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 - Page 883: 20



Work Step by Step

Given: The plane through $(1,2,-2)$ that contains the line $x=2t,y=3-t,z=1+3t$ The direction vector to the line is $a=\lt2,-1,3\gt $ At $t=0$ , the points on the line are $(0,3,1)$. Use these points to make vectors originating at the given point. $b=\lt0-1,3-2,1-(-2)\gt=\lt -1,1,-3\gt$ Use cross product to make a vector perpendicular to these vectors (and to the plane) $n=a\times b=\lt-6,-9,1\gt$ 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 $(0,3,1)$ and has normal vector $\lt -6, -9, 1\gt$ Therefore, $\lt -6, -9, 1\gt$.$(x-0,y-3,z-1)=0$ A scalar equation is: $-6(x-0)-9(y-3)+(z-1)=0$ $-6x-9y+27+z-1=0$ $-6x-9y+z=-26$ Hence, the required equation of the plane is $6x+9y-z=26$
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