Calculus 8th Edition

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

Chapter 12 - Vectors and the Geometry of Space - 12.5 Equations of Lines and Planes - 12.5 Exercises - Page 872: 64


(a) $P(2,0,2)$ (b) $x+y=2$

Work Step by Step

(a) From the given points in the problem, we have $x_1=1+t, y_1=1-t,z_1=2t$ and $x_2=2-s, y_2=s,z_2=2$ Now, arranging $z's$ and $y's$, we have $t=1$ and $s=0$ Putting these values in either parametric equation. $x_1=1+1, y_1=1-1,z_1=2(1)$ Point of intersection: $P(2,0,2)$ (b) Equation of plane is defined as: $a(x-x_0)+b(y-y_0)+c(z-z_0)$ Here, $\lt a,b,c\gt=\lt-2,-2,0\gt$ and $(x_0,y_0,z_0)=(2,0,2)$ $-2(x-2)+(-2)(y-0)+0(z-2)$ $-2x-4-2y=0$ $2x+2y=4$ Hence, $x+y=2$
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.