University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Practice Exercises - Page 638: 8


$\lt -3, -4 \gt$

Work Step by Step

Here, $|v|=\sqrt{(3/5)^2+(4/5)^2}=\sqrt {1}=1$ The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$ Now, $\hat{\textbf{u}}=\dfrac{\lt \dfrac{3}{5},\dfrac{4}{5} \gt}{1}= \lt \dfrac{3}{5},\dfrac{4}{5} \gt$ Thus, $-5 \hat{\textbf{u}}=-5\lt \dfrac{3}{5},\dfrac{4}{5} \gt =\lt -3, -4 \gt$
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