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: 16


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

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: $|v|=\sqrt{(\dfrac{3}{5})^2+0^2+(\dfrac{4}{5})^2}=1$ Thus, $-5 \hat{\textbf{u}}=\dfrac{v}{|v|}=-5 [\dfrac{\lt \dfrac{3}{5},0,\dfrac{4}{5} \gt}{1}]= \lt -3,0,-4 \gt$
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