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: 7


$\lt \dfrac{8}{\sqrt{17}},\dfrac{-2}{\sqrt{17}} \gt$

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, $|v|=\sqrt{(4)^2+(-1)^2}=\sqrt {17}$ Thus, $\hat{\textbf{u}}=\dfrac{v}{|v|}=\dfrac{\lt 4,-1 \gt}{\sqrt {17}}= \lt \dfrac{4}{\sqrt{17}},\dfrac{-1}{\sqrt{17}} \gt$ and $2 \hat{\textbf{u}}=2\lt \dfrac{4}{\sqrt{17}},\dfrac{-1}{\sqrt{17}} \gt=\lt \dfrac{8}{\sqrt{17}},\dfrac{-2}{\sqrt{17}} \gt$
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