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: 11


$2 ; \lt -1,0 \gt$

Work Step by Step

Here, $|v|=\sqrt{(-2)^2+(0)^2}=\sqrt {4}=2$ The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$ Now, $\hat{\textbf{u}}=\dfrac{\lt -2,0 \gt}{2}= \lt -1,0 \gt$ Thus, our final answers are: $2 ; \lt -1,0 \gt$
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