Answer
(a) ${\bf{v}}\cdot{\bf{w}} = - 3$
(b) $||2{\bf{v}} + {\bf{w}}|| = \sqrt {13} $
(c) $||2{\bf{v}} - 3{\bf{w}}|| = \sqrt {133} $
Work Step by Step
(a) By Eq. (1) of Theorem 2:
${\bf{v}}\cdot{\bf{w}} = ||{\bf{v}}||||{\bf{w}}||\cos \theta = 2\cdot3\cdot \cos120^\circ = - 3$
(b)
$||2{\bf{v}} + {\bf{w}}|{|^2} = \left( {2{\bf{v}} + {\bf{w}}} \right)\cdot\left( {2{\bf{v}} + {\bf{w}}} \right)$
$||2{\bf{v}} + {\bf{w}}|{|^2} = 4{\bf{v}}\cdot{\bf{v}} + 2{\bf{w}}\cdot{\bf{v}} + 2{\bf{v}}\cdot{\bf{w}} + {\bf{w}}\cdot{\bf{w}}$
$||2{\bf{v}} + {\bf{w}}|{|^2} = 4||{\bf{v}}|{|^2} + 4{\bf{v}}\cdot{\bf{w}} + ||{\bf{w}}|{|^2}$
Substituting $||{\bf{v}}|| = 2$, $||{\bf{w}}|| = 3$, and the result from part (a): ${\bf{v}}\cdot{\bf{w}} = - 3$ in the last equation, we get
$||2{\bf{v}} + {\bf{w}}|{|^2} = 4\cdot{2^2} + 4\left( { - 3} \right) + {3^2} = 13$
$||2{\bf{v}} + {\bf{w}}|| = \sqrt {13} $
(c)
$||2{\bf{v}} - 3{\bf{w}}|{|^2} = \left( {2{\bf{v}} - 3{\bf{w}}} \right)\cdot\left( {2{\bf{v}} - 3{\bf{w}}} \right)$
$||2{\bf{v}} - 3{\bf{w}}|{|^2} = 4{\bf{v}}\cdot{\bf{v}} - 6{\bf{w}}\cdot{\bf{v}} - 6{\bf{v}}\cdot{\bf{w}} + 9{\bf{w}}\cdot{\bf{w}}$
$||2{\bf{v}} - 3{\bf{w}}|{|^2} = 4||{\bf{v}}|{|^2} - 12{\bf{v}}\cdot{\bf{w}} + 9||{\bf{w}}|{|^2}$
Substituting $||{\bf{v}}|| = 2$, $||{\bf{w}}|| = 3$, and the result from part (a): ${\bf{v}}\cdot{\bf{w}} = - 3$ in the last equation, we get
$||2{\bf{v}} - 3{\bf{w}}|{|^2} = 4\cdot{2^2} - 12\left( { - 3} \right) + 9\cdot{3^2} = 133$
$||2{\bf{v}} - 3{\bf{w}}|| = \sqrt {133} $
