Proving V1 + V2 + V3 + V4 = 0 in a General Tetrahedron

[nikcs123]

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 = \frac{AB x AC}{2}.

Analogously for the other 3 sides, it can be found that:
V₂ = \frac{AB x AD}{2}
V₃ = \frac{AD x AC}{2}
V₄ = \frac{BC x BD}{2}

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.

 

[issacnewton]

your directions of V₂,V₃,V₄ are inward.. for example
V₂ should be
V_2 =\frac{\vec{AD}\times\vec{AB}}{2}

similarly change V₃,V₄... finally use the fact that three vectors
for a triangle put head to tail add to zero... for example
\vec{AB}+\vec{BC}+\vec{CA} = 0

 

[nikcs123]

Ok so fixing the directions on the vectors I find that:

V₂ = \frac{\vec{AD}\times\vec{AB}}{2}
V₃ = \frac{\vec{AC}\times\vec{AD}}{2}
V₄ = \frac{\vec{BD}\times\vec{BC}}{2}

(AB x AC) + (AD x AB) + (AC x AD) + (BD x BC) = 0

So I understand that the 3 vectors of a triangle put together total to 0, but I'm not sure how to apply that in this case.

 

[issacnewton]

consider ,LHS (left hand side)

LHS=V_1+V_2+V_3+V_4

=\frac{1}{2}\left(\vec{AB}\times\vec{AC}+\vec{AD}\times\vec{AB}+\\<br /> \vec{AC}\times\vec{AD}+\vec{BD}\times\vec{BC}\right)......(1)

now look at the second and third terms

\vec{AD}\times\vec{AB}+\vec{AC}\times\vec{AD}

=\vec{AD}\times\vec{AB}-\vec{AD}\times\vec{AC}

=\vec{AD}\times(\vec{AB}-\vec{AC})...(2)

but since \vec{AB}+\vec{BC}+\vec{CA} = 0

we have \vec{AB}-\vec{AC}=-\vec{BC}

plugging this into eq 2 we see that eq 2 becomes

-\vec{AD}\times\vec{BC}......(3)

we can plug above expression in the original equation 1 and LHS becomes

=\frac{1}{2}\left(\vec{AB}\times\vec{AC}-\vec{AD}\times\vec{BC}\\<br /> +\vec{BD}\times\vec{BC}\right)......(4)

from here on we want to eliminate the vertex D. for that use another vector triangle
equation

\vec{AB}+\vec{BD}+\vec{DA} = 0

and continue the manipulation...and you will get all the vectors involving
only vertices A,B and C. again finally use the triangle vector equation to get zero