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: 54



Work Step by Step

The parametric equations are: $x=2; y=3+2t; z=-2-2t$ We have the equation of a plane: $6x+3y-4z=-12$ Now, we plug in all of the parametric equations into the above equation of a plane: $t=\dfrac{-41}{14}$ Thus, our parametric equations become: $x=2; y=3+2(\dfrac{-41}{14})=\dfrac{-20}{7}; z=-2-2(\dfrac{-41}{14})=\dfrac{27}{7}$ Hence, the line will meet at the point: $(2,\dfrac{-20}{7},\dfrac{27}{7})$
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.