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Mathematics 22 Online
OpenStudy (anonymous):

Determine an equation of a sphere given that one of the diameters has the extremities: (2,1,4) and (4,3,10)

OpenStudy (theeric):

Hi! Do you have the equation for a sphere? Just a general sphere?

OpenStudy (theeric):

Also, what kind of coordinate system are you using?

OpenStudy (theeric):

This problem might be beyond me.

OpenStudy (whpalmer4):

The distance formula can be extended to a three-dimensional coordinate system: \[d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\] You can use that to find the diameter, and then the radius. Formula for a sphere is \[(x-h)^2+(y-k)^2+(z-l)^2=r^2\]

OpenStudy (whpalmer4):

you know that the center of the sphere must be at the midpoint of that diameter, too...

OpenStudy (theeric):

\(h\), \(k\), and \(l\) are the center of the sphere, right?

OpenStudy (whpalmer4):

exactly!

OpenStudy (theeric):

\((h,\ k,\ l)\) is the coordinate. Good luck, @liltish007 !

OpenStudy (theeric):

Well, I should probably say \(center\ point\) rather than \(coordinate\)!

OpenStudy (whpalmer4):

and the midpoint of a line is located at the mean of the coordinates of the endpoints.

OpenStudy (whpalmer4):

Actually a pretty easy problem, so long as you know how the familiar equations extend to three dimension!

OpenStudy (theeric):

And the midpoint theorem can be extended, just like the distance formula was. Normally, \(x_{midpoint}=\dfrac{x_2+x_1}{2}\) and \(y\) is similiar. Now \(z\) is also similar :P

OpenStudy (theeric):

I didn't give away too much, did I?

OpenStudy (anonymous):

That was all I need :P Thanks:D

OpenStudy (whpalmer4):

What did you get for your result?

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