# Total Vector Area of Tetrahedron is Zero

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## Theorem

Let $T$ be a tetrahedron whose faces have vector area $\mathbf S_1$, $\mathbf S_2$, $\mathbf S_3$ and $\mathbf S_4$.

Let the positive direction be defined as outward.

Then the total vector area is zero:

- $\mathbf S_1 + \mathbf S_2 + \mathbf S_3 + \mathbf S_4 = \mathbf 0$

## Proof

Let the vector areas be resolved upon the faces of $T$.

Some of the projections will be positive and some negative.

Let $T$ be imagined in a fluid which is in equilibrium under pressure.

Each face experiences a force which is normal to its plane and proportional to its area.

Because the fluid inside is in equilibrium with the fluid outside, the resultant of the forces on its faces will equal zero.

Hence the sum of all the vector areas is also zero, as the pressure is the same on all faces.

Hence the result.

$\blacksquare$

## Sources

- 1951: B. Hague:
*An Introduction to Vector Analysis*(5th ed.) ... (previous) ... (next): Chapter $\text {II}$: The Products of Vectors: $5$. Vector Area