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: 19

Answer

$-4x+3y+z=-14$

Work Step by Step

Given: $(3,-1,1),(4,0,2),(6,3,1)$ Use the points to make two vectors parallel to the plane. $a=\lt 4-3,0-(-1),2-1\gt=\lt 1,1,1\gt$ $b=\lt 6-3,3-(-1),1-1\gt=\lt 3,4,0\gt$ Use cross product to make a vector perpendicular to these vectors (and to the plane) $n=a\times b=\lt-4,3,1\gt$ 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 $(4,0,2)$ and has normal vector $=\lt -4,3,1\gt$ Therefore, $\lt -4,3,1\gt$.$(x-4,y-0,z-2)=0$ A scalar equation is: $-4(x-4)+3y+1(z-2)=0$ $-4x+16+3y+z-2=0$ Hence, the required equation of the plane is $-4x+3y+z=-14$
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