4 vectors perpendicular to four faces of tetrahedron. Show that the sum = zero vector

plexus0208

1. The problem statement, all variables and given/known data
Four vectors are erected perpendicular to the four faces of a general tetrahedron. Each vector is pointing outwards and has a length equal to the area of the face. Show that the sum of these four vectors is zero.

2. Relevant equations

3. The attempt at a solution
Let A, B and C be vectors representing the three edges starting from a fixed vertex. Then, express each of the four vectors in terms of A, B and C.

plexus0208

I figured out how to do the problem.

But there's another part to the problem: Formulate and prove the analogous statement for a plane triangle.

What is meant by a "planar triangle"?

lanedance

i think it probably just means any normal triangle (normal in the sense that it is contained within in a plane in R^3, so planar)

can you show it for such? where i'm guessing the analogy is area to length

LCKurtz

Put your triangle in the xy plane; label the vertices A,B,C clockwise. Make vectors of the sides going from A to B, B to C, C to A (i.e, [itex]\vec a = B - A[/itex] etc.). Those side vectors add to the zero vector. Now rotate the triangle 90 degrees counterclockwise and they become perpendicular to the original sides pointing outward and the right length. Presto!

primoclark

I am trying to solve this same problem. Would the person who asked the question and said he/she figured it out please tell me how its done? Unfortunately, it is just not clicking with me. Thank you.

primoclark

Could you explain how you solved this?