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 734: 13


$7 ; \lt \dfrac{2}{7},\dfrac{-3}{7},\dfrac{6}{7}\gt$

Work Step by Step

As we know that the unit vector $\hat{\textbf{u}}$ is defined as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$ Now, we have: Here, $|v|=\sqrt{(2)^2+(-3)^2+(6)^2}=\sqrt {49}=7$ Thus, $\hat{\textbf{u}}=\dfrac{v}{|v|}=\dfrac{\lt 2,-3,6 \gt}{7}= \lt \dfrac{2}{7},\dfrac{-3}{7},\dfrac{6}{7}\gt$
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