University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 11 - Section 11.1 - Three-Dimensional Coordinate Systems - Exercises - Page 600: 63



Work Step by Step

Since, we are given two equations of a plane: $y=3, y=-1$ The distance between two points can be calculated as: $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ Thus, $\sqrt{(x-x)^2+(y-3)^2+(z-z)^2}=\sqrt{(x-x)^2+(y-(-1))^2+(z-z)^2}$ or, $(y+1)^2=(y-3)^2$ or, $y^2+1+2y=y^2-6y+9$ or, $8y=8$ Hence, $y=1$
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