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 - Page 883: 25



Work Step by Step

Planes $x-z=1$ and $y+2z=3$ Direction vectors of the line of intersection is $ \lt 1-0,3-5, 0-(-1) \gt= \lt 1,-2,1 \gt$ Parametric equations of the line can be $x=1+t,y=3-2t,z=0+t$ $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ $3(x-0)+3(y-5)+3(z-(-1))=0$ $3x+3y+3x=12$ Hence, $x+y+z=4$
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.