Precalculus (10th Edition)

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

Chapter 9 - Polar Coordinates; Vectors - 9.6 Vectors in Space - 9.6 Assess Your Understanding - Page 625: 49



Work Step by Step

If $v=ai+bj+ck$ and $||v||=\sqrt{a^2+b^2+c^2}$, then the unit vector $u$ in the same direction as $v$ is $u=\frac{v}{||v||}$. The magnitude of $v$ is $||v||=\sqrt{1^2+1^2+1^2}\\||v||=\sqrt{1+1+1}=\sqrt{3}.$ Thus the unit vector is: $u=\frac{i+j+k}{\sqrt{3}}\\u=\frac{1}{\sqrt{3}}i+\frac{1}{\sqrt{3}}j+\frac{1}{\sqrt{3}}k.$
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