Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 9 - Polar Coordinates; Vectors - 9.4 Vectors - 9.4 Assess Your Understanding - Page 606: 53

Answer

$\frac{i}{\sqrt{2}}-\frac{j}{\sqrt{2}}.$

Work Step by Step

If $v=ai+bj$ then the unit vector $u$ in the direction of $v$ is $u=\frac{v}{||v||}$, where $||v||=\sqrt{a^2+b^2}.$ $||v||=\sqrt{1^2+(-1)^2}=\sqrt{2}.$ Hence $u=\frac{i-j}{\sqrt{2}}=\frac{i}{\sqrt{2}}-\frac{j}{\sqrt{2}}.$
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