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nikcs123
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Homework Statement
Given a general (not necessarily a rectangular) tetrahedron, let V1, V2, V3, V4 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:
V1 + V2 + V3 + V4 = 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, V2, V3, V4, 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 V1 = [tex]\frac{AB x AC}{2}[/tex].
Analogously for the other 3 sides, it can be found that:
V2 = [tex]\frac{AB x AD}{2}[/tex]
V3 = [tex]\frac{AD x AC}{2}[/tex]
V4 = [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.