Calculate the volume of the parallelepiped I've got three vectors, u, v and w. The vectors have length of 1, 2 respectively 3, and the angles between the vectors u,v are $\pi/4$, u,w are $\pi/3$ and v,w are $\pi/6$. I've to calculate the volume of the parallelepiped spanned by these vectors.
If we denote the matrix A with column vectors u, v and w, we can calculate $AA^T$ to get the following:
$AA^T = |u|^2 + |v|^2 + |w|^2 + 2(u\cdot v + v\cdot w + u\cdot w)$
The determinant can then be calculated as:
$ det A = \sqrt{|u|^2 + |v|^2 + |w|^2 + 2(u\cdot v + v\cdot w + u\cdot w})$
I want to check if my calculations are right. I might be wrong, because when I then crunch the numbers in, I don't get the result I want.
Thank you.
 A: HINT:
The scalar triple product of three vectors u, v, and w
|u$\times$ v$\cdot$ w| expresses the volume  V of parallelepiped also, being equal to:
$$
\begin{vmatrix}
u_1 & v_1 & w_1 \\
u_2 & v_2 & w_2 \\
u_3 & v_3 & w_3 \\
\end{vmatrix} \\
$$
A: The formula is not correct. The correct way of doing it would be
$$
V^2 = (\det(A))^2 = \det(A^T)\det(A) =\det(A^TA) =\begin{vmatrix}
u\cdot u & u\cdot v & u \cdot w \\
v\cdot u & v\cdot v & v \cdot w \\
w\cdot u & w\cdot v & w \cdot w
\end{vmatrix} \\
= (u\cdot u)(v\cdot v)(w\cdot w) + 2(u\cdot v)(u\cdot w)(v\cdot w) \\
- (u\cdot u)(v\cdot w)^2 - (v\cdot v)(u\cdot w)^2 - (w\cdot w)(u\cdot v)^2 \\
=\|u\|^2\|v\|^2\|w\|^2 \left(1 + 2\cos\alpha\cos\beta\cos\gamma 
-\cos^2\alpha -\cos^2\beta -\cos^2\gamma \right)
$$
Taking the square root on both sides:
$$
V= \|u\|\;\|v\|\;\|w\|\;\sqrt{1 + 2\cos\alpha\cos\beta\cos\gamma 
-\cos^2\alpha -\cos^2\beta -\cos^2\gamma}
$$
Note that you can easily find out that your formula is not correct: When you simultaneously change the length of all three vectors by a factor of $a,$ the the volume of the parallelepiped must change by a factor of $a^3$. This does not happen in your formula. In your formula, it would only change by a factor of $a.$
