Proving the Sum of Vector Areas in a Tetrahedron is Zero

plexus0208
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Homework Statement


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.

Homework Equations



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.
 
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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"?
 


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
 


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, \vec a = B - A 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!
 


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.
 


Could you explain how you solved this?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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