Calculus: Early Transcendentals 8th Edition

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

Chapter 12 - Review - Exercises - Page 843: 17



Work Step by Step

Given: The line $(-2,2,4)$ is perpendicular to $2x-y+5z=12$ The direction of the vectors is $<2,-1,5>$ $ r=r_0+tv $ where$ r_0$ is the starting point vector and $v$ is the direction vector. $r=+t<2,-1,5> $ Therefore, the parametric equations of the line are: $x=-2+2t,y=2-t,z=4+5t$
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