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: 8


$\lt -3, -4 \gt$

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