I have a very deficient background in geometry, so I come across questions like these and I'm not sure how to verify my intuition.

Consider three points in $\mathbb{R}^3$, given by position vectors, lying on a sphere centered at the origin. These define both a spherical triangle and a plane. Is the sum of the position vectors normal to the plane? I think this is true but I can't figure out how the details go.

Also, it seems like projecting the triangle onto the plane should put the sum of the vectors in some sort of center for the triangle, would that be the circumcenter?

Thank you in advance!


Call the three vectors $a,b,c$. In order that their sum $a+b+c$ be perpendicular to the plane, its dot product with $a-b$ would have to be $0$; likewise its product with $a-c$ and $b-c$. $$ (a+b+c)\cdot(a-b) = \underbrace{a\cdot a - a\cdot b+b\cdot a-b\cdot b} + c\cdot(a-b). $$ The sum over the $\underbrace{\text{underbrace}}$ is clearly $0$, since $a\cdot a=b\cdot b$. But if you move $c$ around without changing $a$ or $b$, the value of the last term can change, so it won't always be $0$.

Where the sum would go would depend on what kind of projection is involved. If one projects each point $p$ in $\mathbb R^3$ to the intersection between the specified plane of the line through $p$ and the center of the sphere, then the sum would project to the barycenter. The reason is that the barycenter is $(a+b+c)/3$, and that's where the line through $a+b+c$ and the origin intersects the plane. If you mean orthogonal projection, I think you'd usually get a different point.

Here is an amazing reference work: http://faculty.evansville.edu/ck6/encyclopedia/ETC.html

Here's another: http://faculty.evansville.edu/ck6/tcenters/

  • $\begingroup$ That is, in fact, an amazing reference. Thank you so much for this detailed answer, even if it's not what I was hoping for :) $\endgroup$ – Eric Stucky Jun 4 '13 at 22:10

The three points of the triangle define a plane. This intersects the sphere in a circle, which being a circle in the plane passing through the three points, is the circumcircle of the three points.

It is then easy to see (use an isosceles triangle with vertices at either end of a diameter of the circle and the centre of the sphere) that the line from the centre of the sphere to the centre of the circle is perpendicular to the circle at the midpoint of the diameter and hence to the plane which contains the circle.

The "midpoint of the diameter" is the centre of the circle, which is the circumcenter of the triangle.

The centroid of the triangle is at the point $\cfrac{a+b+c}{3}$. So the line to $a+b+c$ passes through the centroid and is perpendicular to the plane of the circle only when the centroid coincides with the circumcenter.


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