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: 38


$${\bf{r}}(t)= \langle -2+6t,3t,-2+9t \rangle.$$

Work Step by Step

To find a vector parametrization for the given line, we have to find the direction vector: $(4,3,7)-(-2,0,-2 )=\langle 6,3,9 \rangle $ Hence, we get the parametrization $${\bf{r}}(t)= \langle -2,0,-2 \rangle +t \langle 6,3,9 \rangle\\ = \langle -2+6t,3t,-2+9t \rangle.$$
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