## Calculus (3rd Edition)

Published by W. H. Freeman

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

#### Answer

$$\frac{1}{3}\lt2, 1, −2\gt$$

#### 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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