relation between planes of a infinitesimal tetrahedron












1












$begingroup$


enter image description hereIn relation to the tetrahedron depicted, the book I'm reading says that this relation between its surfaces holds:



$dsigma_i = dsigma_n cos(mathbf{n}, x_i) = n_i dsigma_m$



I don't understand how to derive it. If n is the unit exterior normal to the surface, (as the book defines it) then $n_i$ should be equal to 1, leading to $dsigma_i = 1cdot dsigma_m$






geometry trigonometry







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  • $begingroup$
    I think $n_i$ is the component of $mathbf{n}$ along axis $x_i$.
    $endgroup$
    – Aretino
    Feb 3 at 10:56


















1












$begingroup$


enter image description hereIn relation to the tetrahedron depicted, the book I'm reading says that this relation between its surfaces holds:



$dsigma_i = dsigma_n cos(mathbf{n}, x_i) = n_i dsigma_m$



I don't understand how to derive it. If n is the unit exterior normal to the surface, (as the book defines it) then $n_i$ should be equal to 1, leading to $dsigma_i = 1cdot dsigma_m$






geometry trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think $n_i$ is the component of $mathbf{n}$ along axis $x_i$.
    $endgroup$
    – Aretino
    Feb 3 at 10:56
















1












1








1





$begingroup$


enter image description hereIn relation to the tetrahedron depicted, the book I'm reading says that this relation between its surfaces holds:



$dsigma_i = dsigma_n cos(mathbf{n}, x_i) = n_i dsigma_m$



I don't understand how to derive it. If n is the unit exterior normal to the surface, (as the book defines it) then $n_i$ should be equal to 1, leading to $dsigma_i = 1cdot dsigma_m$






geometry trigonometry







share|cite|improve this question











$endgroup$




enter image description hereIn relation to the tetrahedron depicted, the book I'm reading says that this relation between its surfaces holds:



$dsigma_i = dsigma_n cos(mathbf{n}, x_i) = n_i dsigma_m$



I don't understand how to derive it. If n is the unit exterior normal to the surface, (as the book defines it) then $n_i$ should be equal to 1, leading to $dsigma_i = 1cdot dsigma_m$






geometry trigonometry




geometry trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 3 at 9:20







Giuliano Malatesta

















asked Feb 3 at 9:12









Giuliano MalatestaGiuliano Malatesta

347




347












  • $begingroup$
    I think $n_i$ is the component of $mathbf{n}$ along axis $x_i$.
    $endgroup$
    – Aretino
    Feb 3 at 10:56




















  • $begingroup$
    I think $n_i$ is the component of $mathbf{n}$ along axis $x_i$.
    $endgroup$
    – Aretino
    Feb 3 at 10:56


















$begingroup$
I think $n_i$ is the component of $mathbf{n}$ along axis $x_i$.
$endgroup$
– Aretino
Feb 3 at 10:56






$begingroup$
I think $n_i$ is the component of $mathbf{n}$ along axis $x_i$.
$endgroup$
– Aretino
Feb 3 at 10:56












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