Relation between surfaces of an infinitesimal tetrahedron

Let $$d\sigma_1,d\sigma_2, d\sigma_3$$ denote the areas of the faces perpendicular to the axes $$x_1,x_2,x_3$$ and let $$d\sigma_n$$ denote the area of the inclined face with unit exterior normal n. My book says that this relation holds:

$$d\sigma_i = d\sigma_n \cos(\mathbf{n},x_i)= n_id\sigma_n \quad for (i=1,2,3)$$

In the limit as the tetrahedron shrinks to the point M. I don't understand how it is derived.

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$$\let\a=\alpha \let\s=\sigma$$ You rightly tagged "geometry" your question. And no limit or infinitesimals are needed as far as the surface is plane. Consider $$\s$$ (finite, not infinitesimal) and $$\s_1$$. They are triangles sharing a base. Can you see what's the ratio of their heights?

• What Do you mean by $\sigma$, $\sigma_n$? I see that all the surfaces $\sigma_i$ are triangles sharing one edge not always a base – matt fick Feb 9 at 10:30
• @matt.fick $\sigma$ is the oblique triangle, $\sigma_1$ is the one on $(x_2,x_3)$ plane etc. These share a side. In a triangle every side may be taken as base to compute area. Given the base, the height follows. – Elio Fabri Feb 9 at 11:00

$$d\sigma_1$$ is the orthogonal projection of $$d\sigma$$ onto $$x_2x_3$$, so that the coefficient of proportionality must be the cosine of the angle between the normals. The latter is the dot product

$$(n_n,n_y,n_z)\cdot(1,0,0).$$

I think that working on a Figure like above you could derive the results you want. And indeed, it's valid for finite quantities and not necessarily infinitesimal.

3D version

$$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$$

EDIT

In above Figure-01 $$$$\mathbf{n}\boldsymbol{=}\left(\rm n_1,n_2,n_3\right)\boldsymbol{=}\left(\cos \theta_1,\cos \theta_2,\cos \theta_3\right) \tag{01}\label{01}$$$$ and $$$$\sigma_{\rm n}\boldsymbol{\equiv} \left[\rm A_1A_2A_3\right]\boldsymbol{=} \left. \begin{cases} \frac12 \left(\rm A_1A_2\right)\left(\rm A_3B_3\right)\\ \frac12 \left(\rm A_2A_3\right)\left(\rm A_1B_1\right)\\ \frac12 \left(\rm A_3A_1\right)\left(\rm A_2B_2\right) \end{cases} \right\} \tag{02}\label{02}$$$$ so $$$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\left. \begin{cases} \sigma_{1}\boldsymbol{\equiv} \left[\rm OA_2A_3\right]\boldsymbol{=}\frac12 \left(\rm A_2A_3\right)\left(\rm OB_1\right)\stackrel{\left(\rm OB_1\right)\boldsymbol{=}\left(\rm A_1B_1\right)\cos\theta_1}{\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!}}\frac12 \left(\rm A_2A_3\right)\left(\rm A_1B_1\right)\cos\theta_1\boldsymbol{=}\rm n_1\sigma_{\rm n}\\ \sigma_{2}\boldsymbol{\equiv} \left[\rm OA_3A_1\right]\boldsymbol{=}\frac12 \left(\rm A_3A_1\right)\left(\rm OB_2\right)\stackrel{\left(\rm OB_2\right)\boldsymbol{=}\left(\rm A_2B_2\right)\cos\theta_2}{\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!}}\frac12 \left(\rm A_3A_1\right)\left(\rm A_2B_2\right)\cos\theta_2\boldsymbol{=}\rm n_2\sigma_{\rm n}\\ \sigma_{3}\boldsymbol{\equiv} \left[\rm OA_1A_2\right]\boldsymbol{=}\frac12 \left(\rm A_1A_2\right)\left(\rm OB_3\right)\stackrel{\left(\rm OB_3\right)\boldsymbol{=}\left(\rm A_3B_3\right)\cos\theta_3}{\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!}}\frac12 \left(\rm A_1A_2\right)\left(\rm A_3B_3\right)\cos\theta_3\boldsymbol{=}\rm n_3\sigma_{\rm n} \end{cases} \right\} \tag{03}\label{03}$$$$

• since the area of a triangle is base $\cdot$ height I think that $d\sigma_i = d\sigma_n \cos(\mathbf{n},x_i)= n_id\sigma_n = height \cdot base$ but I've difficulties in figuring it out – matt fick Feb 10 at 11:53