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: 91

Answer

The pair of vectors is not orthogonal.

Work Step by Step

Step 1: According to vector notation, $\textbf {u}=\langle \sqrt 5,-2 \rangle$ and $\textbf {v}=\langle -5,2\sqrt 5 \rangle$ Step 2: We substitute vectors $\textbf {u}$ and $\textbf {v}$ in the formula for finding the angle between a pair of vectors, $\cos\theta=\frac{\textbf {u}\cdot\textbf {v}}{|\textbf {u}||\textbf {v}|}$ Step 3: $\cos\theta=\frac{\langle \sqrt 5,-2 \rangle\cdot\langle -5,2\sqrt 5 \rangle}{|\langle \sqrt 5,-2 \rangle||\langle -5,2\sqrt 5 \rangle|}$ Step 4: $\cos\theta=\frac{\sqrt 5(-5)-2(2\sqrt 5)}{\sqrt ((\sqrt 5)^{2}+(-2)^{2})\cdot\sqrt ((-5)^{2}+(2\sqrt 5)^{2})}$ Step 5: $\cos\theta=\frac{-5\sqrt 5-4\sqrt 5}{\sqrt (5+4)\cdot\sqrt (25+20)}$ Step 6: $\cos\theta=\frac{-9\sqrt 5}{\sqrt (9)\cdot\sqrt (45)}$ Step 7: $\cos\theta=\frac{-9\sqrt 5}{\sqrt (9)\cdot\sqrt (9\times5)}$ Step 8: $\cos\theta=\frac{-9\sqrt 5}{3\times3\times\sqrt 5}$ Step 9: $\cos\theta=\frac{-9\sqrt 5}{9\sqrt 5}=-1$ Step 10: $\theta=\cos^{-1}(-1)$ Step 11: Solving using the inverse cos function on the calculator, $\theta=\cos^{-1}(-1)=180^{\circ}$ Since the angle between the two vectors is not $90^{\circ}$, the two vectors are not orthogonal.
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