Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Section 12.3 - The Dot Product - 12.3 Exercises: 17

Answer

$cos^{-1}(\frac{-10}{ \sqrt {18}\times \sqrt {8}})\approx 146.4^\circ$

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 $|a|=\sqrt {(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ $a.b=0 \times 1+ 2\times (-4)+ -2 \times 1=0-8-2=-10$ $|a|=\sqrt {(1)^{2}+(-4)^{2}+(1)^{2}}=\sqrt {18}$ $|b|=\sqrt {(0)^{2}+(2)^{2}+(-2)^{2}}=\sqrt {8}$ Angle between two vectors is given by $cos\theta=\frac{a.b}{|a||b|}$ $cos\theta=\frac{-10}{ \sqrt {18}\times \sqrt {8}}$ $\theta=cos^{-1}(\frac{-10}{ \sqrt {18}\times \sqrt {8}})\approx 146.4^\circ$ Hence, $cos^{-1}(\frac{-10}{ \sqrt {18}\times \sqrt {8}})\approx 146.4^\circ$
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