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: 20

Answer

$cos^{-1}(\frac{4}{9\times \sqrt {20}})\approx 84.3^\circ$

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 $|a|=\sqrt {(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ $a.b=8 \times 0+ -1\times (4)+ 4 \times 2=0-4+8=4$ $|a|=\sqrt {(8)^{2}+(-1)^{2}+(4)^{2}}=\sqrt {81}=9$ $|b|=\sqrt {(0)^{2}+(4)^{2}+(2)^{2}}=\sqrt {20}$ Angle between two vectors is given by $cos\theta=\frac{a.b}{|a||b|}$ $cos\theta=\frac{4}{9\times \sqrt {20}}$ $\theta=cos^{-1}(\frac{4}{9\times \sqrt {20}})\approx 84.3^\circ$ Hence, $cos^{-1}(\frac{4}{9\times \sqrt {20}})\approx 84.3^\circ$
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