Question

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]<11,12 – 7|<11}.
​

Answer 1

Answer:

The equation closed sphere with centre (2,3,7) and radius 10 in  is contained in the open cube is

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

∴

$(x-2)^2$  ≤ $10^2$

$(y-3)^{2}$ ≤ $10^2$

$(z-7)^{2}$  ≤  $10^2$

/x-2/ ≤ 10

/y-3/  ≤ 10

/z-7/  ≤ 10

∴/x-2/  < 11

/y-3/ < 11

/z-7/  < 11

Hence proved.

Answer 2

The equation of the sphere with center (2, 3, 7) and radius 10 in R³.

(x - 2)² + (y - 3)² + (z - 7)² = 10²

Now,

(x - 2)² ≥ 0, x ∈ R

0 ≤ (x - 2)² ≤ 10²

|x - 2| ≤ 10

∴ x - 2 < 11, x ∈ R

Now,

(y - 3)² ≥ 0, y ∈ R

0 ≤ (y - 3)² ≤ 10²

|y - 3| ≤ 10

∴ y - 3 < 11, y ∈ R

Now,

(z - 7)² ≥ 0, z ∈ R

0 ≤ (z - 7)² ≤ 10

|z - 7| ≤ 10

∴ z - 7 < 11, z ∈ R

Thus, the closed sphere with center (2,3,7) and radius 10 in R³ is contained in the  open cube P = {(x, y, z):|x - 2| < 11, |y - 3| < 11, |12 - 7| < 11}.