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Question: Given a tetrahedral $\rho$ with faces $\Xi, \Pi, \Gamma, \Delta$ with areas $\xi , \pi, \gamma , \delta$, respectively, assign a normal vector to each face such that $\mid \mid \hat{\xi} \mid \mid = \xi, \mid \mid \hat{\pi} \mid \mid = \pi, \mid \mid \hat{\gamma} \mid \mid = \gamma, \mid \mid \hat{\delta} \mid \mid = \delta$
Prove that the sum of the vectors is the zero using vector cross product.

Edit Talk about over complicating a solution...
Proposed Solution
Let $P_1, P_2,P_3,$ and $P_4$ be vertices of a tetrahedron $\rho$, where $P_4$ is the top of the tetrahedron.
Permute $P_4$ with the other vertices to obtain three vectors: $\hat{a} , \hat{b}, \hat{c}$
To find the normal vector for each side, compute $$\hat{a} \times \hat{b}\, , \,\hat{b} \times \hat{c}\, , \,\hat{c} \times \hat{a}$$
Note that $\mid \mid \hat{u} \times \hat{v}\mid \mid $ gives the area of a parallelogram with sides $\hat{u}, \hat{v}$. Or two times the triangle formed...
So dividing each cross product by two yield the area of the corresponding side.
To find the final normal vector, notice that the plane of the fourth face can be represented by the triangle formed by $\hat{a}-\hat{b}$ and $ \hat{b} - \hat{c}$
$$(\hat{a}-\hat{b}) \times (\hat{b} - \hat{c}) = (\hat{b} \times \hat{a}) + (\hat{b} \times \hat{-b}) +(\hat{-c} \times \hat{a}) + (\hat{c} \times \hat{b}) $$ Since the cross product is anticommutative, and since $u \times -u$ = 0
Dividing this result by two and adding to the previous results clearly yields 0.

Previous post: Tetrahedral Law of Cosines Proof

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Anthony Peter
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  • Wait, why'd you duplicate this question? And ignore the advice you were given? – dfeuer Jul 25 '13 at 00:43
  • I posted this question before he gave the advice. I only just saw it now – Anthony Peter Jul 25 '13 at 00:47
  • You seem to be assuming a regular tetrahedron, but the problem does not—in a regular tetrahedron all the faces would have equal areas. So get that image out of your head and just think about connecting up any four points that don't lie in the same plane, no three of which lie on the same line. – dfeuer Jul 25 '13 at 02:13
  • You don't have to know the components of each vector, using $\mathbf{a}\times \mathbf{b} = -\mathbf{b}\times \mathbf{a}$ is enough. – Shuhao Cao Jul 25 '13 at 02:26

2 Answers2

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Given any triangle $T \subset \mathbb{R}^3$, pick an orientation for it's face. Let $\vec{x}_1(T), \vec{x}_2(T), \vec{x}_3(T)$ be $T$'s vertices ordered in counter-clockwise manner w.r.t the orientation. Let $\vec{\xi}(T)$ be a normal vector of $T$ in the direction of the orientation and $|\vec{\xi}(T)| = \text{Area}(T)$, then

$$\begin{align}\vec{\xi}(T) = & \frac12 ( \vec{x}_2(T) - \vec{x}_1(T) ) \times ( \vec{x}_3(T) - \vec{x}_1(T) )\\ = & \frac12 \left( \vec{x}_1(T) \times \vec{x}_2(T) + \vec{x}_2(T) \times \vec{x}_3(T) +\vec{x}_3(T)\times \vec{x}_1(T)\right) \end{align}$$ The key is $\vec{\xi}(T)$ can be broken down into a sum along its edges.

For any polytope $P$ in $\mathbb{R}^3$, if we subdivide its surface into a bunch of triangles $T_i$, construct the normal $\vec{\xi}(T_i)$ and sum over them. The total sum can be rewritten as a sum over the edges of $T_i$. Each edge of $P$ will appear twice in equal but opposite in sign. As a result,

$$\sum_{T_i \text{ faces of } P} \vec{\xi}(T_i) = \vec{0}$$

Let us use the tetrahedron as an example. Let $\vec{x}_1, \vec{x}_2, \vec{x}_3, \vec{x}_4$ be the vertices of a tetrahedron. Let $T_i$ be the face of the tetrahedron opposite to the vertex $\vec{x}_i$. Assuming the vertices has been labeled as shown in graph at end (the face $T_3$ is in the front) and the $\vec{\xi}(T_i)$ are outward normals, we have:

$$ \begin{cases} 2 \vec{\xi}(T_1) &= \color{red}{\vec{x}_2 \times \vec{x}_3} + \color{orange}{\vec{x}_3 \times \vec{x}_4} +\color{green}{\vec{x}_4\times \vec{x}_2}\\ 2 \vec{\xi}(T_2) &= \color{blue}{\vec{x}_1 \times \vec{x}_4} + \color{orange}{\vec{x}_4 \times \vec{x}_3} +\color{magenta}{\vec{x}_3\times \vec{x}_1}\\ 2 \vec{\xi}(T_3) &= \vec{x}_1 \times \vec{x}_2 + \color{green}{\vec{x}_2 \times \vec{x}_4} +\color{blue}{\vec{x}_4\times \vec{x}_1}\\ 2 \vec{\xi}(T_4) &= \color{magenta}{\vec{x}_1 \times \vec{x}_3} + \color{red}{\vec{x}_3 \times \vec{x}_2} +\vec{x}_2\times \vec{x}_1 \end{cases} \implies \vec{\xi}(T_1) + \vec{\xi}(T_2) + \vec{\xi}(T_3) + \vec{\xi}(T_4) = \vec{0} $$

a tetrahedron

achille hui
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Yowzaz. You don't need any of that insane manipulation. Assume WLOG that one vertex is at the origin. Represent the other three points as vectors. Write those normal vectors you want as cross products. Use the distributive and anticommutative laws.

Edit: A couple facts about cross products that are important for understanding my answer:

For any vectors $a,b\in\Bbb R^3$:

  1. $\Vert a\times b \Vert = \Vert a \Vert \Vert b \Vert |\sin \theta|$, where $\theta$ is the angle between $a$ and $b$. So by the formula for the area of a triangle given two sides and the angle between them, the triangle with vertices $0$, $a$, and $b$ has area $\frac12 \Vert a \times b \Vert$.

  2. $a\times b$ is normal to both $a$ and $b$, so normal to the plane through any point that they determine.

To riff on point 1 a bit more:

You may know that determinants show how volumes transform. Let $a,b$ be non-parallel vectors and let $$n=\frac {a\times b}{\Vert a\times b\Vert}$$ be a unit vector normal to both. Let $M$ be the matrix such that \begin{gather*} M\mathbf i = a \\ M\mathbf j = b \\ M\mathbf k = n. \end{gather*} Then the area of the triangle determined by $a,b$ is the volume of the solid determined by $a,b,n$, which is $\frac 12 \det M$.

dfeuer
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