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: 42


$5 \sqrt 2$

Work Step by Step

Let $A=(-1,1,5)$ and $B=(2,5,0)$ be the two given points. The distance between two points can be calculated as: $|AB|=\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ Thus, we have $|AB|=\sqrt{(2-(-1))^2+(5-1)^2+(0-5)^2}=\sqrt {9+16+25}=\sqrt{50}=5 \sqrt 2$
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