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: 51


$(x - 1)^2+(y -2)^2+(z - 3)^2=14$

Work Step by Step

As we know the standard equation of sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ ...(a) where $(a,b,c)$ represents center and radius of the sphere is $r$ Plug $a=1,b=2,c=3, r=\sqrt{14}$ in equation (a). This implies $(x - 1)^2+(y -2)^2+(z - 3)^2=(\sqrt{14})^2$ or, $(x - 1)^2+(y -2)^2+(z - 3)^2=14$
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