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: 15


$\sqrt {33}; \lt \dfrac{8}{\sqrt {33}},\dfrac{-2}{\sqrt {33}},\dfrac{8}{\sqrt {33}}\gt$

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: $|v|=\sqrt{(4)^2+(-1)^2+(4)^2}=\sqrt {33}$ Thus, $2 \hat{\textbf{u}}=\dfrac{v}{|v|}=2 [\dfrac{\lt 4,-1,4 \gt}{\sqrt {33}}]= \lt \dfrac{8}{\sqrt {33}},\dfrac{-2}{\sqrt {33}},\dfrac{8}{\sqrt {33}}\gt$
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