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distance between pairs of parallel lines, select two from list of length four, symbolically
*To*: mathgroup at smc.vnet.net
*Subject*: [mg48106] distance between pairs of parallel lines, select two from list of length four, symbolically
*From*: Kai Gauer <gauerkk at uregina.ca>
*Date*: Thu, 13 May 2004 00:08:44 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
Hi fellow newsgroup folks!
(** apologies for the lengthy post; thanks in **)
(** advance for approaches at solutions **)
(** mathematica_version=3.0 **)
I can express a series of 4 linear_equations for plotting
after assignment of coefficients as:
{
2/161*(-2093 + 1/2*1*Sqrt[13*(-2 +
Sqrt[533])]*(-648 - 25*Sqrt[533] + (+577 +
24*Sqrt[533])*x - 23*(+22 + Sqrt[533])*y)) == 0,
2/161*(-2093 + 1/2*1*Sqrt[13*(-2 +
Sqrt[533])]*(+648 + 25*Sqrt[533] - (+577 +
24*Sqrt[533])*x + 23*(+22 + Sqrt[533])*y)) == 0,
2/161*(-2093 - 1/2*I*Sqrt[13*(+2 +
Sqrt[533])]*(+648 - 25*Sqrt[533] + (-577 +
24*Sqrt[533])*x - 23*(-22 + Sqrt[533])*y)) == 0,
2/161*(-2093 + 1/2*I*Sqrt[13*(+2 +
Sqrt[533])]*(+648 - 25*Sqrt[533] + (-577 +
24*Sqrt[533])*x - 23*(-22 + Sqrt[533])*y)) == 0
}. I can measure the distance between any two of these
lines by selecting an (x1,y1) on one line, and computing
the following:
mydistance:=Abs[u*x1+v*y1+w]/Abs[Sqrt[u^2+v^2]]
(** I would greatly prefer to later peel off the Abs[] from
the denominator term, but in numeral form, I have to: first
calculate the denominator, and then rationalize it, and
then cancel anything left from the new denominator with the
old numerator, having FullSimplify 'ied - I am also not
even sure that I need to attach the Abs[], after reading
some Linear Algebra texbooks **)
The question is: how to use a similar process to get a
rationalized denominator symbolic form of
mydistance[a,b,c,f,g,h] (I have to re-write the u, v, w in
terms of the variables above)?
The general form of linear_equations would start off looking like:
{
(4*h^2*(-(a*b) + h^2)*
(a*(-(b*c) + f^2) + b*g^2 + h*(-2*f*g + c*h)) +
Sqrt[2]*E^(I/2*Arg[(a*b*c - a*f^2
- b*g^2 + 2*f*g*h - c*h^2)*(-a + b + Sqrt[(a - b)^2 +
4*h^2])]*Sign[-(a*b*h) + h^3]^2)*
Sqrt[(-(a*b*h) + h^3)^2*Sqrt[(a*(-(b*c) + f^2)
+ b*g^2 + h*(-2*f*g + c*h))^2]]*(-(b*g) + 2*f*h
+ g*Sqrt[(a - b)^2 + 4*h^2] + a^2*x + 2*h^2*x + b*h*y
+ h*Sqrt[(a - b)^2 + 4*h^2]*y + a*(g - b*x
+ Sqrt[(a - b)^2 + 4*h^2]*x + h*y))*Sqrt[Abs[-a + b
+ Sqrt[(a - b)^2 + 4*h^2]]])/(4*(-(a*b*h) + h^3)^2) == 0,
(4*h^2*(-(a*b) + h^2)*
(a*(-(b*c) + f^2) + b*g^2 + h*(-2*f*g + c*h)) -
Sqrt[2]*E^(I/2*Arg[(a*b*c - a*f^2
