Relation between surfaces of an infinitesimal tetrahedron

Question

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.

Answer 1

$\let\a=\alpha \let\s=\sigma$ 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?

Answer 2

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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.

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3D version

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

EDIT

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In above Figure-01 \begin{equation} \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} \end{equation} and \begin{equation} \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} \end{equation} so \begin{equation} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\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} \end{equation}

Answer 3

$d\sigma_1$ 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

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