Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 11 - Three-Dimensional Space; Vectors - 11.1 Rectangular Coordinates In 3-Space; Spheres; Cylindrical Surfaces - Exercises Set 11.1 - Page 771: 3

Answer

$-6+\sqrt{38}$

Work Step by Step

The center of this sphere is $(1,-2,0)$ Distance between $(4,0,5)$ and $(1,-2,0)$ is \[ \sqrt{(-0-2)^{2}+(-5+0)^{2}+(-4+1)^{2}}=\sqrt{(-5)^{2}+(-3)^{2}+(-2)^{2}}=\sqrt{25+9+4}=\sqrt{38} \] Notice that radius of this sphere is $6,$ and $\sqrt{38}>6$ Thus, the point $(4,0,5)$ is outside the sphere And, therefore, the shortest distance to the sphere is $-6+\sqrt{38}$.
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