Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 8 - Polar Coordinates; Vectors - Section 8.6 Vectors in Space - 8.6 Assess Your Understanding - Page 646: 59


$v=7[\cos 64.6^{\circ} i+\cos 149.0^{\circ} j+\cos 106.6^{\circ}k ]$

Work Step by Step

The magnitude of any vector (let us say $v$) can be determined using the formula $||v||=\sqrt{p^2+q^2+r^2} $ Let $\alpha, \beta, \gamma $ be the three direction angles with the following formulas: $\alpha =\cos^{-1} \dfrac{p}{||v||}$ and $\beta =\cos^{-1} \dfrac{q}{||v||}$ and $\gamma =\cos^{-1} \dfrac{r}{||v||}$ In our case: $||v||=\sqrt {(3)^2+(-6)^2+(-2)^2}=\sqrt {49}=7$ and $p=3; q=-6 ; r=-2$ Therefore, $\alpha =\cos^{-1} \dfrac{3}{7} \approx 64.6^{\circ} $ and $\beta =\cos^{-1} \dfrac{-6}{7} \approx 149.0^{\circ} $ and $\gamma =\cos^{-1} \dfrac{2}{7}\approx 106.6^{\circ}$ Thus, we get $v=||v|| [\cos \alpha i+ \cos \beta j+ \cos \gamma k ] =7[\cos 64.6^{\circ} i+\cos 149.0^{\circ} j+\cos 106.6^{\circ}k ]$
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