Answer to Question #105479 in Analytic Geometry for khushi

Question

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}..

Answer 1

We can parametrize a sphere as the set of all points $(x,y,z)$ such that $\sqrt{(x-2)^2+(y-3)^2+(z-7)^2}\leq 10$

$\\ \implies (x-2)^2+(y-3)^2+(z-7)^2\leq 100$

As each term here is a square, they are nonnegative, so we get $3$ simultaneous inequalities

$(x-2)^2\leq 100<121\\ (y-3)^2\leq 100<121\\ (z-7)^2\leq 100<121\\ \implies |x-2|<11\ \&\ |y-3|<11\ \&\ |z-7|<11\$

Thus, any $(x,y,z)$ belonging to the closed sphere belongs to $P$, the open cube.

Thus, the closed sphere is contained in the open cube.