Scale vector to zero First time poster, so I'll do my best to keep my question clear.
I have two vectors: 
$v_1 = (100, 4, 25)$
$v_2 = (0, 2, 3)$
I want to calculate the distance between $v_1$ and $v_2$.
Here's the catch: All $v_2$ dimensions are on a [0, 3] scale, while $v_1$'s dimensions are on arbitrary scales.
How do I scale $v_2$'s dimensions so that the minimum value of a $v_2$ dimension corresponds to 0, and the maximum value of a $v_2$ dimension corresponds to 3?
My objective is to make the distance calculation meaningful based on the relative value of the $v_1$ dimensions.
 A: COMMENT: (This is a comment, not an answer; I'm writing it here because it's more convenient.)
It's very unclear what you want. Can you clarify? Here are some questions/observations on which it would be helpful for you to elaborate.
(1) It sounds like you have several samples of 12-tuples, say $N$ samples
$$S^{(1)}=(s^{(1)}_1,s^{(1)}_2,s^{(1)}_3,\cdots,s^{(1)}_{12})$$
$$S^{(2)}=(s^{(2)}_1,s^{(2)}_2,s^{(2)}_3,\cdots,s^{(2)}_{12})$$
$$\cdots$$
$$S^{(n)}=(s^{(n)}_1,s^{(n)}_2,s^{(n)}_3,\cdots,s^{(n)}_{12})$$
where the superscripts indicate the sample ID and the subscripts distinguish among the 12 components (coordinates) of the tuples. Is this accurate?
(2) Is it true that the possible range of values for each slot may be different? For example, maybe the first coordinate values for all the samples, $s^{(1)}_1, s^{(2)}_1, s^{(3)}_1,\cdots, s^{(12)}_1$, are all in the range $[2,5]$, while (say) the second coordinate values for all the samples, $s^{(1)}_2, s^{(2)}_2, s^{(3)}_2,\cdots, s^{(12)}_2$, are all in the range $[0,1]$.
(3) If the answer to (2) is "yes", then do you want to scale all of the coordinates so that all of the scaled coordinates take value in the range $[0,3]$?
MORE: If these are correct, then suppose the $k$th coordinate has a range of values $[a_k,b_k]$. You seek a linear map for that coordinate $f_k:[a_k,b_k] \to[c,d]$ (here $c=0$ and $d=3$, so the target range is $[0,3]$; I present a more general map, so you can use any $c$ and $d$ you wish). Such a map is just
$$f_k(x) = c + \left(\frac{d-c}{b_k-a_k}\right)(x-a_k)$$
so in the case $c=0$ and $d=3$ this would be
$$f_k(x) = \left(\frac{3}{b_k-a_k}\right)(x-a_k)$$
The transformed sample 12-tuples would be
$$\hat S^{(n)}=(f_1(s^{(n)}_1),f_2(s^{(n)}_2),f_3(s^{(n)}_3),\cdots,f_{12}(s^{(n)}_{12}))$$
But it's not at all clear if this is what you want.
EVEN MORE: To see why the linear map $$f(x) = c + \left(\frac{d-c}{b-a}\right)(x-a)$$ maps the interval $[a,b]$ onto the interval $[c,d]$, note that it is the composition of the following maps:
$x\mapsto x-a$ shifts the interval $[a,b]$ to the origin, so the image is $[0,b-a]$.
$x\mapsto \left(\frac{d-c}{b-a}\right)x$ scales the interval $[0,b-a]$ onto the interval $[0,d-c]$.
$x\mapsto x + c$ shifts the interval $[0,d-c]$ onto the interval $[c,d]$.
The composition of these maps gives the desired map $f$, which maps the interval $[a,b]$ linearly onto the interval $[c,d]$.
