Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Section 12.5 - Equations of Lines and Planes - 12.5 Exercises - Page 831: 6

Answer

$x=4+4t,y=3+3t,z=-1-t$ or: $x=4t,y=3t,z=-t$ and $\frac{x-4}{4}=\frac{y-3}{3}=\frac{z+1}{-1}$ or: $\frac{x}{4}=\frac{y}{3}=\frac{z}{-1}$

Work Step by Step

The direction vector for a line through the origin $(0,0,0)$ and the point $(4,3,-1)$ is $\lt 4,3,-1 \gt$ . Parametric equations defined by: $x=x_0+at$, $y=y_0=bt$ and $z=z_0+ct$ Thus, the parametric equations are: $x=4+4t,y=3+3t,z=-1-t$ The symmetric equations are defined by: $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ Hence, the symmetric equations are: $\frac{x-4}{4}=\frac{y-3}{3}=\frac{z+1}{-1}$
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