University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.5 - Lines and Planes in Space - Exercises - Page 631: 59


$x=4; y=3+6t; z=1+3t$

Work Step by Step

Let us take $z=1$ We have the equation of a plane: $x-2y+4=2$ and $x+y-2=5$ $\implies y=3$ and $x=7-y=7-3=4$ Thus, $r_0=\lt 4,3,1 \gt$ We know that $r=r_0+tv$ and $v=\lt 0,6,3 \gt$ Thus, our parametric equations become: $x=4+(0)t; y=3+6t; z=1+3t$ Hence, our parametric equations are: $x=4; y=3+6t; z=1+3t$
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