Question

On the exterior of a tetrahedron, one vector is erected perpendicularly to each face, pointing outwards, and its length is equal to the area of the face. Show that the sum of these four vectors is 0.
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Can anyone show me how to start this? There is a hit on google for Yahoo "Vectors on a Tetrahedron" that I found but I can't decipher it.

Answer 1

The 4 vectors will be of equal magnitude but each of the two will be going in the opposite direction, hence they cancel each other out. 2 vectors cancel each other out and the other 2 also cancel each other out which gives us a sum of 0.

Answer 2

How can I show this?

Answer 3

If you look at a tetrahedron, and draw a vector going through the center of face at 90 degrees and do this for each face, it will will be easier to see that the vectors cancel each other out.

Answer 4

Try constructing vectors on each edge of the tetrahedron. take cross product of the any two vector defining any face of the object. you shall get 4 vectors normal to each face. Now since magnitude of cross product gives the area of the parallelogram, in our case the vector will have the magnitude half of that of cross product. if n1 is the normal AND

Answer 5

if n1 is a normal vector to face 1 and p is cross product of vectors definning the face, then n1=1/2 P.. So you find all normal vectors and sum them. YOu may have to deal with some coordinates