Deriving volume of parallelepiped as a function of edge lengths and angles between the edges

Question

In Wikipedia it is stated that the volume of the parallelepiped given its edge lengths $a,b,c$, and the internal angles between the edges $\alpha ,\beta ,\gamma $ is:

$V=abc\sqrt{1+2\cos \alpha \cos \beta \cos \gamma -\cos ^{2}\alpha - \cos^{2}\beta - \cos ^{2}\gamma }\qquad$(*).

I was not able to derive it by using the determinant formula and expressing $\cos \alpha ,\cos \beta ,\cos \gamma $ in terms of $a,b,c$ and $\alpha ,\beta ,\gamma $. For instance

$a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}=ab\cos \alpha $.

Question: Could you give a hint on how can the formula (*) be proved?

Answer 1

The volume satisfies $V=|\det D|$ where $D$ is the matrix $$\pmatrix{a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&b_2&c_3}$$ where $\mathbf{a}=(a_1,a_2,a_3)$ is the vector corresponding to $a$ etc. Thus $$V^2=\det(DD^t)=\det \pmatrix{\mathbf{a}\cdot\mathbf{a} &\mathbf{a}\cdot\mathbf{b} &\mathbf{a}\cdot\mathbf{c}\\ \mathbf{b}\cdot\mathbf{a} &\mathbf{b}\cdot\mathbf{b} &\mathbf{b}\cdot\mathbf{c}\\ \mathbf{c}\cdot\mathbf{a} &\mathbf{c}\cdot\mathbf{b} &\mathbf{c}\cdot\mathbf{c}}.$$

Now you can express $V^2$ in terms of $a$, $b$, $c$ and the various cosines.

Answer 2

Please see this link, under trigonometric approach: http://www.ehow.com/how_6545076_prove-formula-volume-cuboids.html