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: 25



Work Step by Step

Here, $v=2i+j-2k$ and $|v|=\sqrt{2^2+1^2+(-2)^2}=\sqrt 9 =3$ The unit vector $\hat{\textbf{v}}$ can be calculated as: $\hat{\textbf{v}}=\dfrac{v}{|v|}$ Now, $\hat{\textbf{v}}=\dfrac{2i+j-2k}{3}$ Thus, $v=|v|\hat{\textbf{v}}=3(\dfrac{2}{3}i+\dfrac{1}{3}k-\dfrac{2}{3}k)$
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