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: 62


$\sqrt 6$

Work Step by Step

Formula to find the distance between two points is $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ We will have to verify that $|PA|=|PB|$ $|PA|=\sqrt{(3-2)^2+(1+1)^2+(2-3)^2}=\sqrt{(1)^2+(2)^2+(-1)^2}=\sqrt{6}$ and $|PB|=\sqrt{(3-4)^2+(1-3)^2+(2-1)^2}=\sqrt{(-1)^2+(-2)^2+(1)^2}=\sqrt{6}$ Hence, $|PA|=|PB|=\sqrt 6$
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