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: 28



Work Step by Step

Since, we have $t=2s+2$ and $-t+2=s+3$ After solving, we get $s=-1$ and $t=2s +2=2(-1)+2=0$ The normal to the plane is $n=\lt -6,-3,3 \gt$ 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 $(0,2,1)$, we have $-6(x-0)-3(y -2)+3(z-1)=0$ or, $-6x-3y+3z=-3$
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