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: 38

Answer

$6x-22y-29z+101=0$ or, $6x-22y-29z=-101$

Work Step by Step

The plane passes through the points $(0,-2,5)$ and $(-1,3,1)$ and is perpendicular to the plane $ 2z=5x +4y$. Normal vector is: $\lt 5,4,-2\gt $ Points are $P=(0,-2,5)$ and $Q=(-1,3,1)$ and Thus, $PQ\times N=\lt 6,-22, -29\gt$ 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 $\lt 6,-22, -29\gt$, we get $6x-22(y+2)-29(z-5)=0$ After simplification, we get $6x-22y-29z+101=0$ or, $6x-22y-29z=-101$
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