Answer
$ a.\qquad$ proof given below.
$ b.\qquad$
(1) When one (or both) vectors is a zero vector,
or
(2) 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 $|\cos\alpha | \leq 1,$
$|{\bf u}\cdot{\bf v}|\leq |{\bf u}|\cdot|{\bf v}|$
${\bf (\mathrm{b})}$
$|{\bf u}\cdot{\bf v}|$ = $|{\bf u}|\cdot|{\bf v}|\cdot|\cos\alpha |$=$|{\bf u}|\cdot|{\bf v}|$
This occurs when:
(1)
Either ${\bf u}$ or ${\bf v}$ is a zero vector, (we have 0=0).
or
(2)
If neither ${\bf u}$ nor ${\bf v}$ is a zero vector, the equation stands true when $|\cos\alpha |=1$, that is, if $\alpha=0$ or $\alpha=\pi$.
In other words, when ${\bf u}$ and ${\bf v}$ are parallel.
