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: 15


$\sqrt {33}; \lt \dfrac{8}{\sqrt {33}},\dfrac{-2}{\sqrt {33}},\dfrac{8}{\sqrt {33}}\gt$

Work Step by Step

Here, $|v|=\sqrt{(4)^2+(-1)^2+(4)^2}=\sqrt {33}$ The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$ Now, $2 \hat{\textbf{u}}=2 [\dfrac{\lt 4,-1,4 \gt}{\sqrt {33}}]= \lt \dfrac{8}{\sqrt {33}},\dfrac{-2}{\sqrt {33}},\dfrac{8}{\sqrt {33}}\gt$ Thus, our final answers are: $\sqrt {33}; \lt \dfrac{8}{\sqrt {33}},\dfrac{-2}{\sqrt {33}},\dfrac{8}{\sqrt {33}}\gt$
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