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: 57

Answer

$x=1-t; y=1+t; z=-1$

Work Step by Step

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