Results 1 to 7 of 7

Math Help - Sphere...

  1. #1
    Member
    Joined
    May 2007
    Posts
    150

    Sphere...

    The point P=(11,2,5) is on the sphere x^2 + y^2 + z^2 - 10x - 2y - 14z = -26. What are the cooridantes of the point on this sphere that is farthest from P?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by stones44 View Post
    The point P=(11,2,5) is on the sphere x^2 + y^2 + z^2 - 10x - 2y - 14z = -26. What are the cooridantes of the point on this sphere that is farthest from P?
    Hint: It is a point on the other side of the diameter.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by stones44 View Post
    The point P=(11,2,5) is on the sphere x^2 + y^2 + z^2 - 10x - 2y - 14z = -26. What are the cooridantes of the point on this sphere that is farthest from P?
    Hello,

    there must be a typo because the point P is not on the sphere:

    11 + 2 + 5 - 110 - 4 - 70 = -34
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Eventhough the point is not on the sphere we can still find the maximum distance from that point. The first step is to connect the line segment with the center of the sphere and the point. The intersection will be a point on the sphere that is closest to the point. But on the other side of the diameter will be the furtherest point.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    778
    Hello, stones44!

    There must be a typo in the problem.
    I'll take a guess and try to fix it . . .


    The point P\,(11,{\color{red}-2},5) is on the sphere: . x^2 + y^2 + z^2 - 10x - 2y - 14z \:=\: -26

    What are the coordinates of the point on this sphere
    that is farthest from P?

    We have: . x^2 - 10x + y^2 - 2y + z^2 - 14z \:=\:-26

    Complete the square:
    . . x^2 - 10x + {\color{blue}25} + y^2 - 2y + {\color{blue}1} + z^2 - 14z + {\color{blue}49} \:=\:-26 + {\color{blue}25} + {\color{blue}1} + {\color{blue}49}

    . . (x-5)^2 + (y-1)^2 + (x-7)^2 \:=\:49

    The sphere has center C\,(5,1,7) and radius r=7.


    We want the point Q diametrically opposite P\,(11,-2,5)

    Here is a very sneaky way to find point Q.


    Consider the "movement" from the center C to point P.

    We have: . C(5,1,7) \to P(11,-2,5)

    . . The x-change is: . 5\to11\:=\:+6 . . . . 6 units forward.

    . . The y-change is: . 1 \to \text{-}2\:=\:-3 . . . . 3 units to the left.

    . . The z-change is: . 7 \to 5 \:=\:-2 . . . . 2 units down.


    To locate point Q, we start at center C and move "backwards".

    . . x-coordinate: .move 6 units back: . x \:=\:5 - 6 \:=\:-1

    . . y-coordinate: .move 3 units to the right: . y \:=\:1 + 3 \:=\:4

    . . z-coordinate: .move 2 units up: . z \:=\:7 + 2\:=\:9


    Therefore: . Q\,(\text{-}1,\,4,\,9)

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Well, I regrouped and came back. I will go with my original assessment.

    We have the sphere:

    (x-5)^{2}+(y+1)^{2}+(z-7)^{2}=49

    It is centered at C(5,-1,7) and has radius 7.

    We need a point antipodal to P(11,2,5)

    f(x,y,z)=(x-11)^{2}+(y-2)^{2}+(z-5)^{2}

    Subject to the constraint: (x-5)^{2}+(y+1)^{2}+(z-7)^{2}=49

    {\nabla}f(x,y,z)={\lambda}{\nabla}g(x,y,z)

    2(x-11)i+2(y-2)j+(2(z-5)k={\lambda}\left[(2(x-5)i+2(y+1)j+2(z-7)k\right]

    x-11={\lambda}(x-5), \;\ y-2={\lambda}(y+1), \;\ z-5={\lambda}(z-7)

    {\lambda}=\frac{x-11}{x-5}, \;\ {\lambda}=\frac{y-2}{y+1}, \;\ {\lambda}=\frac{z-5}{z-7}

    Equate equation one and two and solve for y:

    \frac{x-11}{x-5}=\frac{y-2}{y+1}, \;\ y=\frac{x-7}{2}

    Equate one and three and solve for z: \frac{x-11}{x-5}=\frac{z-5}{z-7}, \;\ z=\frac{26-x}{3}

    Sub these into the constraint and we get:

    x^{2}-10x-11=(x-11)(x+1)=0

    The solution is clearly x = -1. Because the other solution gives us our P(11,2,5)

    By resubbing, this gives y=-4 and z=9

    The point on the sphere opposed to P(11,2,5) is

    \boxed{P(-1, \;\ -4, \;\ 9)}

    Let's check:

    \sqrt{(11-(-1))^{2}+(2-(-4))^{2}+(5-9)^{2}}=14

    14 is the diameter of the sphere.

    Check.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by stones44 View Post
    The point P=(11,2,5) is on the sphere x^2 + y^2 + z^2 - 10x - 2y - 14z = -26. What are the cooridantes of the point on this sphere that is farthest from P?
    Hello,

    there must be a typo as I've pointed out in my previous post.

    There are two possible cases:
    a) The coordinates of the point are wrong. This case was done by Soroban.
    b) the equation of the sphere is wrong. This case was done by galactus

    Here is another attempt to b):

    If the equation of the sphere is:

     x^2 + y^2 + z^2 - 10x + 2y - 14z = -26 then the given point is on it. Rearrange this equation by completing the squares:

    x^2-10x{\color{red}+25}+y^2+2y{\color{red}+1}+ z^2-14z{\color{red}+49} =-26{\color{red}+25+1+49}

    (x-5)^2+(y+1)^2+(z-7)^2=7^2 . That's the equation of a sphere with the centre C(5,-1, 7) and the radius of 7.

    As Soroban pointed out calculate the difference d from P to C, double this difference and add it to the coordinates of P - and that's it:

    d=\left \{\begin{array}{l}x=-6\\y=-3\\z=2\end{array}\right. Double these values and add them to the coordinates of P(11, 2, 5)~\longrightarrow~P'(-1, -4, 9)

    I wasn't sure if you are allowed to use vectors. Therefore I described the procedure without using this expression and without using the usual notation for vectors.

    EDIT: Even though galactus has already done this problem I'll not delete this post because you have another way to solve this problem.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. sphere
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: July 26th 2011, 10:43 AM
  2. Sphere
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 23rd 2010, 09:03 AM
  3. Sphere
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 23rd 2010, 03:30 AM
  4. Sphere...
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: November 24th 2009, 07:08 PM
  5. Sphere
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 11th 2007, 06:37 PM

Search Tags


/mathhelpforum @mathhelpforum