Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 10 - Section 10.6 - The Three-Dimensional Coordinate System - Exercises - Page 482: 13

Answer

$d = 3$

Work Step by Step

$P_1 = (1,-1,1)$ $P_2 = (2,1,3)$ We can find the distance between these two points: $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ $d = \sqrt{(2-1)^2+(1-(-1))^2+(3-1)^2}$ $d = \sqrt{(1)^2+(2)^2+(2)^2}$ $d = \sqrt{1+4+4}$ $d = \sqrt{9}$ $d = 3$
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