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Path: utzoo!linus!decvax!dartvax!chuck
From: chuck@dartvax.UUCP (Chuck Simmons)
Newsgroups: net.math
Subject: Re: Some Math problems
Message-ID: <2503@dartvax.UUCP>
Date: Tue, 23-Oct-84 03:42:09 EDT
Article-I.D.: dartvax.2503
Posted: Tue Oct 23 03:42:09 1984
Date-Received: Wed, 24-Oct-84 07:53:05 EDT
References: <689@ihuxt.UUCP>
Organization: Dartmouth College, Hanover, NH
Lines: 91

<Hi!  I'm Judy.  Fly me to Hawaii!>

>1.Given 4 points on a plane, (3,0,0), (0,0,0), (3,0,5) and (0,0,5),
>  (i.e. a plane on the xz plane), how do you determine the equation
>  of that plane???

I'm not sure I understand this question.  Let me assume that you have
been given a number of points (purportedly all lying in the same plane)
and you want to discover an equation for this plane.  

First, since a plane is determined by 3 points, choose 3 points from
the set of points that you have been given which are not colinear.
(If all triplets of points are colinear, then all your points lie
on the same line and you won't be able to find a unique equation for
the plane.)

[3 points are colinear if a linear combination of them adds up to zero.
Let our 3 selected points be P1=(x1,y1,z1), P2=(x2,y2,z2), and P3=(x3,y3,z3).
If there exist some a,b,c in R such that aP1+bP2+cP3=0, then the points
are colinear.  So if we can solve the system of equations:
    ax1+bx2+cx3 = 0
    ay1+by2+cy3 = 0
    az1+bz2+cz3 = 0
then the points are colinear.]

The equation of a plane is given by ax+by+cz+d=0. 
So we need to solve the system of equations:
    ax1+by1+cz1+d=0
    ax2+by2+cz2+d=0
    ax3+by3+cz3+d=0
The astute reader will notice there will usually be more than one solution.
Presumably these solutions will be scalar multiples of each other.

Example:  Given points (3,0,0), (0,0,0), (3,0,5) and (0,0,5), determine
the equation of the plane on which they lie.

Choose (3,0,0), (0,0,0), and (3,0,5) as 3 points that determine the
plane.  Then, we solve:
    3a+0b+0c+d=0
    0a+0b+0c+d=0
    3a+0b+5c+d=0

This gives a=c=d=0.  Choose b=1 and the desired equation pops 
out as y=0.


>2. Given a matrix A where
>
>		.68	-.6928	.24	0
>		.6928	 .5	-.52	0
>	A = 	.24	 .52	 .82	0
>		 0	  0	  0	1
>
>find a transformation B such that BB=A.

This one seems a lot harder.  Does anyone know if there is a general
method of taking the square-root of a square matrix?  One approach that
I'm investigating uses the Newton-Raphson method.  This method appears
to have promise.  I used it on a 2x2 matrix and came up with an answer.
Our Honeywell went to bed for the night before I could try my algorithm
out on the above matrix.

The Newton-Raphson method is used to take the square root of a real
number.  Suppose we want to know the square root of X.  We compute
successive approximations to the square root.  These approximations
converge rather quickly to the square root.  The following pl1 routine
implements the Newton-Raphson method.

sqrt: procedure (x, epsilon) returns (float);
declare x       float, /* number to find root of */
        epsilon float, /* how close we need to get */
        guess   float; /* our approximation */

guess = 1;                                /* initial approximation */
do while (abs (x-guess*guess) > epsilon); /* loop til we get close */
guess = (guess + x/guess) / 2;            /* get a new guess */
end;

return  (guess);
end sqrt;

I am attempting to approximate the square root of a matrix using this
algorithm but using a square matrix instead of a floating point number.
I don't really know that my sequence of approximations will converge for
matrices.  Also, there is the possibility that you may run into an
approximation which is non-invertible.

Anyway...  I hope this hasn't been too tedious.

-- Chuck
dartvax!chuck