Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 12 - Vectors and the Geometry of Space - Review - Exercises - Page 860: 17

Answer

$x=-2+2t,y=2-t,z=4+5t$

Work Step by Step

Given: The line $(-2,2,4)$ 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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