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: 14


$\sqrt 6 ; \lt \dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt 6},\dfrac{-1}{\sqrt 6}\gt$

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: $|v|=\sqrt{(1)^2+(2)^2+(-1)^2}=\sqrt {6}$ Thus, $\hat{\textbf{u}}=\dfrac{v}{|v|}=\dfrac{\lt 1,2,-1 \gt}{\sqrt 6}= \lt \dfrac{1}{\sqrt 6},\dfrac{2}{\sqrt 6},\dfrac{-1}{\sqrt 6}\gt$
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