Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - 12.3 The Dot Product - 12.3 Exercises - Page 852: 15


$cos^{-1}(\frac{1}{\sqrt 5})\approx 63^\circ$

Work Step by Step

The dot product of $a=a_{1}i+a_{2}j$ and $b=b_{1}i+b_{2}j$ is defined as $a.b=a_{1}\times b_{1}+a_{2}\times b_{2}$ The magnitude of a vector is given as $|a|=\sqrt {(a_{1})^{2}+(a_{2})^{2}}$ $a.b=4\times 2+ 3\times (-1)=8-3=5$ $|a|=\sqrt {(4)^{2}+(3)^{2}}=5$ $|b|=\sqrt {(2)^{2}+(-1)^{2}}=\sqrt 5$ Angle between two vectors is given by $cos\theta=\frac{a.b}{|a||b|}$ $cos\theta=\frac{5}{5\times \sqrt 5}=\frac{1}{\sqrt 5}$ $\theta=cos^{-1}(\frac{1}{\sqrt 5})\approx 63^\circ$ Hence, $cos^{-1}(\frac{1}{\sqrt 5})\approx 63^\circ$
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