Answer to Question #107178 in Analytic Geometry for Nikesh gautam pandit ji

Question #107178
Show that the closed sphere with centre (2,3,7) 3and radius 10 in 3 R is contained in the
open cube P = {(x, y,z :) x − 2 <11, y − 3 <11, z − 7 <11}.
1
Expert's answer
2020-03-31T13:59:44-0400

Show that the closed sphere with centre (2,3,7) and radius 10 in "R^3" is contained in the

open cube

"P=\\{(x,y,z): x-2<11, \\\\\ny-3<11, z-7<11\\}".

Solution

The equation of the sphere with centre (2, 3, 7) and radius 10 in "R^3"  is

"(x-2)^2+(y-3)^2+(z-7)^2=10^2\\\\\n(x-2)^2\\geq 0, x\\in R\\\\\n(y-3)^2\\geq 0, y\\in R\\\\\n(z-7)^2\\geq 0, z\\in R\\\\"

Then

"0\\leq (x-2)^2\\leq 10^2\\\\\n|x-2|\\leq 10\\\\\n0\\leq (y-3)^2\\leq 10^2\\\\\n|y-3|\\leq 10\\\\\n0\\leq (z-7)^2\\leq 10^2\\\\\n|z-7|\\leq 10\\\\"

Hence

"x-2<11\\\\\ny-3<11\\\\\nz-7<11\\\\\nx,y,z\\in R"

This means that the closed sphere with centre (2,3,7) and radius 10 in "R^3" is contained in the

open cube

"P=\\{(x,y,z): x-2<11, \\\\\ny-3<11, z-7<11\\}"


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