Calculus (3rd Edition)

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

Chapter 14 - Calculus of Vector-Valued Functions - 14.4 Curvature - Preliminary Questions - Page 734: 1


$$\frac{1}{3}\lt2, 1, −2\gt$$

Work Step by Step

The unit tangent vector of a line with direction vector $v = \lt2, 1, −2\gt$ is$$\frac{v}{\|v\|}=\frac{\lt2, 1, −2\gt}{\sqrt{4+1+4}}=\frac{1}{3}\lt2, 1, −2\gt$$
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