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: 28

Answer

$1(\dfrac{3}{5}i+\dfrac{4}{5}k)$

Work Step by Step

Formula to find the unit vector $\hat{\textbf{v}}$ is: $\hat{\textbf{v}}=\dfrac{v}{|v|}$ Given:$v=\dfrac{3}{5}i+\dfrac{4}{5}k$; $|v|=\sqrt{(\dfrac{3}{5})^2+(\dfrac{4}{5})^2}=\sqrt{\dfrac{9}{25}+\dfrac{16}{25}}=1$ Thus, $\hat{\textbf{v}}=\dfrac{(\dfrac{3}{5}i+\dfrac{4}{5}k)}{1}=(\dfrac{3}{5}i+\dfrac{4}{5}k)$ and, $v=|v|\hat{\textbf{v}}=1(\dfrac{3}{5}i+\dfrac{4}{5}k)$
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