University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.5 - Lines and Planes in Space - Exercises - Page 631: 30

Answer

$x+y-2z=7$

Work Step by Step

Since, we have $t=1+s$ and $3-3t=4+s$ After solving, we get $s=-1$ and $t=1+s=1-1=0$ The normal to the plane is $n=\lt -2,-2,4 \gt$ We know that the standard equation of a plane passing through the point $(x_0,y_0,z_0)$ is written as: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ Then for point $(0,3,-2)$, we have $-2(x-0)-2(y -3)+4(z+2)=0$ or, $-2x-2y+4z=-14$ or, $x+y-2z=7$
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