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 853: 19


$cos^{-1}(\frac{7}{ \sqrt {130}})\approx 52^\circ$

Work Step by Step

The dot product of $a=a_{1}i+a_{2}j+a_{3}k$ and $b=b_{1}i+b_{2}j+b_{3}k$ is defined as $a.b=a_{1}\times b_{1}+a_{2}\times b_{2}+a_{3}\times b_{3}$ The magnitude of a vector is given as $|a|=\sqrt {(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ $a.b=4 \times 2+ -3\times (0)+ 1 \times -1=8-0-1=7$ $|a|=\sqrt {(4)^{2}+(-3)^{2}+(1)^{2}}=\sqrt {26}$ $|b|=\sqrt {(2)^{2}+(0)^{2}+(-1)^{2}}=\sqrt {5}$ Angle between two vectors is given by $cos\theta=\frac{a.b}{|a||b|}$ $cos\theta=\frac{7}{ \sqrt {26}\times \sqrt {5}}=\frac{7}{ \sqrt {130}}$ $\theta=cos^{-1}(\frac{7}{ \sqrt {130}})\approx 52^\circ$ Hence, $cos^{-1}(\frac{7}{ \sqrt {130}})\approx 52^\circ$
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