University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.3 - The Dot Product - Exercises - Page 616: 15

Answer

Both parts shown below.

Work Step by Step

Let ${\bf v}=\langle a,b,c\rangle$. The angle between ${\bf i}$ and ${\bf v}$ is $\displaystyle \cos\alpha=\cos\angle({\bf i,v})=\frac{{\bf i}\cdot{\bf v}}{|{\bf i}||{\bf v}|}$, For $\beta$ and $\gamma$, replace ${\bf i}$ with ${\bf j}$ and ${\bf k}$. ${\bf (a)}$ $\displaystyle \cos\alpha=\frac{(1)(a)+(0)(b)+(0)(c)}{1\cdot|{\bf v}|}=\frac{a}{|{\bf v}|}$ $\displaystyle \cos\beta=\frac{{\bf j}\cdot{\bf v}}{|{\bf j}||{\bf v}|}=\frac{(0)(a)+(1)(b)+(0)(c)}{1\cdot|{\bf v}|}=\frac{b}{|{\bf v}|}$ $\displaystyle \cos\gamma=\frac{{\bf k}\cdot{\bf v}}{|{\bf k}||{\bf v}|}=\frac{(0)(a)+(0)(b)+(1)(c)}{1\cdot|{\bf v}|}=\frac{c}{|{\bf v}|}$ ${\bf (b)}$ If $|{\bf v}|=1$, from the above results, it follows that $\displaystyle \cos\alpha=\frac{a}{|{\bf v}|}=a,$ $\displaystyle \cos\beta=\frac{b}{|{\bf v}|}=b$ $\displaystyle \cos\gamma=\frac{c}{|{\bf v}|}=c$ (a, b, and c are the direction cosines of ${\bf v}$).
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