Answer
Since ${\bf{v}}\cdot{\bf{w}} = 0$. Therefore,
$||{\bf{v}} + {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} + ||{\bf{w}}|{|^2}$.
Work Step by Step
Since $||{\bf{v}} + {\bf{w}}|{|^2} = \left( {{\bf{v}} + {\bf{w}}} \right)\cdot\left( {{\bf{v}} + {\bf{w}}} \right)$, so
$||{\bf{v}} + {\bf{w}}|{|^2} = {\bf{v}}\cdot{\bf{v}} + {\bf{w}}\cdot{\bf{v}} + {\bf{v}}\cdot{\bf{w}} + {\bf{w}}\cdot{\bf{w}}$
$||{\bf{v}} + {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} + 2{\bf{v}}\cdot{\bf{w}} + ||{\bf{w}}|{|^2}$
By Theorem 2 of Section 13.3:
${\bf{v}}\cdot{\bf{w}} = ||{\bf{v}}||||{\bf{w}}||\cos \theta $,
where $\theta$ is the angle between two nonzero vectors ${\bf{v}}$ and ${\bf{w}}$.
If the vectors ${\bf{v}}$, ${\bf{w}}$ are orthogonal, then $\theta = \frac{\pi }{2}$. Thus, ${\bf{v}}\cdot{\bf{w}} = 0$. Therefore, $||{\bf{v}} + {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} + ||{\bf{w}}|{|^2}$.
