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: 61

Answer

$6+\sqrt{17}+\sqrt{33}$

Work Step by Step

Apply the distance formula, $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$ $|AB|=\sqrt{(1-(-1))^{2}+(-1-2)^{2}+(3-1)^{2}}=\sqrt{4+9+4}=\sqrt{17}$ $|AC|=\sqrt{(3-(-1))^{2}+(4-2)^{2}+(5-1)^{2}}=\sqrt{16+4+16}=\sqrt{36}=6$ $|BC|=\sqrt{(3-1)^{2}+(4-(-1))^{2}+(5-3)^{2}}=\sqrt{4+25+4}=\sqrt{33}$ The perimeter is the sum of lengths of the sides $P=6+\sqrt{17}+\sqrt{33}$
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