Answer to Question #288862 in Analytic Geometry for Pankaj

Question

Show that the closed sphere with centre (1,3,5) and radius 8 in R³ is contained in the

open cube

P ={(x, y, z): | x -1| <10, | y - 3 | < 10, | z - 5 | <10}.

Answer 1

The equation of the sphere with centre $(1, 3, 5)$ and radius $8$ in $\R^3$ is

$$(x-1)^2+(y-3)^2+(z-5)^2=8^2$$

$(x-1)^2\geq0, x\in \R$

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

$(z-5)^2\geq0, z\in \R$

Then

$$0\leq(x-1)^2\leq8^2$$ $$|x-1|\leq8$$ $$0\leq(y-3)^2\leq8^2$$ $$|y-3|\leq8$$ $$0\leq(z-5)^2\leq8^2$$ $$|z-5|\leq8$$

Hence

$$|x-1|<10, |y-3|<10, |z-5|<10, x,y,z\in \R$$

This means that the closed sphere with centre $(1,3,5)$ and radius $8$ in $\R^3$ is contained in the open cube

$$P = \{(x, y, z): |x − 1| <10, |y − 3| <10, |z − 5| <10\}.$$