Answer
$3$
Work Step by Step
Apply the distance formula, $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$
$d=\sqrt{(3-1)^{2}+(3-1)^{2}+(0-1)^{2}}$
$d=\sqrt{4+4+1}=\sqrt{9}$
$d=3$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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: 41
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