Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.4 Vectors, Operations, and the Dot Product - 7.4 Exercises - Page 331: 79

Answer

$90^{\circ}$

Work Step by Step

Step 1: We let $\textbf {u}=\langle 1,2 \rangle$ and $\textbf {v}=\langle -6,3 \rangle$ 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: $\cos\theta=\frac{\langle 1,2 \rangle\cdot\langle -6,3 \rangle}{|\langle 1,2 \rangle||\langle -6,3 \rangle|}$ Step 4: $\cos\theta=\frac{1(-6)+2(3)}{\sqrt (1^{2}+2^{2})\cdot\sqrt ((-6)^{2}+3^{2})}$ Step 5: $\cos\theta=\frac{-6+6}{\sqrt (1+4)\cdot\sqrt (36+9)}$ Step 6: $\cos\theta=\frac{0}{\sqrt (5)\cdot\sqrt (45)}$ Step 7: $\cos\theta=0$ Step 8: $\theta=\cos^{-1}(0)$ Step 9: Solving using the inverse cos function on the calculator, $\theta=\cos^{-1}(0)=90^{\circ}$
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