Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 12 - Vector-Valued Functions - 12.1 Exercises - Page 821: 16

Answer

$$x = - 4t + 1,{\text{ }}y = 4t - 6,{\text{ }}z = - 3t + 8$$

Work Step by Step

$$\eqalign{ & P\underbrace {\left( {1, - 6,8} \right)}_{{x_1},{y_1},{z_1}},{\text{ }}Q\underbrace {\left( { - 3, - 2,5} \right)}_{{x_2},{y_2},{z_2}} \cr & {\text{Find }}{\bf{v}} \cr & {\bf{v}} = \left\langle {{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}} \right\rangle \cr & {\bf{v}} = \left\langle { - 3 - 1, - 2 + 6,5 - 8} \right\rangle \cr & {\bf{v}} = \left\langle { - 4,4, - 3} \right\rangle \cr & {\text{Then, }} \cr & {\bf{r}}\left( t \right) = {\bf{v}}t + \left\langle {{x_1},{y_1},{z_1}} \right\rangle \cr & {\bf{r}}\left( t \right) = \left\langle { - 4,4, - 3} \right\rangle t + \left\langle {1, - 6,8} \right\rangle \cr & {\bf{r}}\left( t \right) = \left\langle { - 4t + 1,4t - 6, - 3t + 8} \right\rangle \cr & x = - 4t + 1,{\text{ }}y = 4t - 6,{\text{ }}z = - 3t + 8 \cr} $$
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