Question

Show that the open sphere S with centre at (3,1,4) and radius 5 in R3 is contained in the open cube
P1={(x,y,z) : |x-3|<5 , |y-1|<5 , |z-4|<5} and P1 is contained in the sphere with centre (3,1,4) and radius 8.66 .

Answer 1

Answer on Question # 41398 – Math - Integral Calculus

Show that the open sphere $S$ with centre at $(3,1,4)$ and radius 5 in R3 is contained in the open cube $P1 = \{(x,y,z): |x-3|<5, |y-1|<5, |z-4|<5\}$ and $P1$ is contained in the sphere with centre $\{3,1,4\}$ and radius 8.66.

**Solution:**

As, we can see P1 is the cube with the center $(3,1,4)$ and lateral length 10. And, the sphere with radius 5 has diameter 10, so it is contained in such a cube.

The cube P1 has diagonal $10\sqrt{3} \approx 17.32$. And to put this cube into the sphere, its diameter must not be less than 17.32. So, P1 is contained in sphere with centre $(3,1,4)$ and radius 8.66, because both two figures have the same center and sphere diameter is $8.66*2 = 17.32$.

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