Question

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}.

Answer 1

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

open cube

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

Solution

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

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

Then

$0\leq (x-2)^2\leq 10^2\\ |x-2|\leq 10\\ 0\leq (y-3)^2\leq 10^2\\ |y-3|\leq 10\\ 0\leq (z-7)^2\leq 10^2\\ |z-7|\leq 10\\$

Hence

$x-2<11\\ y-3<11\\ z-7<11\\ x,y,z\in R$

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

open cube

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