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$$\frac{v}{\|v\|}=\frac{\lt2, 1, −2\gt}{\sqrt{4+1+4}}=\frac{1}{3}\lt2, 1, −2\gt$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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