Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Practice Exercises - Page 735: 53


$ \lt \dfrac{10}{\sqrt{35}},\dfrac{-2}{\sqrt{35}},\dfrac{-6}{\sqrt{35}} \gt$

Work Step by Step

The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$ Now, $v=\lt 5,-1,-3 \gt$ and $|v|=\sqrt{(5)^2+(-1)^2+(-3)^2}=\sqrt {35}$ Now, $2 \hat{\textbf{u}}=\dfrac{v}{|v|}=2 [\dfrac{\lt 5,-1,-3 \gt}{\sqrt {35}}]= \lt \dfrac{10}{\sqrt{35}},\dfrac{-2}{\sqrt{35}},\dfrac{-6}{\sqrt{35}} \gt$
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