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: 17

Answer

$\sqrt{33}$

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: $d=\sqrt{(0-(-1))^2+(-2-2)^2+(1-(-3))^2}\\ =\sqrt{1+(-4)^2+(4)^2}\\=\sqrt {1+16+16}\\ =\sqrt{33}$
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