Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 727: 58

Answer

$x=1+14t; y=2t; z=15t$

Work Step by Step

We are given that the equation of plane are $3x-2z=3$ and $2x-2z=2$ Since, $r=r_0+tv$ and $v=\lt 14,2,15 \gt$ Now, consider $y=0$ Then, $x=1$ and $z=x-1=1-1=0$ Thus, $r_0=\lt 1,0,0 \gt$ Now, the parametric equations are $x=1+(14)t; y=0+2t; z=0+15t=15t$ or, $x=1+14t; y=2t; z=15t$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.