Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - 13.5 Planes in 3-Space - Exercises - Page 684: 15


$\langle 3,-8,11\rangle $

Work Step by Step

$3(x-4)-8(y-1)+11z=0$ can be rewritten as $3x-8y+11z=4$. The normal vector of a plane in the form $ ax+by+cz=d $ is $\vec n=\langle a,b,c\rangle $ Therefore, the normal vector of $3x-8y+11z=4$ is given as: $\vec n=\langle 3,-8,11\rangle $
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.