Calculus 8th Edition

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

Chapter 12 - Vectors and the Geometry of Space - Review - Exercises: 18

Answer

$x+4y-3z=6$

Work Step by Step

Let $(x_{0},y_{0},z_{0})$ be a point on the plane and $\lt a,b,c\gt$ be a normal vector to the plane. Then a vector equation of the plane is $\lt a,b,c\gt$.$(x_{0},y_{0},z_{0})=0$ A scalar equation is: $a(x-x_{0}),b(y-y_{0}),c(z-z_{0})$ Since, the plane $(2,1,0)$ and has normal vector $=\lt 1,4,-3\gt$ (the same normal vector as its parallel plane $x+4y-3z=1$, Therefore, $\lt 1,4,-3\gt$.$(x-2,y-1,z)=0$ A scalar equation is: $x-2+4y-4-3z=0$ Hence, the required equation of the plane is $x+4y-3z=6$
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