Results 1 to 7 of 7

Math Help - Help - Equation of a sphere in 3D

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    11

    Help - Equation of a sphere in 3D

    the problem:
    Find the equation of the sphere that contains the points A = (2,4,4) , B= (4,2,0) , and C = (4,0,2), if its center lies on the plane y=x.


    My attempt:

    |AB|= sqroot(24)
    |BC|= sqroot(8)
    |AC|= sqroot(24)

    from this i guessed that the center of the circle was point A, since points B and C were both equal distances away from them.
    which gives me an r = sqroot(24)

    Thus, the equation of the sphere is:

    (x-2)^2 + (y-4)^2 + (z-4)^2 = 24



    My problem is... the center (2,4,4) is not on the plane y=x

    I'm confused. any help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Black's Avatar
    Joined
    Nov 2009
    Posts
    105
    Let (x_0, x_0, z_0) be the center of the sphere. Choosing point B, the radius of the sphere is

    r=\sqrt{(x_0-4)^2+(x_0-2)^2+z_0^2}.

    Then the equation of the sphere is

    (x-x_0)^2+(y-x_0)^2+(z-z_0)^2=(x_0-4)^2+(x_0-2)^2+z_0^2.

    Plug in points A and C to get the system of equations

    (2-x_0)^2+(4-x_0)^2+(4-z_0)^2=(x_0-4)^2+(x_0-2)^2+z_0^2
    (4-x_0)^2+x_0^2+(2-z_0)^2=(x_0-4)^2+(x_0-2)^2+z_0^2.

    Then just solve for (x_0,z_0) and put the values into the equation of the sphere to get your answer.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    11
    Quote Originally Posted by Black View Post
    Let (x_0, x_0, z_0) be the center of the sphere. Choosing point B, the radius of the sphere is

    r=\sqrt{(x_0-4)^2+(x_0-2)^2+z_0^2}.

    Then the equation of the sphere is

    (x-x_0)^2+(y-x_0)^2+(z-z_0)^2=(x_0-4)^2+(x_0-2)^2+z_0^2.

    Plug in points A and C to get the system of equations

    (2-x_0)^2+(4-x_0)^2+(4-z_0)^2=(x_0-4)^2+(x_0-2)^2+z_0^2
    (4-x_0)^2+x_0^2+(2-z_0)^2=(x_0-4)^2+(x_0-2)^2+z_0^2.

    Then just solve for (x_0,z_0) and put the values into the equation of the sphere to get your answer.

    thank you.

    was point B chosen to find the radius because there was no value for z?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Black's Avatar
    Joined
    Nov 2009
    Posts
    105
    Yeah, I chose B because of the of the 0 in the z-coordinate. Choosing A or C to find the radius will still lead you to the same answer.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,421
    Thanks
    1856
    Quote Originally Posted by rain21 View Post
    the problem:
    Find the equation of the sphere that contains the points A = (2,4,4) , B= (4,2,0) , and C = (4,0,2), if its center lies on the plane y=x.


    My attempt:

    |AB|= sqroot(24)
    |BC|= sqroot(8)
    |AC|= sqroot(24)

    from this i guessed that the center of the circle was point A, since points B and C were both equal distances away from them.
    which gives me an r = sqroot(24)
    Well, there is your problem! You are told that the point A is on the sphere. It can't be the center of the sphere! Also, obviously, A does NOT "lie on the plane y=x". If you must assume things, don't assume things that contradict the information you are given!

    You need to find the center. Do you remember how to find the center of a circle given three points on the circle? The perpendicular bisector of each segment through to points must intersect at the center. For three dimensions, the corresponding idea is that the planes orthogonal to the line segments. The segment from A = (2,4,4) to B= (4,2,0) has midpoint (3, 3, 2) and "direction vector" 2\vec{i}- 4\vec{k} so the plane orthogonal to that segment and containing (3, 3, 2) is 2(x- 3)- 4(z- 2)= 0. The segment from A= (2, 4, 4) to C = (4,0,2) has midpoint (3, 2, 3) and direction vector 2\vec{i}- 4\vec{j}- 2\vec{k} so the plane orthogonal to the segment and containing (3, 2, 3) is 2(x- 3)- 4(y- 2)- 2(z- 3)= 0. That is not enough to determine the center of the sphere (using the segment from B to C will not give you anything new) which is why they told you "its center lies on the plane y=x".

    Solve the three equations 2(x- 3)- 4(z- 2)= 0, 2(x- 3)- 4(y- 2)- 2(z- 3)= 0, and y= x for (x, y, z), the center of the sphere.

    Thus, the equation of the sphere is:

    (x-2)^2 + (y-4)^2 + (z-4)^2 = 24



    My problem is... the center (2,4,4) is not on the plane y=x

    I'm confused. any help?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jan 2010
    Posts
    11
    Quote Originally Posted by Black View Post
    Yeah, I chose B because of the of the 0 in the z-coordinate. Choosing A or C to find the radius will still lead you to the same answer.
    I followed everything you advised, but wouldn't the radius be: r=\sqrt{(4-x_0)^2+(2-x_0)^2+z_0^2}.


    I got a center of (2,2,2) and a radius = \sqrt{8}

    =/
    Last edited by rain21; January 10th 2010 at 09:08 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member Black's Avatar
    Joined
    Nov 2009
    Posts
    105
    Quote Originally Posted by rain21 View Post
    I got a center of (2,2,2) and a radius = \sqrt{8}

    =/
    Yeah, that's it.

    You can always check your answer by plugging in the three points into the equation.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. What is the standard equation of a sphere?
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: February 23rd 2011, 12:59 PM
  2. equation of a sphere
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: October 2nd 2010, 03:44 AM
  3. [SOLVED] Sphere equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 23rd 2010, 10:07 PM
  4. Find the equation of a sphere
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 6th 2009, 05:44 AM
  5. find the equation of the sphere
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 27th 2007, 03:57 PM

Search Tags


/mathhelpforum @mathhelpforum