Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - Chapter Review Exercises - Page 701: 13


$$ u=\frac{1}{6}\langle 2,-11,7 \rangle.$$

Work Step by Step

Since $ w=\langle 2,-2,1 \rangle $ and $ v=\langle4,5,-4 \rangle $, we have \begin{align*}v+5u =3w-u&\Longrightarrow 6u=3w-v \\ &\Longrightarrow u=\frac{1}{6}(3w-v)\\ &\Longrightarrow u=\frac{1}{6}(3\langle 2,-2,1 \rangle-\langle 4,5,-4 \rangle)\\ &\Longrightarrow u=\frac{1}{6}\langle 2,-11,7 \rangle. \end{align*}
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