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. Thanks, Imola
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 Check out our web site! http://mathforum.org/dr.math/
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