Question #288862

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}.


1
Expert's answer
2022-01-24T15:34:54-0500

The equation of the sphere with centre (1,3,5)(1, 3, 5) and radius 88 in R3\R^3 is



(x1)2+(y3)2+(z5)2=82(x-1)^2+(y-3)^2+(z-5)^2=8^2

(x1)20,xR(x-1)^2\geq0, x\in \R

(y3)20,yR(y-3)^2\geq0, y\in \R

(z5)20,zR(z-5)^2\geq0, z\in \R

Then


0(x1)2820\leq(x-1)^2\leq8^2x18|x-1|\leq80(y3)2820\leq(y-3)^2\leq8^2y38|y-3|\leq80(z5)2820\leq(z-5)^2\leq8^2z58|z-5|\leq8

Hence


x1<10,y3<10,z5<10,x,y,zR|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)(1,3,5) and radius 88 in R3\R^3 is contained in the open cube

P={(x,y,z):x1<10,y3<10,z5<10}.P = \{(x, y, z): |x − 1| <10, |y − 3| <10, |z − 5| <10\}.


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