Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Section 12.5 - Equations of Lines and Planes - 12.5 Exercises - Page 831: 10

Answer

$x=2+t,y=1-t,z=t$ and $x-2=\frac{y-1}{-1}=z$

Work Step by Step

The direction vector for a line through the $(2,1,0)$ and is perpendicular to $i+j+k$ Equation of the line is given by $r=r_0+tv$ $r_0=(2,1,0) $ and $v= \lt 1,-1,1 \gt$ Parametric equations are defined by: $x=x_0+at$, $y=y_0=bt$ and $z=z_0+ct$ Thus, the parametric equations are: $x=2+t,y=1-t,z=t$ The symmetric equations are defined by: $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ Hence, the symmetric equations are: $x-2=\frac{y-1}{-1}=z$
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.