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 609: 33

Answer

$\dfrac{7}{13}(12i-5k)$

Work Step by Step

Here, $u=12i +0j-5k$ and $|u|=\sqrt{(12)^2+(0)^2+(-5)^2}=\sqrt {1}=\sqrt {169}=13$ The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{u}{|u|}$ Now, $\hat{\textbf{u}}=\dfrac{12i +0j-5k}{13}=(\dfrac{12}{13}i +0j -\dfrac{5}{13}k)$ Thus, $7\hat{\textbf{u}}=\dfrac{7}{13}(12i-5k)$
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