Answer
$a.\qquad $proof given below.
$ b.\qquad$ When one (or both) vectors is a zero vector,
and if both are nonzero, when they are parallel.
Work Step by Step
${\bf (a)}$
$|{\bf u}\cdot{\bf v}|=\left||{\bf u}|\cdot|{\bf v}|\cdot\cos\alpha \right|$
$|{\bf u}\cdot{\bf v}|=(|{\bf u}|\cdot|{\bf v}|)\cdot|\cos\alpha |$
... and since $|\cos\alpha | \leq 1,$
$|{\bf u}\cdot{\bf v}|\leq |{\bf u}|\cdot|{\bf v}|$
${\bf (\mathrm{b})}$
$|{\bf u}\cdot{\bf v}|$ equals $|{\bf u}|\cdot|{\bf v}|\cdot|\cos\alpha |$=$|{\bf u}|\cdot|{\bf v}| $
when
(1)
if either of ${\bf u},\ {\bf v}$ is a zero vector,
(we have 0=0)
or
(2)
if neither is a zero vector, when $|\cos\alpha |=1$, that is,
if $\alpha=0$ or $\alpha=\pi$, that is, when ${\bf u}$ and ${\bf v}$ are parallel.
