Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 9 - Polar Coordinates; Vectors - 9.6 Vectors in Space - 9.6 Assess Your Understanding - Page 625: 20

Answer

$2\sqrt6$

Work Step by Step

The distance $d$ from $P_1(x_1,y_1,z_1)$ to $P_2(x_2,y_2,z_2)$ is given by the formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. Hence here $d=\sqrt{(4-2)^2+(1-(-3))^2+(-1-(-3))^2}=\sqrt{2^2+4^2+2^2}=\sqrt{4+16+4}=\sqrt{24}=2\sqrt6$
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