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Finding the Equation of a Globe

Date: 6/29/96 at 0:43:49
From: jae berry
Subject: Finding the Equation of a Globe

I am Imola Lajos, an 18 years old Hungarian exchange student.  In 18 
days I have to take a math exam of two years material I never saw 
before. There are some problematic examples from 3D analytic geometry 
(I had a course already, so it is not totally new):

1) There is a globe that has 2 points given, P(-14/3/8) , Q(2/3/-16);
the center of the globe C is on the line g...
X = (3  -2  13) + t(-2  1  -5)    (this is the only way I could write 
down the vectors)  I know that the center is where lines g and s cross 
where s is the line of symmetry of PQ in the point of H(-6/3/-4), I 
know that the normal vector is (-2  0  -1) and I can put up an 
equation for line s, but than I am stuck, I can't figure out the 
equation of the globe.

2) 3 points of the globe are given:
P(1/5/9),  Q(4/14/-3) , r(13/2/0)  radius is also given  r=21
I know I have to use the equation (x-cx)2+(y-cy)2+(z-cz)2=r2  
(2=square), (c indicates center)...but finding the equation of the 
globe is a miracle again.

3) Radius r = square root of 27    The globe has a point A(X/0/0) with 
the plane e...x+y-z=4 together.  The question is once again the 
equation of the globe.  I could only come up with 2 equations for 3 
unknown, so it didn't help.

Thank you for your help and I am sorry if my English language in math 
is so poor, but my math book is written in German, that is my second 
language anyhow.  


Date: 8/16/96 at 14:16:40
From: Doctor James
Subject: Re: Finding the Equation of a Globe

Just to make it all easier to see, I will denote all vectors as
R=(x_1,x_2,x_3), where "x_1" is x sub one (or x-one).
Also, x^2 means x to the power of 2.

1. If you have the equation for the line s, this can be solved by
setting the two lines equal to each other. That is, if
s = (a_0,b_0,c_0) + t(a,b,c), then set up the equation
(a_0,b_0,c_0) + t(a,b,c) = (3,-2,13) + t(-2,1,-5), and solve
for t. Then the point (3,-2,13) + t(-2,1,-5) is the center. You then
need to find the radius, which you can do by finding the euclidean
distance between the center point and a point on the surface you are
given. if the center c = (x_0,y_0,z_0), the radius r is
( (x_0-2)^2 + (y_0-3)^2 + (z_0+16) )^(1/2). Then the equation for
the circle is (x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 = r^2

2. Again, let the center be c = (x_0,y_0,z_0). Then the center must
be the same distance from each point, so
(x_0-1)^2 + (y_0-5)^2 + (z_0-9)^2   =
(x_0-4)^2 + (y_0-14)^2 + (z_0+3)^2  =
(x_0-13)^2 + (y_0-2)^2 + (z_0-0)^2  = Radius^2
There you have three equations for three unknowns. To solve these, you 
can expand out each line. The x_0^2's should cancel, with the y_0^2's 
and z_0^2's. The equation for the circle is then
(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 = Radius^2

3. I'm not quite sure what you are asking here. Specifically, I don't 
know what the plane you stated is for. If the center lies on the 
plane, then you are looking for an (x_0,y_0,z_0) such that
x_0-y_0+z_0=4, and (X-x_0)^2 + (y_0)^2 + (z_0)^2 = 27. However, unless 
X=sqrt(27), you'll either end up with no possible spheres or an 
infinite number of spheres. (In fact, there will be a circle on the 
plane of possible centers of the sphere.) So either I'm completely 
wrong, I didn't understand the question, or the question is likely to 
be unsolvable.

-Doctor James,  The Math Forum
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Associated Topics:
College Higher-Dimensional Geometry

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