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: 16


$1; \lt -3,0,-4 \gt$

Work Step by Step

Here, $|v|=\sqrt{(\dfrac{3}{5})^2+0^2+(\dfrac{4}{5})^2}=1$ The unit vector $\hat{\textbf{u}}$ can be calculated as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$ Now, $-5 \hat{\textbf{u}}=-5 [\dfrac{\lt \dfrac{3}{5},0,\dfrac{4}{5} \gt}{1}]= \lt -3,0,-4 \gt$ Thus, our final answers are: $1; \lt -3,0,-4 \gt$
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