Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 5 - Inner Product Spaces - 5.1 Length and Dot Product in Rn - 5.1 Exercises: 16

Answer

$v=(0, \sqrt 6, \frac{\sqrt 6}{2}, -\frac{\sqrt 6}{2})$

Work Step by Step

The magnitude of vector, $u$ is given by $\sqrt {(0^{2}+(2)^{2}+(1)^{2}+(-1)^{2}}=\sqrt {6}$ Hence a unit vector in the same direction as $u$ is given by $\frac{1}{\sqrt {6}}\times(02,1,-1)$ $=(0,\frac{2}{\sqrt {6}}, \frac{1}{\sqrt {6}}, - \frac{1}{\sqrt {6}})$ Vector, $v$ has length 3, so must be equal to $3\times(0,\frac{2}{\sqrt {6}}, \frac{1}{\sqrt {6}}, - \frac{1}{\sqrt {6}})$ $=(0, \frac{6}{\sqrt {6}}, \frac{3}{\sqrt {6}} - \frac{3}{\sqrt {6}})$ Therefore $v=(0, \sqrt 6, \frac{\sqrt 6}{2}, -\frac{\sqrt 6}{2})$
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