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 - Section 12.2 - Vectors - Exercises 12.2 - Page 704: 30


$1(\dfrac{1}{\sqrt 3}i +\dfrac{1}{\sqrt 3}j +\dfrac{1}{\sqrt 3}k)$

Work Step by Step

Formula to find the unit vector $\hat{\textbf{v}}$ is: $\hat{\textbf{v}}=\dfrac{v}{|v|}$ Given:$v=\dfrac{1}{\sqrt 3}i +\dfrac{1}{\sqrt 3}j +\dfrac{1}{\sqrt 3}k$; $|v|=\sqrt{(\dfrac{1}{\sqrt 3})^2+(\dfrac{1}{\sqrt 3})^2+(\dfrac{1}{\sqrt 3})^2}=1$ Thus, $\hat{\textbf{v}}=\dfrac{(\dfrac{1}{\sqrt 3}i +\dfrac{1}{\sqrt 3}j +\dfrac{1}{\sqrt 3}k)}{(1)}=(\dfrac{1}{\sqrt 3}i +\dfrac{1}{\sqrt 3}j +\dfrac{1}{\sqrt 3}k)$ and $v=|v|\hat{\textbf{v}}=1(\dfrac{1}{\sqrt 3}i +\dfrac{1}{\sqrt 3}j +\dfrac{1}{\sqrt 3}k)$
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