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: 29



Work Step by Step

Since, we have $-1+t=1-4s$ and $2+t=1+2s$ After solving, we get $s=\dfrac{1}{2}$ and $t=2-4s=2-4(1/2)=0$ The normal to the plane is $n=\lt 0,6,6 \gt$ 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 $(-1,2,1)$, we have $0(x+1)+6(y -2)+6(z-1)=0$ or, $6y+6z=18$ or, $y+z=3$
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