Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.5 Lines and Curves in Space - 11.5 Exercises: 17

Answer

$\textbf{r} = \langle 0,0,0\rangle + t\langle-2,8,-4 \rangle$ = $t\langle-2,8,-4 \rangle$

Work Step by Step

Equation of a Line: $\textbf{r} = \textbf{r}_0 + t\textbf{v}$ $\textbf{r}_0 = \langle 0,0,0\rangle$ We can rewrite $\textbf{r}(t) = \langle 3-2t, 5+8t,7-4t\rangle$ to $\textbf{r}(t) = \langle 3,5,7\rangle + t\langle -2,8,-4\rangle$ We simply have to take the $\textbf{v}$ component of the given $\textbf{r}(t)$ to obtain a parallel line since the $\textbf{v}$ gives the direction of the line. $\textbf{v} = \langle -2,8,-4\rangle$ $\textbf{r} = \langle 0,0,0\rangle + t\langle-2,8,-4 \rangle$
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