Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.1 - Three-Dimensional Coordinate Systems - Exercises 12.1 - Page 696: 45


$2 \sqrt 3$

Work Step by Step

Formula to find the distance between two points is $|PQ|=\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ Let us consider $P=(0,0,0)$ and $Q=(2,-2,-2)$ Now, $|PQ|=\sqrt{(2-0)^2+(-2-0)^2+(-2-0)^2}=\sqrt{(2)^2+(-2)^2+(-2)^2}=\sqrt {4+4+4}$ or, $|PQ|=\sqrt{12}=2 \sqrt 3$
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.