Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 727: 21



Work Step by Step

When the plane passing through the point $P(a,b,c)$ and normal to vector $n=pi+qj+rk$ then we have a component equation as follows: $a(x-a)+b(y-b)+c(z-c)=0$ ...(x) Here $P(0,2,-1)$ and $n=3i-2j-k$ From Equation (x), we have $3(x-0)+(-2)(y-2)+(-1)(z-(-1))=0$ $\implies 3x-2y+4-z-1=0$ Thus, $3x-2y-z=-3$
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.