- #1

nikcs123

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## Homework Statement

Given a general (not necessarily a rectangular) tetrahedron, let V

_{1}, V

_{2}, V

_{3}, V

_{4}denote vectors whose lengths are equal to the areas of the four faces, and whose directions are perpendicular to these faces and point outward. Show that:

V

_{1}+ V

_{2}+ V

_{3}+ V

_{4}= 0.

## The Attempt at a Solution

So I have a tetrahedron ABCD pictured above (ignore PQSR, only good image i could find).

So in order to define the vectors

_{1}, V

_{2}, V

_{3}, V

_{4}, I start with one of the sides, side ABC.

To find a vector with a direction perpendicular to side ABC, I take the cross product of 2 of the edges (AB X AC). That cross product produces a vector with a direction normal to the side, but with an area that is 2x the area of side ABC. So V

_{1}= [tex]\frac{AB x AC}{2}[/tex].

Analogously for the other 3 sides, it can be found that:

V

_{2}= [tex]\frac{AB x AD}{2}[/tex]

V

_{3}= [tex]\frac{AD x AC}{2}[/tex]

V

_{4}= [tex]\frac{BC x BD}{2}[/tex]

From here I simplify and find that: (AB x AC) + (AB x AD) + (AD x AC) + (BC x BD) = 0.

I'm stuck at this point, I think from here I might have to apply properties of cross products to deduce that the left side also equals 0, but am unsure of how to accomplish that.