Answer
$\sqrt {21}$
Work Step by Step
In this case $P_{1}$ is at origin. So the distance formula reduces to d= $\sqrt {x^{2}+y^{2}+z^{2}}$.
Substituting $x= 4, y= 1,$ and $z=2$, we obtain
$d= \sqrt {4^{2}+1^{2}+2^{2}}= \sqrt {21}$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
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: 15
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