Answer
(a) Since $||{\bf{v}} + {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} + 2{\bf{v}}\cdot{\bf{w}} + ||{\bf{w}}|{|^2}$,
so, $||{\bf{v}} + {\bf{w}}|{|^2} = {3^2} + {5^2} + 2{\bf{v}}\cdot{\bf{w}}$
(b) $||{\bf{v}} + {\bf{w}}|| = 7$
Work Step by Step
(a) We have $||{\bf{v}} + {\bf{w}}|{|^2} = \left( {{\bf{v}} + {\bf{w}}} \right)\cdot\left( {{\bf{v}} + {\bf{w}}} \right)$. So,
$||{\bf{v}} + {\bf{w}}|{|^2} = \left( {{\bf{v}} + {\bf{w}}} \right)\cdot\left( {{\bf{v}} + {\bf{w}}} \right) = {\bf{v}}\cdot{\bf{v}} + {\bf{v}}\cdot{\bf{w}} + {\bf{w}}\cdot{\bf{v}} + {\bf{w}}\cdot{\bf{w}}$
Since ${\bf{v}}\cdot{\bf{v}} = ||{\bf{v}}|{|^2}$, ${\bf{w}}\cdot{\bf{w}} = ||{\bf{w}}|{|^2}$, and ${\bf{v}}\cdot{\bf{w}} = {\bf{w}}\cdot{\bf{v}}$, we have
$||{\bf{v}} + {\bf{w}}|{|^2} = ||{\bf{v}}|{|^2} + 2{\bf{v}}\cdot{\bf{w}} + ||{\bf{w}}|{|^2}$.
Since $||{\bf{v}}|| = 3$ and $||{\bf{w}}|| = 5$, so $||{\bf{v}} + {\bf{w}}|{|^2} = {3^2} + {5^2} + 2{\bf{v}}\cdot{\bf{w}}$.
(b) Since the angle between ${\bf{v}}$ and ${\bf{w}}$ is $\theta = \frac{\pi }{3}$. By Eq. (1) of Theorem 2:
${\bf{v}}\cdot{\bf{w}} = ||{\bf{v}}||||{\bf{w}}||\cos \theta = 3\cdot5\cdot\cos \frac{\pi }{3}$
${\bf{v}}\cdot{\bf{w}} = \frac{{15}}{2}$
From the result in part (a) we have $||{\bf{v}} + {\bf{w}}|{|^2} = {3^2} + {5^2} + 2{\bf{v}}\cdot{\bf{w}}$. So,
$||{\bf{v}} + {\bf{w}}|{|^2} = {3^2} + {5^2} + 2{\bf{v}}\cdot{\bf{w}} = 9 + 25 + 15 = 49$
$||{\bf{v}} + {\bf{w}}|| = 7$
