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: 26

Answer

$x-2y+z=6$

Work Step by Step

The standard equation of plane passing through the point $(a,b,c)$ is given as $p(x-a)+q(y-b)+r(z-c)=0$ The equation of normal to the plane; $n=\lt 1,-2,1 \gt$ For point $(1,-2,1)$, we get $1(x-1)-2(y+2)+1(z-1)=0$ or, $x-1-2y-4+z-1=0$ Thus, $x-2y+z=6$
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