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 329: 36

Answer

$16; 315^{\circ}$

Work Step by Step

The magnitude of a vector $\textbf{u}=\langle a,b \rangle$ is given as $|\textbf{u}|=\sqrt (a^{2}+b^{2})$. Since $\textbf{u}=\langle 8\sqrt 2,-8\sqrt 2 \rangle$, the magnitude is: $|\textbf{u}|=\sqrt ((8\sqrt 2)^{2}+(-8\sqrt 2)^{2})=\sqrt (128+128)=\sqrt (256)=16$ The direction angle $\theta$ can be found through the equation $\tan\theta=\frac{b}{a}$. Substituting the values of $a$ and $b$ in the formula and solving using a calculator, $\theta=\tan^{-1} (\frac{8\sqrt 2}{-8\sqrt 2})=\tan^{-1} (-1)=-45^{\circ}$ The vector has a positive horizontal component and a negative vertical component which places it in the fourth quadrant. Since the direction angle is supposed to be the positive angle between the x-axis and the position vector, we need to add $360^{\circ}$ to $-45^{\circ}$ to yield the direction angle $\theta$. Therefore, the direction angle $\theta=315^{\circ}$.
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