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: 4

Answer

$r=\lt2t,14-3t,-10+9t\gt$ $x=2t,y=14-3t,z=-10+9t $

Work Step by Step

Equation of the line is given by $r=r_0+tv$ $r=(0,14,-10) $ Thus, the vector equation of the line is $r=(0,14,-10)+t\lt2,-3,9\gt$ $r=\lt2t,14-3t,-10+9t\gt$ Hence, the parametric equations are: $x=2t,y=14-3t,z=-10+9t $
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