Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Section 12.1 - Three-Dimensional Coordinate Systems - 12.1 Exercises - Page 796: 10

Answer

$|PQ|=3$ $|QR|=3\sqrt{5}$ $|RP|=6$ The triangle is not isosceles.

Work Step by Step

The side lengths can be calculated using the distance formula: $d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2)}$ $|PQ|=\sqrt{(4-2)^2+(1+1)^2+(1-0)^2)}=3$ $|QR|=\sqrt{(4-4)^2+(-5-1)^2+(4-1)^2)}=\sqrt{45}=3\sqrt{5}$ $|RP|=\sqrt{(2-4)^2+(-1+5)^2+(0-4)^2)}=6$ Since none of the legs are the same length, the triangle is not isosceles.
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