- b*g^2 + 2*f*g*h - c*h^2)*(-a + b + Sqrt[(a - b)^2 +
4*h^2])]*Sign[-(a*b*h) + h^3]^2)*
Sqrt[(-(a*b*h) + h^3)^2*Sqrt[(a*(-(b*c) + f^2)
+ b*g^2 + h*(-2*f*g + c*h))^2]]*(-(b*g) + 2*f*h
+ g*Sqrt[(a - b)^2 + 4*h^2] + a^2*x + 2*h^2*x + b*h*y
+ h*Sqrt[(a - b)^2 + 4*h^2]*y + a*(g - b*x
+ Sqrt[(a - b)^2 + 4*h^2]*x + h*y))*Sqrt[Abs[-a + b
+ Sqrt[(a - b)^2 + 4*h^2]]])/(4*(-(a*b*h) + h^3)^2) == 0,
(4*h^2*(-(a*b) + h^2)*
(a*(-(b*c) + f^2) + b*g^2 + h*(-2*f*g + c*h)) +
Sqrt[2]*E^(I/2*Arg[(a*(-(b*c) + f^2)
+ b*g^2 + h*(-2*f*g + c*h))*(+a - b + Sqrt[(a - b)^2
+ 4*h^2])]*Sign[-(a*b*h) + h^3]^2)*
Sqrt[(-(a*b*h) + h^3)^2*Sqrt[(a*(-(b*c) + f^2)
+ b*g^2 + h*(-2*f*g + c*h))^2]]*(-(b*g) + 2*f*h
- g*Sqrt[(a - b)^2 + 4*h^2] + a^2*x + 2*h^2*x + b*h*y
- h*Sqrt[(a - b)^2 + 4*h^2]*y + a*(g - b*x
- Sqrt[(a - b)^2 + 4*h^2]*x + h*y))*Sqrt[Abs[+a - b
+ Sqrt[(a - b)^2 + 4*h^2]]])/(4*(-(a*b*h) + h^3)^2) == 0,
(4*h^2*(-(a*b) + h^2)*
(a*(-(b*c) + f^2) + b*g^2 + h*(-2*f*g + c*h)) +
Sqrt[2]*E^(I/2*Arg[(a*(-(b*c) + f^2)
+ b*g^2 + h*(-2*f*g + c*h))*(+a - b + Sqrt[(a - b)^2
+ 4*h^2])]*Sign[-(a*b*h) + h^3]^2)*
Sqrt[(-(a*b*h) + h^3)^2*Sqrt[(a*(-(b*c) + f^2)
+ b*g^2 + h*(-2*f*g + c*h))^2]]*(+(b*g) - 2*f*h
+ g*Sqrt[(a - b)^2 + 4*h^2] - a^2*x - 2*h^2*x - b*h*y
+ h*Sqrt[(a - b)^2 + 4*h^2]*y + a*(-g + b*x
+ Sqrt[(a - b)^2 + 4*h^2]*x - h*y))*Sqrt[Abs[+a - b
+ Sqrt[(a - b)^2 + 4*h^2]]])/(4*(-(a*b*h) + h^3)^2) == 0
}
But in this case, my only assignment of coefficients
[a,b,c,f,g,h] occur at the mydistance function. Since
mydistance can be any one (possibly more) of six operations
between the 2 lines in question, how do I choose the unique
positive real value of mydistance when the Abs is taken off
the denominator in the form similar to the above numerical
operation and assign ExpandNumerator.
Notice that ExpandNumerator[3+Sqrt[5]/Sqrt[7]] reduces a
rationalized denominator, but how to do this in a symbolic
form? It may also help to notice that when the two real
coefficient lines are returned, they appear to be parallel
to one another. I also need to leave the formula in a form
similar to this, and not looking like a TrigReduce[] type
of equation.
I'd also like to keep whatever symmetries which you see
here as stable as possible.
A normal approach is to sometimes subtract line1-line2
where line1: Ax+By+C==0, line2: Ax+By+D==0 to come up with
|D-C|, but I tried this approach, and in some cases,
wouldn't remove the x and y terms (this is "harder" to work
with because of the rank 1 returmn value using linear
algebra methods).
Kai G. Gauer
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