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.
Best Answer
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: $$ \begin{vmatrix} u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u_3 & v_3 & w_3 \\ \end{vmatrix} \\ $$