Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 10 - Section 10.6 - The Three-Dimensional Coordinate System - Exercises - Page 490: 14

Answer

$d = 5.39$

Work Step by Step

$P_1 = (1,0,2)$ $P_2 = (3,4,-1)$ We can find the distance between these two points: $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ $d = \sqrt{(3-1)^2+(4-0)^2+(-1-2)^2}$ $d = \sqrt{(2)^2+(4)^2+(-3)^2}$ $d = \sqrt{4+16+9}$ $d = \sqrt{29}$ $d = 5.39$
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