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: 62



Work Step by Step

Mid point of points $(2,5,5)$ and $(-6,3,1)$ is $M=(-2,4,3)$ Normal vector: $n=\lt -8,-2,-4\gt $ Plug these points and the vector components of the normal vector in the equation of the plane. Thus, $-8(x+2)-2( y-4)-4(z-3)=0 $ $-8x-16-2y+8-4z+12=0$ $8x+2y+4z=4$ Hence, the equation of plane is: $4x+y+2z=2$
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