Thursday, March 3, 2011

3D to 2D vector?

I was just curious to see some mathematical approach to justify the transformation of a 3D vector to 2D vector in the following way:
x' = y-x
y' = z-x ... ... (1)

This is quite simple and obvious, that the mapping can be performed (as shown below). Now the question is, why do we do that? As the exact answer is unknown to me, my answer is, it will be easier to handle linear systems of certain type, i.e. involving 3D vectors, as we can change them to systems of 2D vectors easily (from a programming perspective).

Here, I will denote [x, y, z] format as a 3D vector with components x, y, z and [x', y'] as a 2D vector with components x', y' where, [x, y, z] and [x', y'] satisfies the transformation showed in (1).

Lets assume we have a linear system consisting of 2 such 3D vectors:
α[x1, y1, z1] + β[x2, y2, z2] = [γ1, γ2, γ3] ... (2)

Now if we apply the transformation from (1), we get a system of 2D vectors as follows:
α[y1-x1, z1-x1] + β[y2-x2, z2-x2] = [γ21, γ31] ... (3)

Now lets see if we can show (2) and (3) to be equivalent:

From (2), we get 3 equations:
αx1 + βx2 = γ1 ... (I)
αy1 + βy2 = γ2 ... (II)
αz1 + βz2 = γ3 ... (III)

Similarly from (3) we get 2 equations:
α(y1-x1) + β(y2-x2) = γ21
α(z1-x1) + β(z2-x2) = γ31

It is quite obvious to see that, we can get the later 2 equations from the first set, by subtracting (I) from both (II) and (III).

Similarly, this can be extended to a more general format:
α1[x1, y1, z1] + α2[x2, y2, z2] + ... ... + αn[xn, yn, zn] = [γ1, γ2, γ3]

Thus, we can conclude, (2) and (3) represents the same system. And, following this way, we can always reduce the dimensions of vectors involved in a linear system.

Additionally, if γ1 = γ2 = γ3 = γ then clearly, this is more simple, as in 2D, the vector in right side will be just [0, 0].

Actually, one of my desperate friend hit me today with this, as, if we can't get the answer, he will suicide... So, this is what I told him. Unfortunately, I am not from a mathematics background, and therefore, I may have abused some mathematical terms, sorry for that, and if anyone can tell me something more details and more general about it (what should I call this? I don't know either), or, if you find this post is totally wrong, please, save my friend from committing suicide by letting me know :p

Thank you!

1 comment:

  1. dost...two words for another method: linear operators :D