anurag said:
Dear jk86,
I don't think so. Consider a non-orthogonal coordinate system like in which angle between any two axis in less than 90 degree. and two vector along x and y-axis [1 0 0] & [0 1 0], then the normal cross product is [0 0 1] which is along z-direction but for this coordinate system, z is not perpendicular to x and y axis. and you know cross product of two vector should be perpendicular to both vector.
Ah, OK I'm sorry I should have read your post more carefully. If you are calculating the dot product of \vec{a}\cdot\vec{b}, you can expand each in terms of its contravariant components. As an example, define a coordinate system (u,v,w) via the Cartesian coordinates (x,y,z) using some relations:
<br />
\begin{align}<br />
x &= u + v\\<br />
y &= u - v\\<br />
z &= 3uv + 2w<br />
\end{align}<br />
If the basis vectors for the non-orthogonal (u,v,w) coordinate system are \vec{e}_u,\vec{e}_v,\vec{e}_{w} (and they are \hat{e}_x,\hat{e}_y,\hat{e}_z for the Cartesian basis) then you can write a general vector \vec{r}=x\hat{e}_x + y\hat{e}_y+z\hat{e}_z=(u+v)\hat{e}_x + (u-v)\hat{e}_y + (3uv + 2w)\hat{e}_z. You can then find the non-orthogonal basis vectors by:
<br />
\begin{align}<br />
\vec{e}_u &= \frac{\partial \vec{r}}{\partial u} = \hat{e}_x + \hat{e}_y + 3v\hat{e}_z\\<br />
\vec{e}_v &= \frac{\partial \vec{r}}{\partial u} = \hat{e}_x - \hat{e}_y + 3u\hat{e}_z\\<br />
\vec{e}_w &= \frac{\partial \vec{r}}{\partial u} = 2\hat{e}_z<br />
\end{align}<br />
You can verify that the example is indeed non-orthogonal by computing dot products such as \vec{e}_u\cdot\vec{e}_w = (3v\hat{e}_z)\cdot(2\hat{e}_z)=6v. To compute more general dot products, and make all this simpler, you should first find the metric tensor:
<br />
g_{ij}\equiv \vec{e}_i\cdot\vec{e}_j = \begin{bmatrix}2+9v^2 & 9uv & 6v\\ 9uv & 2+9u^2 & 6u\\ 6v & 6u & 4\end{bmatrix}<br />
where i,j refer to u,v,w basis indices. Then for some vectors \vec{a} and \vec{b}, you get \vec{a}\cdot\vec{b}=(\sum_i a^i\vec{e}_i)\cdot (\sum_j b^j\vec{e}_j)=\sum_{ij}g_{ij}a^ib^j. You are then simply sticking a matrix g_{ij} in between the vectors---a matrix which is diagonal in an orthogonal coordinate system. As for the cross product you should be able to do something similar using the orthonormal basis definition [\vec{a}\times\vec{b}]_i = \epsilon_{ijk}\vec{a}^j\vec{b}^k. I think it just becomes [\vec{a}\times\vec{b}]_i = g^{ij}\epsilon_{jkl}a^{k}b^{l}, where \epsilon_{jkl} is the Levi-Civita symbol.
