Vector weighting functions

Orthogonal but not Orthonormal Basis Sets

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

${\vec {\mathbf {u}}}\bullet {\vec {\mathbf {v}}}=\sum _{k=1}^{3}u_{k}{\vec {\mathbf {a}}}_{k}\bullet \sum _{m=1}^{3}v_{m}{\vec {\mathbf {a}}}_{m}=\sum _{k=1}^{3}u_{k}\sum _{m=1}^{3}v_{m}{\vec {\mathbf {a}}}_{k}\bullet {\vec {\mathbf {a}}}_{m}=\sum _{k=1}^{3}u_{k}\sum _{m=1}^{3}v_{m}\delta _{k,m}=\sum _{k=1}^{3}v_{k}u_{k}$

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

${\vec {\mathbf {a}}}_{k}\bullet {\vec {\mathbf {a}}}_{n}=w_{k}\delta _{k,n}$

where the $w_{k}$ is the square of the length of ${\vec {\mathbf {a}}}_{k}$ and the symbol $\delta _{k,n}$ is one when k = n and zero otherwise? Well the general inner product of ${\vec {\mathbf {u}}}$ and ${\vec {\mathbf {v}}}$ becomes

${\vec {\mathbf {u}}}\bullet {\vec {\mathbf {v}}}=\sum _{k=1}^{3}u_{k}{\vec {\mathbf {a}}}_{k}\bullet \sum _{m=1}^{3}v_{m}{\vec {\mathbf {a}}}_{m}=\sum _{k=1}^{3}u_{k}\sum _{m=1}^{3}v_{m}{\vec {\mathbf {a}}}_{k}\bullet {\vec {\mathbf {a}}}_{m}=\sum _{k=1}^{3}u_{k}\sum _{m=1}^{3}v_{m}w_{k}\delta _{k,m}=\sum _{k=1}^{3}w_{k}v_{k}u_{k}$.

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 $w_{z}=10^{-6}$.