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I have a maths question where the equation of a sphere $S$ is said to be $x^2+y^2+z^2-12x-6y-4z = 0$, and I'm asked to show that the centre is outside the pyramid whose vertices are $A(12,0,0)$, $B (0,6,0)$, $C (0,0,4)$ and the origin. Edit: These 4 points are also contained within the sphere.

I know how to find the origin and radius etc, but I've spent a good hour on this question and I still don't see how you can do it.

Sebastian
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  • Find the plane equations for the 4 faces of the tetrahedron, then show that the center point is outside one of the faces. – NovaDenizen Jan 05 '14 at 12:37
  • How can you show that a point is outside a plane? Either it's contained within the plane or it isn't, and showing that the centre isn't contained within one of the faces of the tetrahedron doesn't really help. – Sebastian Jan 05 '14 at 12:49
  • You can write a plane equation like $ax + by + cz - d = 0$. The function $f(x,y,z) = ax + by + cz - d$ can tell you where a point is relative to a plane. If $f(x,y,z) = 0$, it's on the plane. If $f$ is negative, the point is on one side of the plane, and if $f$ is positive, it's on the other side. – NovaDenizen Jan 05 '14 at 13:18

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By completing the square, we find that the sphere has standard form $(x - 6)^2 + (y-3)^2 + (z-2)^2 = 49$, so it has centre $(6, 3, 2)$ and radius $7$.

The face of the tetrahedron containing the vertices $A$, $B$, and $C$ is a subset of the plane $x + 2y + 3z = 12$. The plane splits $\mathbb{R}^3$ into two regions which are defined by $x + 2y + 3z \leq 12$ and $x + 2y + 3z > 12$.

As $0 + 2(0) + 3(0) = 0 \leq 12$, the origin is contained in the region defined by $x + 2y + 3z \leq 12$. As this region contains all four vertices of the tetrahedron it must contain the entire tetrahedron.

As $3(6) + 2(3) + 3(2) = 30 > 12$, the centre of the sphere is not contained in the region defined by $x + 2y + 3z \leq 12$. In particular, it is not contained in the tetrahedron.

Michael Albanese
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