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: 37


$${\bf{r}}(t)= \langle 1+2t,1-6t,1+t \rangle.$$

Work Step by Step

To find a vector parametrization for the given line, we first find the direction vector: $(3,-5,2)-(1,1,1)=\langle 2,-6,1 \rangle $ Hence the parametrization is given by $${\bf{r}}(t)= \langle 1,1,1 \rangle +t \langle 2,-6,1 \rangle\\ = \langle 1+2t,1-6t,1+t \rangle.$$
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