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: 7


$\lt \dfrac{8}{\sqrt{17}},\dfrac{-2}{\sqrt{17}} \gt$

Work Step by Step

Here, $|v|=\sqrt{(4)^2+(-1)^2}=\sqrt {17}$ The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$ Now, $\hat{\textbf{u}}=\dfrac{\lt 4,-1 \gt}{\sqrt {17}}= \lt \dfrac{4}{\sqrt{17}},\dfrac{-1}{\sqrt{17}} \gt$ Thus, $2 \hat{\textbf{u}}=2\lt \dfrac{4}{\sqrt{17}},\dfrac{-1}{\sqrt{17}} \gt=\lt \dfrac{8}{\sqrt{17}},\dfrac{-2}{\sqrt{17}} \gt$
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