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: 30


$1(\dfrac{1}{\sqrt 3}i +\dfrac{1}{\sqrt 3}j +\dfrac{1}{\sqrt 3}k)$

Work Step by Step

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