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: 10


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

Work Step by Step

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