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: 62


$|PA|=|PB|=\sqrt 6$

Work Step by Step

The distance between two points can be calculated as: $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ We need to verify that $|PA|=|PB|$ Thus, $|PA|=\sqrt{(3-2)^2+(1+1)^2+(2-3)^2}=\sqrt{6}$ and $|PB|=\sqrt{(3-4)^2+(1-3)^2+(2-1)^2}=\sqrt{6}$ Hence, the perimeter of the triangle is: $|PA|=|PB|=\sqrt 6$
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