Sum of normal vectors to the surfaces of a tetrahedron is zero.

Question

Show that the sum of some outer-pointing vectors perpendicular on the faces of a tetrahedron which are proportional to the areas of the faces is the zero vector.

Can somebody give me some advice about how to start? I don't this this is something hard to prove with the cross product.

Answer 1

Hint: Yes, you may use the cross products of vectors representing the four edges. Note that only three of the vectors are independent and the fourth is determined once any three are given.

For the surface with two of its edge vectors $\vec a$ and $\vec b$, its contribution to the sum is

$$\frac12 \vec a \times \vec b$$

Then, sum up the contributions from the four surfaces to arrive at zero using the vector operation.

Answer 2

The statement can be generalized to any polyhedron.

For any polyherdron, triangulate its surface so that all faces are triangles. For each triangle, let $\vec{a}, \vec{b}, \vec{c}$ be its vertices, ordered counterclockwisely when you view them from outside. The area weighted outward pointing normal of this face equals to

$$\frac12 (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) = \frac12(\vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c} \times \vec{a})$$

As one can see, RHS is a sum of 3 pieces, one for each edge of the triangle. When you sum over the triangular faces of the polyhedron, each edge will contribute twice but the contribution from the two triangles adjacent to an edge cancel each other. So the sum of all area weighted normal vanishes.

For a concrete example, let's say we have a tetrahedron. By a suitable choose of coordinate system, we can assume one of its vertex is $\vec{0}$. Let $\vec{a},\vec{b},\vec{c}$ be the other $3$ vertices ordered as described above. The other $3$ faces (again ordered as described above) will be $( \vec{0},\vec{c},\vec{b})$, $(\vec{0},\vec{a},\vec{c})$ and $(\vec{0},\vec{b},\vec{a})$. The sum of the area weighted normal becomes

$$\begin{align} & \frac12( \color{red}{\vec{a}\times\vec{b}} + \color{green}{\vec{b}\times\vec{c}} + \color{blue}{\vec{c} \times \vec{a}}) + \frac12 \color{green}{\vec{c}\times\vec{b}} + \frac12 \color{blue}{\vec{a}\times\vec{c}} + \frac12 \color{red}{\vec{b}\times\vec{a}} \\ = & \frac12(\color{red}{\vec{a}\times\vec{b}} + \color{red}{\vec{b}\times\vec{a}}) + \frac12(\color{green}{\vec{b}\times\vec{c}} + \color{green}{\vec{c}\times\vec{b}}) + \frac12(\color{blue}{\vec{c} \times \vec{a}} + \color{blue}{\vec{a} \times \vec{c}})\\ = & \color{red}{\vec{0}} + \color{green}{\vec{0}} + \color{blue}{\vec{0}}\\ = & \vec{0}\end{align}$$

This statement can be further generalized to any region $\Omega$ bounded by a regular enough surface $\partial \Omega$. We have

$$\int_{\partial \Omega} d\vec{S} = \vec{0}\tag{*1}$$

To show this, it just suffice to notice for any constant vector $\vec{k}$, divergence theorem tell us

$$\vec{k} \cdot \left( \int_{\partial \Omega} d\vec{S}\right) = \int_{\partial \Omega} \vec{k}\cdot d\vec{S} = \int_{\Omega} (\vec{\nabla}\cdot \vec{k}) dV = \int_{\Omega} 0 dV = 0$$ Since this is true for all $\vec{k}$, $(*1)$ follows.

Answer 3

If our tetrahedron has vertices at the origin and at point $a,b,c,$ with $a,b,c$ arranged counter-clockwise when looking down on the vertex at the $O.$

$a\times b$ give the vector normal the the face with vertices $a,b,O$ and has magnitude proportional to the area.

$b\times c$ and $c\times a$ are found similarly

the vector normal to the fourth face... $(c-a)\times (b-a)$

Now you need to show that $a\times b + b\times c + c\times a + (c-a)\times (b-a) = 0$