Answer to Question #103434 in Analytic Geometry for Avinash

Question

Show that the closed sphere with centre )7,3,2( and 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

The equation of the closed sphere with the center $(2, 3, 7)$ and radius $10$ is

$$(x-2)^2+(y-3)^2+(z-7)^2=10^2$$ $$| x-2|\leq 10, |y-3| \leq 10, |z-7|\leq 10$$ $$--10\leq x-2\leq 10, -10\leq y-3\leq 10, -10\leq z-7\leq 10$$

Therefore the closed sphere with the center $(2,3,7)$ and radius $10$ is contained in the

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