University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.2 - Vectors - Exercises - Page 608: 26

Answer

$11(\dfrac{9}{11}i-\dfrac{-2}{11}j+\dfrac{6}{11}k)$

Work Step by Step

Here, $v=9i-2j+6k$ and $|v|=\sqrt{9^2+(-2)^2+(6)^2}=\sqrt {121}=11$ The unit vector $\hat{\textbf{v}}$ can be calculated as: $\hat{\textbf{v}}=\dfrac{v}{|v|}$ Now, $\hat{\textbf{v}}=\dfrac{9i-2j+6k}{11}=(\dfrac{9}{11},\dfrac{-2}{11},\dfrac{6}{11})$ Thus, $v=|v|\hat{\textbf{v}}=11(\dfrac{9}{11}i-\dfrac{-2}{11}j+\dfrac{6}{11}k)$
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