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: 5

Answer

$r=\lt1+t,3t,6+t\gt$ and $x=1+t,y=3t,z=6+t $

Work Step by Step

Equation of the line is given by $r=r_0+tv$ $r_0=(1,0,6) $ and $v= \lt 1,3,1 \gt$ Thus, the vector equation of the line is $r=(1,0,6) +t \lt 1,3,1 \gt$ $r=\lt1+t,3t,6+t\gt$ Hence, the parametric equations are: $x=1+t,y=3t,z=6+t $
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