RE: [Algorithms] decompose onto non-orthogonal vectors

[Tom F. <to...@mu...>]

Throw one of the dimensions away (any one) - then you have only two
equations. NOW can you solve it? :-)

The third dimension will just give the same answer if p lies in the plane of
a and b. If it isn't, then it's not a well-formed question, and you'll need
another axis (c) to define a suitable basis. Then you have three dimensions
and three unknowns, which you solve as you stated.

Tom Forsyth - Muckyfoot bloke.
Whizzing and pasting and pooting through the day.

> -----Original Message-----
> From: Klaus Hartmann [mailto:k_h...@os...]
> Sent: 15 July 2000 15:01
> To: gda...@li...
> Subject: Re: [Algorithms] decompose onto non-orthogonal vectors
> 
> 
> 
> ----- Original Message -----
> From: Scott Justin Shumaker <sjs...@um...>
> To: <gda...@li...>
> Sent: Saturday, July 15, 2000 9:15 AM
> Subject: Re: [Algorithms] decompose onto non-orthogonal vectors
> 
> 
> > Okay, let me try and explain this as simply as possible.  
> You want to find
> > scalars u,v such that u*a + v*b = p.  Since these are 3-dimensional
> > vectors, you have 3 equations and 2 unknowns:
> >
> > u*a1 + v*b1 = p1
> > u*a2 + v*b2 = p2
> > u*a3 + v*b3 = p3
> >
> > where p1,p2,p3 are the coordinates of the point p, likewise 
> for a1,a2,a3
> > and a, and b1,b2,b3 and b.
> >
> > How do you solve a system of 3 linear equations with 2 
> unknowns?  If you
> > don't know Gaussian elimination, this can be done with some simple
> > algebraic rules and substituion.
> 
> Okay, this may become very embarrassing for me now... :) How 
> can you solve a
> system of 3 linear equations in two unknows, u, v, if they 
> cannot be reduced
> to two linear equations (for example, a2 = a3, b2 = b3, and 
> p2 = p3)???
> 
> Let's assume that we change the 3 linear equations as follows:
> 
> u*a1 + v*b1 + w*c1 = p1
> u*a2 + v*b2 + w*c2 = p2
> u*a3 + v*b3 + w*c3 = p3
> 
> Then there's a unique solution, if
> 
>       | a1 b1 c1 |
> det | a2 b2 c2 | != 0
>       | a3 b3 c3 |
> 
> If we set c1 = c2 = c3 = 0, then we have the same system as 
> above (yours),
> but with 3 unknowns:
> 
> u*a1 + v*b1 + w*0 = p1
> u*a2 + v*b2 + w*0 = p2
> u*a3 + v*b3 + w*0 = p3
> 
> Then
>                | a1 b1 0 |
> Dn = det | a2 b2 0  | = 0
>                | a3 b3 0 |
> 
> Since Dn = 0, we get a division by zero, and thus there's an 
> infinite number
> of solutions.
> 
> ... or did I just miss something???
> 
> Niki
> 
> 
> 
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