Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Review Exercises - Page 352: 45


$90^{\circ}$, orthogonal

Work Step by Step

Step 1: We let $\textbf {u}=\langle 5,-3 \rangle$ and $\textbf {v}=\langle 3,5 \rangle$ Step 2: The formula for finding the angle between a pair of vectors is $\cos\theta=\frac{\textbf {u}\cdot\textbf {v}}{|\textbf {u}||\textbf {v}|}$ Step 3: $\cos\theta=\frac{\langle 5,-3 \rangle\cdot\langle 3,5 \rangle}{|\langle 5,-3 \rangle||\langle 3,5 \rangle|}$ Step 4: $\cos\theta=\frac{5(3)-3(5)}{\sqrt (5^{2}+(-3)^{2})\cdot\sqrt (3^{2}+5^{2})}$ Step 5: $\cos\theta=\frac{15-15}{\sqrt (25+9)\cdot\sqrt (9+25)}$ Step 6: $\cos\theta=\frac{0}{\sqrt (34)\cdot\sqrt (34)}$ Step 7: $\cos\theta=0$ Step 8: $\theta=\cos^{-1}(0)$ Step 9: Solving using the inverse cos function on the calculator, $\theta=\cos^{-1}(0)=90^{\circ}$ Since the angle between the vectors is $90^{\circ}$, the vectors are orthogonal.
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