# Vector weighting functions

### Orthogonal but not orthonormal basis sets

Suppose we have two vectors from an orthonormal system, $\vec \bold u$ and $\vec \bold v$. Taking the inner product of these vectors, we get

$\vec \bold u \bullet \vec \bold v = \sum_{k=1}^3 u_k \vec \bold a_k \bullet \sum_{m=1}^3 v_m \vec \bold a_m = \sum_{k=1}^3 u_k \sum_{m=1}^3 v_m \vec \bold a_k \bullet \vec \bold a_m = \sum_{k=1}^3 u_k \sum_{m=1}^3 v_m \delta_{k,m} = \sum_{k=1}^3 v_k u_k$

What if they aren't from a normalized system, so that

$\vec \bold a_k \bullet \vec \bold a_n = w_k \delta_{k,n}$

where the $w_k$ is the square of the length of $\vec \bold a_k$ and the symbol $\delta_{k,n}$ is one when k = n and zero otherwise? Well the general inner product of $\vec \bold u$ and $\vec \bold v$ becomes

$\vec \bold u \bullet \vec \bold v = \sum_{k=1}^3 u_k \vec \bold a_k \bullet \sum_{m=1}^3 v_m \vec \bold a_m = \sum_{k=1}^3 u_k \sum_{m=1}^3 v_m \vec \bold a_k \bullet \vec \bold a_m = \sum_{k=1}^3 u_k \sum_{m=1}^3 v_m w_k\delta_{k,m} = \sum_{k=1}^3 w_k v_k u_k$.

You can interpret the $w_k$ as a weighting factor between the different directions so that different directions all end up in the units you would like. For example, suppose that the x and y directions were measured in meters, and the z direction was measured in centimeters, and you would like to use meters as your base unit. You could either convert the z dimensions to meters (probably simpler) or use a weighting function $w_x = 1$, $w_y = 1$ and $w_z = 10^{-6}$. In this sense, the system could be considered orthonormal with these units and this weighting arrangement.

Principle author of this page: Rob Frohne