University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 11 - Practice Exercises - Page 638: 14

Answer

$\sqrt 6 ; \lt \dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt 6},\dfrac{-1}{\sqrt 6}\gt$

Work Step by Step

Here, $|v|=\sqrt{(1)^2+(2)^2+(-1)^2}=\sqrt {6}$ The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$ Now, $\hat{\textbf{u}}=\dfrac{\lt 1,2,-1 \gt}{\sqrt 6}= \lt \dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt 6},\dfrac{-1}{\sqrt 6}\gt$ Thus, our final answers are: $\sqrt 6 ; \lt \dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt 6},\dfrac{-1}{\sqrt 6}\gt$
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