## Thomas' Calculus 13th Edition

$a.\qquad$proof given below. $b.\qquad$ When one (or both) vectors is a zero vector, and if both are nonzero, when they are parallel.
${\bf (a)}$ $|{\bf u}\cdot{\bf v}|=\left||{\bf u}|\cdot|{\bf v}|\cdot\cos\alpha \right|$ $|{\bf u}\cdot{\bf v}|=(|{\bf u}|\cdot|{\bf v}|)\cdot|\cos\alpha |$ ... 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.