Answer to Question #104654 in Analytic Geometry for Akshay Kumar

Question

Show that the closed sphere with centre )7,3,2( and radius 10 in R³ 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 R³ 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 R³ is

$$(x-2)^2+(y-3)^2+(z-7)^2=10^2$$

$(x-2)^2\geq0, x\in \R$

$(y-3)^2\geq0, y\in \R$

$(z-7)^2\geq0, z\in \R$

Then

$$0\leq(x-2)^2\leq10^2$$ $$|x-2|\leq10$$

$$0\leq(y-3)^2\leq10^2$$ $$|y-3|\leq10$$

$$0\leq(z-7)^2\leq10^2$$ $$|z-7|\leq10$$

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 R³ is contained in the

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