Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 10 - Section 10.6 - The Three-Dimensional Coordinate System - Exercises - Page 491: 47

Answer

(a) The sphere has symmetry with respect to the point $P$. (b) The sphere has symmetry with respect to this line.

Work Step by Step

We can write the general equation of a sphere: $(x-a)^2+(y-b)^2+(z-c)^2 = r^2$ where $(a,b,c)$ is the center of the sphere and $r$ is the radius The equation of the sphere is: $(x-1)^2+(y-2)^2+(z+5)^2 = 49$ The center of the sphere is $(1,2,-5)$ (a) The only point that a sphere has symmetry with respect to is the sphere's center. Since the point $P = (1,2,-5)$ is the sphere's center, the sphere has symmetry with respect to the point $P$. (b) A sphere has symmetry with respect to any line that passes through the sphere's center. $l: (x,y,z) = (-5,4,-13)+ n(3,-1,4)$ If the line passes through the sphere's center, we can find the required value for $n$: $x-coordinate: -5+n(3) = 1$ $3n = 6$ $n = 2$ We can find the point the line passes through when $n=2$: $l: (x,y,z) = (-5,4,-13)+ n(3,-1,4)$ $(x,y,z) = (-5,4,-13)+ (2)(3,-1,4)$ $(x,y,z) = (-5+6,4-2,-13+8)$ $(x,y,z) = (1,2,-5)$ Since the line passes through the sphere's center, the sphere has symmetry with respect to this line.
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