Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - Review - Exercises: 16

Answer

$x=1+3t,y=2t,z=−1+t$

Work Step by Step

Given: $x=4+3t,y=2t,z=−2+t$ The direction of the vectors is $<3,2,1>$ Formula for a line: $ r=r_0+tv $ where$ r_0$ is the starting point vector and $v$ is the direction vector. $r=<1,0,−1>+t<3,2,1> $ Therefore, the parametric equations of the line are: $x=1+3t,y=2t,z=−1+t$
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