Cosine law for tetrahedron

During dinner tonight, Hon Leung, Ken and I were discussing some possible exercises for the bright secondary school students. I suggested generalizing the cosine law

\displaystyle c^2=a^2+b^2-2ab \cos C

for a triangle {\triangle ABC} to a tetrahedron or even an {n}-dimensional simplex. I think this would be a good exercise for them. Anyway, after discussing with them, we’ve come up with the following

Theorem 1 (Hon Leung-Ken’s theorem, 2011 (Cosine law for tetrahedron))
For a tetrahedron {ABCD\subset\mathbb{R}^3}, the area of the triangle {\triangle BCD} is given by

\displaystyle  \begin{array}{rcl}  \text{Area}(BCD)^2 &=& \text{Area}(ABC)^2 +\text{Area}(ABD)^2 +\text{Area}(ACD)^2\\ &&-2\text{Area}(ABC)\text{Area}(ABD)\cos b\\ &&-2\text{Area}(ABC)\text{Area}(ACD)\cos c\\ &&-2\text{Area}(ABD)\text{Area}(ACD)\cos d \end{array}

where {b} is the angle between {\triangle ABC} and {\triangle ABD}, {c} is the angle between {\triangle ACD} and {\triangle ABC}, and {d} is the angle between {\triangle ABD} and {\triangle ACD}.


Proof: We can w.l.o.g. assume that {A=0}, then regarding the points {B, C, D} as vectors in {\mathbb{R}^3}, we have

\displaystyle \text{Area}(BCD)= \frac{1}{2}|(B-C)\times (D-C)|.

Thus

\displaystyle  \begin{array}{rcl}  \text{Area}(BCD)^2&=&\frac{1}{4}|(B-C)\times (D-C)|^2\\ &=&\frac{1}{4}|B\times D+D\times C+C\times B|^2\\ &=&\frac{1}{4}(|B\times D|^2+|D\times C|^2+|C\times B|^2)\\ &&-\frac{1}{2}( \langle B\times D, C\times D\rangle+ \langle D\times B, C\times B\rangle+ \langle D\times C, B\times C\rangle )\\ &=&\text{Area}(ABD)^2 +\text{Area}(ACD)^2+\text{Area}(ABC)^2 \\ &&-\frac{1}{2}( \langle B\times D, C\times D\rangle+ \langle D\times B, C\times B\rangle+ \langle D\times C, B\times C\rangle ). \end{array}

But

\displaystyle \frac{1}{2}B\times D= \text{Area}(ABD) \vec n_{ABD}

where {\vec n_{ABD}} is the normal of {\triangle {ABD}}, and

\displaystyle \langle \vec n_{ABD}, \vec n_{ACD}\rangle =\cos d,

we thus have

\displaystyle  \begin{array}{rcl}  \text{Area}(BCD)^2 &=& \text{Area}(ABC)^2 +\text{Area}(ABD)^2 +\text{Area}(ACD)^2\\ &&-2\text{Area}(ABC)\text{Area}(ABD)\cos b\\ &&-2\text{Area}(ABC)\text{Area}(ACD)\cos c\\ &&-2\text{Area}(ABD)\text{Area}(ACD)\cos d. \end{array}

\Box

Remark 1 Obviously the theorem can be generalized to an {n}-dimensional simplex in {\mathbb{R}^n}, note also that the {(n-1)}-dimensional volume of the simplex spanned by {0, A_1, \cdots, A_{n-1}} is

\displaystyle \frac{1}{(n-1)!}|A_1\times \cdots \times A_{n-1}|.

(This also recovers the classical cosine law when {n=2}. )

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3 Responses to Cosine law for tetrahedron

  1. Anonymous says:

    Google dihedral angle and Feng Luo, you may find very interesting versions Cosine Laws for Euclidean, Hyperbolic, and Spherical tetrahedra.

  2. KKK says:

    Thank you very much for your information. And indeed I’ve attended several of his lectures in the past (but it’s long time ago already so I’ve forgotten much about their details). Nevertheless I remember he’s talked about these cosine laws on simplices one or two times, I found it very interesting. ☺

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