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: 61

Answer

$x+2y+z=5$

Work Step by Step

Mid point of points $(3,4,0)$ and $(1,0,-2)$ is $M=(2,2,-1)$ Normal vector: $n=\lt 1,2,1\gt $ Transform $M$ into vector form as: $M=\lt x-2, y-2,z+1\gt $ $n \cdot M= \lt 1,2,1\gt \cdot \lt x-2, y-2,z+1\gt =0 $ $1(x-2)+2( y-2)+1(z+1)=0 $ Hence, the equation of plane is: $x+2y+z=5$
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