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: 8


$x=1+1.6t,y=2.4-1.2t,z=4.6-4.3t$ and $\frac{x-1}{1.6}=\frac{y-2.4}{-1.2}=\frac{z-4.6}{-4.3}$

Work Step by Step

The direction vector for a line through the $(1,2.4,4.6)$ and the point $(2.6,1.2,0.3)$ is $\lt 1.6,-1.2,-4.3 \gt$ . Parametric equations defined by: $x=x_0+at$, $y=y_0=bt$ and $z=z_0+ct$ Thus, the parametric equations are: $x=1+1.6t,y=2.4-1.2t,z=4.6-4.3t$ 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: $\frac{x-1}{1.6}=\frac{y-2.4}{-1.2}=\frac{z-4.6}{-4.3}$
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