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

In summary: D.In summary, the vectors 1, V2, V3, V4, and D all have equal lengths and directions perpendicular to the tetrahedron's four faces, but the sum of these vectors is zero, due to the presence of vertex D.
  • #1
nikcs123
4
0

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



pic6.gif


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

similarly change V3,V4... finally use the fact that three vectors
for a triangle put head to tail add to zero... for example
[tex]\vec{AB}+\vec{BC}+\vec{CA} = 0[/tex]
 
  • #3
Ok so fixing the directions on the vectors I find that:

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

(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.
 
  • #4
consider ,LHS (left hand side)

[tex]LHS=V_1+V_2+V_3+V_4[/tex]

[tex]=\frac{1}{2}\left(\vec{AB}\times\vec{AC}+\vec{AD}\times\vec{AB}+\\
\vec{AC}\times\vec{AD}+\vec{BD}\times\vec{BC}\right)......(1)[/tex]

now look at the second and third terms

[tex]\vec{AD}\times\vec{AB}+\vec{AC}\times\vec{AD}[/tex]

[tex]=\vec{AD}\times\vec{AB}-\vec{AD}\times\vec{AC}[/tex]

[tex]=\vec{AD}\times(\vec{AB}-\vec{AC})...(2)[/tex]

but since [tex]\vec{AB}+\vec{BC}+\vec{CA} = 0[/tex]

we have [tex]\vec{AB}-\vec{AC}=-\vec{BC}[/tex]

plugging this into eq 2 we see that eq 2 becomes

[tex]-\vec{AD}\times\vec{BC}......(3)[/tex]

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

[tex]=\frac{1}{2}\left(\vec{AB}\times\vec{AC}-\vec{AD}\times\vec{BC}\\
+\vec{BD}\times\vec{BC}\right)......(4)[/tex]

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

[tex]\vec{AB}+\vec{BD}+\vec{DA} = 0[/tex]

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
 
Last edited:

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

1. What is a general tetrahedron?

A general tetrahedron is a three-dimensional shape with four triangular faces, six edges, and four vertices. It is a type of pyramid where all four faces are triangles and no two edges are parallel.

2. How is V1 + V2 + V3 + V4 equal to 0 in a general tetrahedron?

In a general tetrahedron, the sum of the vertices (V) must equal 0. This is because the sum of the vectors from any vertex to the opposite face will be equal and opposite to the sum of the vectors from the other three vertices to the opposite face. This results in all the vectors cancelling each other out and the final sum being 0.

3. Can you provide an example of a general tetrahedron where V1 + V2 + V3 + V4 = 0?

One example of a general tetrahedron where V1 + V2 + V3 + V4 = 0 is a regular tetrahedron, where all four vertices are equidistant from each other and the center of the shape. In this case, the vectors from each vertex to the opposite face will have the same magnitude and direction, resulting in their sum being 0.

4. What is the significance of proving V1 + V2 + V3 + V4 = 0 in a general tetrahedron?

Proving V1 + V2 + V3 + V4 = 0 in a general tetrahedron is important in understanding the geometry and relationships within this shape. It also has practical applications in fields such as engineering and physics, where the properties of tetrahedrons are utilized in various calculations and designs.

5. Are there any exceptions to the rule of V1 + V2 + V3 + V4 = 0 in a general tetrahedron?

No, there are no exceptions to this rule. In a general tetrahedron, the sum of the vertices will always equal 0 due to the nature of the shape and the relationship between the vectors from the vertices to the opposite face. This is a fundamental property of tetrahedrons and cannot be altered.

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