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: 14

Answer

$$x = 4t,{\text{ }}y = 5t + 2,{\text{ }}z = 3t - 1$$

Work Step by Step

$$\eqalign{ & P\underbrace {\left( {0,2, - 1} \right)}_{{x_1},{y_1},{z_1}},{\text{ }}Q\underbrace {\left( {4,7,2} \right)}_{{x_2},{y_2},{z_2}} \cr & {\text{Find }}{\bf{v}} \cr & {\bf{v}} = \left\langle {{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}} \right\rangle \cr & {\bf{v}} = \left\langle {4 - 0,7 - 2,2 - \left( { - 1} \right)} \right\rangle \cr & {\bf{v}} = \left\langle {4,5,3} \right\rangle \cr & {\text{Then, }} \cr & {\bf{r}}\left( t \right) = {\bf{v}}t + \left\langle {{x_1},{y_1},{z_1}} \right\rangle \cr & {\bf{r}}\left( t \right) = \left\langle {4,5,3} \right\rangle t + \left\langle {0,2, - 1} \right\rangle \cr & {\bf{r}}\left( t \right) = \left\langle {4t,5t + 2,3t - 1} \right\rangle \cr & x = 4t,{\text{ }}y = 5t + 2,{\text{ }}z = 3t - 1 \cr} $$
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