Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 7 - Section 7.5 - Vectors: An Algebraic Approach - 7.5 Problem Set - Page 405: 56

Answer

$\qquad|U|=5 \sqrt{2} , 315^{\circ} = \theta$

Work Step by Step

\[ V=\langle 5,-5\rangle \] Here the $x$ component of the vector is $5=V_{x}$, and the $y$ component of the vector is \[ -5=V_{y} \] The magnitude of the vector $\mathrm{V}$ is \[ \sqrt{V_{y}^{2}+V_{x}^{2}}=|V| \] Replacing $V_{y}$ with $V_{x}$ values \[ \begin{aligned} &\sqrt{5^{2}+(-5)^{2}}=|V| \\ &=\sqrt{25+25} \\ &=\sqrt{50} \\ &=5 \sqrt{2} \end{aligned} \] $\theta$ is the angle made with the positive $x-$ axis and the vector. \[ \tan \theta=\frac{V_{x}}{V_{y}} \] \[ =\frac{-5}{5} \] $ \tan \theta=-1 $ $ \theta=\tan ^{-1}(-1) $ $ =315^{\circ} $
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