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: 21



Work Step by Step

The plane passing through the point $P(x_0,y_0,z_0)$ and normal to vector $n=ai+bj+ck$ is represented by a component equation such as: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0~~~~~$ (1) Since, we have $P(0,2,-1)$ and $n=3i-2j-k$ Thus, from the equation (1), we have $3(x-0)+(-2)(y-2)+(-1)(z-(-1))=0$ or, $3x-2y+4-z-1=0$ Hence, $3x-2y-z=-3$
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