Answer
(a) ${{\bf{u}}_{||{\bf{v}}}}$ is in $ - {\bf{v}}$ direction.
${{\bf{v}}_{||{\bf{u}}}}$ is in $ - {\bf{u}}$ direction.
(b) ${{\bf{u}}_{||{\bf{v}}}}$ has the greater magnitude.
Work Step by Step
(a) Since the angle between ${\bf{u}}$ and ${\bf{v}}$ is obtuse we have ${\bf{u}}\cdot{\bf{v}} = {\bf{v}}\cdot{\bf{u}} < 0$. So, the projection of ${\bf{u}}$ along ${\bf{v}}$, that is, the vector ${{\bf{u}}_{||{\bf{v}}}}$ is in $ - {\bf{v}}$ direction. Likewise, the projection of ${\bf{v}}$ along ${\bf{u}}$, that is, the vector ${{\bf{v}}_{||{\bf{u}}}}$ is in $ - {\bf{u}}$ direction.
(b) By Eq. (4) of Theorem 3, the projection of ${\bf{u}}$ along ${\bf{v}}$ is the vector ${{\bf{u}}_{||{\bf{v}}}}$:
${{\bf{u}}_{||{\bf{v}}}} = \left( {\frac{{{\bf{u}}\cdot{\bf{v}}}}{{||{\bf{v}}||}}} \right){{\bf{e}}_{\bf{v}}}$,
where ${{\bf{e}}_{\bf{v}}}$ is the unit vector along ${\bf{v}}$.
The projection of ${\bf{v}}$ along ${\bf{u}}$ is the vector ${{\bf{v}}_{||{\bf{u}}}}$:
${{\bf{v}}_{||{\bf{u}}}} = \left( {\frac{{{\bf{v}}\cdot{\bf{u}}}}{{||{\bf{u}}||}}} \right){{\bf{e}}_{\bf{u}}} = \left( {\frac{{{\bf{u}}\cdot{\bf{v}}}}{{||{\bf{u}}||}}} \right){{\bf{e}}_{\bf{u}}}$
where ${{\bf{e}}_{\bf{u}}}$ is the unit vector along ${\bf{u}}$.
Since $||{{\bf{e}}_{\bf{u}}}|| = 1$ and $||{{\bf{e}}_{\bf{v}}}|| = 1$, The magnitudes of ${{\bf{u}}_{||{\bf{v}}}}$ and ${{\bf{v}}_{||{\bf{u}}}}$ are
$||{{\bf{u}}_{||{\bf{v}}}}|| = ||\frac{{{\bf{u}}\cdot{\bf{v}}}}{{||{\bf{v}}||}}||$
$||{{\bf{v}}_{||{\bf{u}}}}|| = ||\frac{{{\bf{u}}\cdot{\bf{v}}}}{{||{\bf{u}}||}}||$
From the figure we see that $||{\bf{v}}||$ is smaller in magnitude, hence from the equations above we conclude that ${{\bf{u}}_{||{\bf{v}}}}$ has the greater magnitude. This fact is also depicted in the figure.
