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: 11

Answer

$2 ; \lt -1,0 \gt$

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: $|v|=\sqrt{(-2)^2+(0)^2}=\sqrt {4}=2$ Thus, $\hat{\textbf{u}}=\dfrac{v}{|v|}=\dfrac{\lt -2,0 \gt}{2}= \lt -1,0 \gt$
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