Solving the Equation of a Sphere with Diameter Endpoints

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Homework Statement



Write the equation of the sphere whose diameter has endpoints at (-1,1,-3) and (3,-1,2).



Homework Equations





The Attempt at a Solution



(x-x1)2+ (y-y1)2+ (z-z1)2= r2

the center of the sphere is : (1,0,-1/2)
r= \sqrt{}5/4

then the equation is: (x-1)2+(y-0)2+(z+1/2)2=5/4

Please someone check my answer and if I am not right, help me to solve it.
 
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mamma_mia66 said:
the center of the sphere is : (1,0,-1/2)
r= \sqrt{}5/4

Hi mamma_mia66! :smile:

Centre is right. :biggrin:

But r is wrong. :cry:

(you've calculated the distance from the centre to the origin! :wink:)
 
I am not sure that I know how to find the r ? too old to remember it after so many years not being in school.
 
Yes you do! :biggrin:

they've given you a diameter:wink:
 
r= sq. rt. (3-1)2+ (1-0)2+(2-1/2)2= 21/4
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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