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 - Page 235: 14


$v=(2\sqrt 2, -2\sqrt 2)$

Work Step by Step

The magnitude of vector, $u$ is given by $\sqrt {(1)^{2}+(-1)^{2}}=\sqrt 2$ Hence a unit vector in the same direction as $u$ is given by $\frac{1}{\sqrt 2}\times(1,-1)$ $=(\frac{1}{\sqrt 2}, -\frac{1}{\sqrt 2})$ Vector, $v$ has length 4, so must be equal to $4\times(\frac{1}{\sqrt 2}, -\frac{1}{\sqrt 2})$ $=(\frac{4}{\sqrt 2}, -\frac{4}{\sqrt 2})$ Therefore, $v=(2\sqrt 2, -2\sqrt 2)$
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