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Brian science forum beginner
Joined: 16 May 2005
Posts: 45
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Posted: Sun Jun 11, 2006 6:48 pm Post subject:
Distance from a Sphere
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Hi everyone,
can anyone tell me how to locate the point on the sphere with the equation
x^2 + y^2 + z^2 + 2x - 2y - 4z - 3 = 0 which is closest to the origin ??
Any help much appreciated !!!
Thanks
Brian |
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G.E. Ivey science forum Guru
Joined: 29 Apr 2005
Posts: 308
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Posted: Sun Jun 11, 2006 6:58 pm Post subject:
Re: Distance from a Sphere
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| Quote: | Hi everyone,
can anyone tell me how to locate the point on the
sphere with the equation
x^2 + y^2 + z^2 + 2x - 2y - 4z - 3 = 0 which is
closest to the origin ??
Any help much appreciated !!!
Thanks
Brian
|
Do you know how to find the center of the sphere?
Write this as (x^2+ 2x+ )+ (y^2- 2y+ )+ (z^2-4z+ )= 3 and complete the square in each part so you get
(x- x_0)^2+ (y- y_0)^2+ (z- z_0)^2= R^2.
The line from the center, (x_0,y_0,z_0), to (0,0,0) passes through the sphere at the point closest to (0,0,0). |
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Paul Sperry science forum Guru
Joined: 08 May 2005
Posts: 371
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Posted: Sun Jun 11, 2006 9:43 pm Post subject:
Re: Distance from a Sphere
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In article
<2545127.1150051755803.JavaMail.jakarta@nitrogen.mathforum.org>, Brian
<fdbsj453fgjer@yahoo.com> wrote:
| Quote: | Hi everyone,
can anyone tell me how to locate the point on the sphere with the equation
x^2 + y^2 + z^2 + 2x - 2y - 4z - 3 = 0 which is closest to the origin ??
Any help much appreciated !!!
Thanks
Brian
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If you know the technique, this is a more or less classic Lagrange
multiplier problem:
Minimize x^2 + y^2 + z^2 subject to the constraint
x^2 + y^2 + z^2 + 2x - 2y - 4z - 3 = 0.
G. E. Ivey's suggestion also works fine.
--
Paul Sperry
Columbia, SC (USA) |
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Brian science forum beginner
Joined: 16 May 2005
Posts: 45
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Posted: Mon Jun 12, 2006 11:09 am Post subject:
Re: Distance from a Sphere
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Thank you for your replies.
They have helped !
Brian |
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