RE: [Algorithms] decompose onto non-orthogonal vectors

[Discoe, B. <ben...@in...>]

> From: ro...@do... [mailto:ro...@do...]
>
> Chapter 1 in any linear algebra book.

I assure i looked in chapters 1 though 18 of my book ("calculus and analytic
geomtry", thomas/finney) and nothing resemling a formula for decomposing a
vector is anywhere in it.  In general, math textbooks tend to be concerned
with axioms and definitions, not with formulas that actually prove to be
useful.

> That is A SYSTEM OF TWO LINEAR EQUATIONS IN TWO UNKNOWNS u, v,
> which you ought to have learned to solve in intermediate algebra,

I was asking if somebody knew the solution, not for how to do the math,
which i'm willing to leave to people who enjoy re-deriving solutions to math
problems.

> 	u = (p1 b2 - p2 b1)/(a1 b2 - a2 b1)
> 	v = (a1 p2 - a2 p1)/(a1 b2 - a2 b1)

OK, i'll believe that after solving for v and substituting twice (a page or
so of algebra) you get that 2d solution.

> If the denominator (a1 b2 - a2 b1) is zero then it means that a and b
> are linearly dependent (i.e. collinear) and either it has no solution
> (if p is not also collinear with a and b) or infinitely many different
> solutions (if p is collinear with a and b)

The picture i drew showed a nondegenerate case.

> Now suppose that you are given this as a 3D problem, say
> 	a = (a1,  a2,  a3) and b = (b1, b2, b3)

Yes.  I'm sorry i forgot to state that i do have 3d vectors, and p is known
to lie in the plane formed by a and b.

> It may or may not have solutions..
> Cramer's rule does not apply here, but Gaussian elimination
> still does.

Is this your way of saying you don't know the answer?

It would be pretty sad if not a single person on the list knows the answer.
I was kinda hoping *you* would, Ron :)

> No, I will not belabor the list with a review of Gaussian elimination
> algorithms.  You will find one in Chapter 1 of your favorite linear
> algebra book.

The index in my textbook does not list "Gaussian elimination".

Thanks,
Ben
http://vterrain.org/