Precalculus (10th Edition)

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

Chapter 9 - Polar Coordinates; Vectors - Chapter Review - Review Exercises - Page 635: 38

Answer

$\sqrt{43}$

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 $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ Use the formula above to obtain: $d=\sqrt{(4-1))^2+(-2-3)^2+(1-(-2))^2}\\ d=\sqrt{9+25+9}\\ d=\sqrt{43}$
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