Solving the Equation of a Sphere with Diameter Endpoints

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Homework Help Overview

The problem involves finding the equation of a sphere given the endpoints of its diameter. The specific points provided are (-1,1,-3) and (3,-1,2).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the sphere's center and radius, with one participant attempting to derive the radius from the diameter endpoints. Questions arise regarding the correctness of the radius calculation and the interpretation of the distance formula.

Discussion Status

Some participants have confirmed the center of the sphere while expressing uncertainty about the radius. There is an ongoing exploration of the correct method to calculate the radius based on the given endpoints, with differing interpretations noted.

Contextual Notes

Participants are navigating the challenge of recalling mathematical concepts related to spheres and distances, with one expressing a lack of recent experience in school. The discussion reflects a mix of confidence and uncertainty regarding the calculations involved.

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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:

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