Answer
The angle between ${\bf{v}}$ and ${\bf{w}}$:
$\theta = \frac{{2\pi }}{3}$
Work Step by Step
Since $||{\bf{v}} + {\bf{w}}|{|^2} = \left( {{\bf{v}} + {\bf{w}}} \right)\cdot\left( {{\bf{v}} + {\bf{w}}} \right)$, so
$||{\bf{v}} + {\bf{w}}|{|^2} = {\bf{v}}\cdot{\bf{v}} + {\bf{w}}\cdot{\bf{v}} + {\bf{v}}\cdot{\bf{w}} + {\bf{w}}\cdot{\bf{w}}$
(1) ${\ \ }$ $||{\bf{v}} + {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} + 2{\bf{v}}\cdot{\bf{w}} + ||{\bf{w}}|{|^2}$
If $||{\bf{v}} + {\bf{w}}|| = ||{\bf{v}}|| = ||{\bf{w}}||$, then $||{\bf{v}} + {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} = ||{\bf{w}}|{|^2}$.
Write $||{\bf{v}} + {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} = ||{\bf{w}}|{|^2} = a$. Thus, equation (1) becomes
$||{\bf{v}} + {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} + 2{\bf{v}}\cdot{\bf{w}} + ||{\bf{w}}|{|^2}$
$a = a + 2{\bf{v}}\cdot{\bf{w}} + a$
$2{\bf{v}}\cdot{\bf{w}} = - a$
${\bf{v}}\cdot{\bf{w}} = - \frac{a}{2}$
By Theorem 2 of Section 13.3:
${\bf{v}}\cdot{\bf{w}} = ||{\bf{v}}||||{\bf{w}}||\cos \theta $,
where $\theta$ is the angle between the vectors ${\bf{v}}$ and ${\bf{w}}$.
So,
$ - \frac{a}{2} = \sqrt a \sqrt a \cos \theta $, ${\ \ }$ $\cos \theta = - \frac{1}{2}$
For $0 \le \theta \le \pi $, we obtain $\theta = \frac{{2\pi }}{3}$.
