Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.3 - The Dot Product - Exercises 12.3 - Page 713: 25


$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.
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