Calculus (3rd Edition)

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

Chapter 13 - Vector Geometry - 13.2 Vectors in Three Dimensions - Exercises - Page 659: 39


$${\bf{r}}(t)= \langle 4t, t,t \rangle.$$

Work Step by Step

To find a vector parametrization for the given line, we have to find the direction vector: $(4,1,1 )-(0,0,0)=\langle 4,1,1 \rangle $ Hence, we get the parametrization $${\bf{r}}(t)= \langle 0,0,0 \rangle +t \langle 4,1,1 \rangle\\ = \langle 4t, t,t \rangle.$$
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