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: 16

Answer

$cos^{-1}(\frac{50}{\sqrt {4901}})\approx 44^\circ$

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 $|a|=\sqrt {(a_{1})^{2}+(a_{2})^{2}}$ $a.b=-2\times 5+ 5\times 12=-10+60=50$ $|a|=\sqrt {(-2)^{2}+(5)^{2}}=\sqrt {29}$ $|b|=\sqrt {(5)^{2}+(12)^{2}}=\sqrt {169}=13$ Angle between two vectors is given by $cos\theta=\frac{a.b}{|a||b|}$ $cos\theta=\frac{50}{ \sqrt {29}\times \sqrt {169}}=\frac{50}{\sqrt {4901}}$ $\theta=cos^{-1}(\frac{50}{\sqrt {4901}})\approx 44^\circ$ Hence, $cos^{-1}(\frac{50}{\sqrt {4901}})\approx 44^\circ$
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