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: 9


$x=-8+11t,y=1-3t,z=4$ and $\frac{x+8}{11}=\frac{y-1}{-3},z=4$

Work Step by Step

The direction vector for a line through the $(-8,1,4)$ and the point $(3,-2,4)$ is $\lt 11,-3,0 \gt$ . Parametric equations defined by: $x=x_0+at$, $y=y_0=bt$ and $z=z_0+ct$ Thus, the parametric equations are: $x=-8+11t,y=1-3t,z=4-(0)t=4$ 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: $\frac{x+8}{11}=\frac{y-1}{-3},z=4$
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