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: 19

Answer

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

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, $cos^{-1}(\frac{7}{ \sqrt {130}})\approx 52^\circ$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.