Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 8 - Polar Coordinates; Vectors - Section 8.6 Vectors in Space - 8.6 Assess Your Understanding - Page 645: 20

Answer

$2 \sqrt 6$

Work Step by Step

The distance $d$ between two points $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ can be calculated as: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. Here, we have: $d=\sqrt{(4-2)^2+(1-(-3))^2+(-1-(-3)))^2}\\ =\sqrt{(2)^2+(4)^2+(2)^2}\\=\sqrt {4+16+4}\\ =\sqrt{24}\\=2 \sqrt 6$
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