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: 27



Work Step by Step

Since, we have $2t+1=s+2$ and $3t+2=2s+4$ After solving, we get $t=0$ and $s=-1+2t=-1$ The normal to the plane is $n=\lt -20,12,1 \gt$ 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 $(1,2,3)$, we have $-20(x-1)+12(y -2)+1(z-3)=0$ or, $-20x+20+12y-24+z-3=0$ or, $-20x+12y+z=7$
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