Elementary Geometry for College Students (7th Edition)

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: 12

Answer

$d = 7.81$

Work Step by Step

$P_1 = (-1,2,3)$ $P_2 = (2,-2,9)$ We can find the distance between these two points: $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ $d = \sqrt{(2-(-1))^2+((-2)-2)^2+(9-3)^2}$ $d = \sqrt{(3)^2+(-4)^2+(6)^2}$ $d = \sqrt{9+16+36}$ $d = \sqrt{61}$ $d = 7.81$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.