Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - 12.5 Equations of Lines and Planes - 12.5 Exercises - Page 872: 39



Work Step by Step

The plane passes through the points $(1,5,1)$ and is perpendicular to the planes $ 2x+y-2z=2$ and $x+3z=4$. Normal vector is: $\lt 3,-8,-1\gt $ The general form of the equation of the plane is: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ or, $ax+by+cz=ax_0+by_0+cz_0$ Plugging in the values ,we get $3x-8y-z=3-40-1$ After simplification, we get $3x-8y-z=-38$
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