|^2 + |a_\perp|^2$ (which we can derive by expanding $\langle a_\parallel + a_\perp, a_\parallel + a_\perp\rangle$) we can also deduce that $|a_\perp| = \frac{\sqrt3}2 |a|$. Similarly, $|c_\perp| = \frac{\sqrt3}2 
|c|$. Therefore
\begin{align}
\langle a,c\rangle &= \langle a_\parallel + a_\perp, c_\parallel + c_\perp\rangle \\
&= \langle a_\parallel, c_\parallel\rangle + \langle a_\parallel, c_\perp\rangle + \langle a_\perp, c_\parallel\rangle + \langle a_\perp, c_\perp\rangle \\
&= |a_\parallel| |c_\parallel| + 0 + 0 + |a_\perp| |c_\perp| \cos \angle(a_\perp,c_\perp) \\
&\ge |a_\parallel| |c_\parallel| - |a_\perp| |c_\perp| \\
&= \frac14 |a| |c| - \frac34 |a| |c| \\
&= -\frac12 |a||c|.
\end{align}
This tells us that $\cos \theta_3 = \frac{\langle a,c\rangle}{|a||c|} \ge -\frac12$. Since $\cos$ is monotone decreasing on $[0,\pi]$, we know that $\theta_3 \le \arccos (-\frac12) = \frac{2\pi}{3}$. (We have equality exactly when $\cos \angle(a_\perp, c_\perp) = -1$: when $a_\perp$ and $c_\perp$ point in parallel but opposite directions.)
By taking $a=c$, we can also see that $\theta_3 = 0$ is achievable, so $0$ is the minimum possible value.
