Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.1 - Three-Dimensional Coordinate Systems - Exercises 12.1 - Page 696: 63



Work Step by Step

Here, $y=3, y=-1$ Formula to find the distance between two points is: $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ Now, $\sqrt{(x-x)^2+(y-3)^2+(z-z)^2}=\sqrt{(x-x)^2+(y-(-1))^2+(z-z)^2}$ $\implies (y+1)^2=(y-3)^2$ or, $y^2+1+2y=y^2-6y+9$ Thus, $8y=8$ Hence, $y=1$
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