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Find eqn of sphere tangent to 2 planes

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the equation of a sphere of radius 3 which is tangent to both the planes x-2y+2z=3 and 3x+4z=8


    2. Relevant equations

    The only eqn's I need are cross product and MAYBE the distance formula.

    3. The attempt at a solution

    Initially I shifted each normal vector by 3 units and took the cross product, but that just left me with the same answer as if I crossed the initial two normals.

    I know that you have to do something with a 3 unit shift, but I'm honestly pretty lost.

    Cheers.
     
  2. jcsd
  3. Oct 11, 2011 #2

    SammyS

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    What do you get for that cross-product?
     
  4. Oct 11, 2011 #3
    When I take that cross product I get (-8,2,6). I don't know if that is the right way to go forward with the problem though.
     
  5. Oct 11, 2011 #4

    SammyS

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    Can you find the equations for two more planes, parallel to each of the given planes and each 3 units from the given planes?
     
  6. Oct 11, 2011 #5
    I was thinking of doing that but I am unsure how (very new to 3 dimensional problems)

    Do you mean that each plane is parallel to both planes?

    i.e new plane 1 is parallel to both or just the first one.


    I don't know how you would find a parallel plane with knowing where it passes through?..

    Edit-I understand what you are saying, we want a parallel plane that is shifted 3 units, I just don't know how to mathematically do that. I was thinking (3,3,3), but if I take the dot product of that I just find the same plane..
     
    Last edited: Oct 11, 2011
  7. Oct 11, 2011 #6

    SammyS

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    No, that can't be. The given planes are not parallel to each other. New plane one is parallel to plane 1, but a distance of 3 units away.

    By the Way, there may be a more direct way than what I'm suggesting.

    Pick a value for each of two of the variables, & stick them into the equation for plane1. I picked y=0, z=0 so x=3. ∴ the point (3,0,0) is a point on plane1. Use the vector-parametric form of the line perpendicular to plane1 & passing through (3,0,0). Find a point on that line that's 3 units from (3,0,0). Plug that into x-2y+2z to find the value of the constant.

    Do similar for plane2.
     
  8. Oct 11, 2011 #7
    I'm honestly clueless at this point, could you please clarify a little bit. I understand the gist of what you are attempting to do but don't know how to arrive there.

    I have x=3+t y=-2t z=2t

    now what
     
    Last edited: Oct 11, 2011
  9. Oct 11, 2011 #8

    SammyS

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    What you have looks right.

    I thought about this over supper. I knew there was a bit more direct way. It is accomplishing the same thing.

    The idea is to get the equations for the two new planes.

    The equation we're given for plane1 is x -2y +2z = 3. Any point (x0, y0, z0) on the plane satisfies the equation, x0 - 2y0 + 2z0 = 8. Fortunately, the normal (perpendicular) vector to the plane is given by the coefficients, and is in this case (1, -2, 2). So. we need to go 3 units along this vector from the point (x0, y0, z0) in plane1 to find a point in new plane1. Fortunately the magnitude (length) of the normal vector is 3. ∴ , the point (x0 + 1, y0 - 2, z0 + 2) is 3 units from plane1 so it's in the new plane1. Any point in the new plane1 satisfies x -2y +2z = ? What's the constant? Plug in our point & see.

    x0 + 1 - 2(y0 - 2) + 2(z0 + 2) = x0 - 2y0 + 2z0 + 1 + 4 + 4.

    But we know that x0 - 2y0 + 2z0 = 3, so we have 3 + 9 = 12. The equation for new plane1 is

    x -2y +2z = 12 .

    Do similar for plane2, except the length of the normal vector (3, 0, 4) is 5. So, multiply by 3/5 to get the normal vector (3/5, 0, 4/5) of length 3.

    You will have equations for two new planes, each parallel to one of the old planes and each a distance of 3 from the old planes.

    Now consider the line of interaction of these two new planes. this line is a distance of 3 units from both the old planes. ∴ , a sphere of radius 3 with its center on this line is tangent to both of the old planes.

    So, pick a point on the line of intersection. (That's what you need the vector, (-8,2,6), for.)
     
  10. Oct 11, 2011 #9
    i follow you for plane one but not at all for plane two..confused with the multiplication. the new second plane is no longer parallel because it has different coefficients.



    would we be correct in saying, you find both planes, and then you need to find a new intersection of (x,y,z) which will be the points of the center of the circle.

    I understand the logic, just not how to get there.
     
    Last edited: Oct 11, 2011
  11. Oct 11, 2011 #10

    SammyS

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    Plane2 is given by 3x+4z=8 . Find a new plane2 in a similar manner. This new plane2 needs to be parallel to old plane2 and a distance of 3 from old plane2.
     
  12. Oct 11, 2011 #11
    i just did 3/5*(x0 +3/5)+4/5*(zo+4/5) this gives me a total of 4. So my new plane is 3x+4z=4. Is this right? If so I now have

    3x+4z=4 and x-2y+2z=12. I know we must now use cross product, but not totally sure with what.

    Thanks in advance, Last step!!
     
    Last edited: Oct 11, 2011
  13. Oct 11, 2011 #12

    SammyS

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    No, It's 3*(x0 +3/5)+4*(z0+4/5) = 3x0 + 9/5 + 4*z0 + 16/5 = 8 + 25/5 = 13 , b/c 3x0 + 4*z0 = 8
     
  14. Oct 11, 2011 #13

    SammyS

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    For last step. You know the cross product from previous post (But that doesn't matter.). Find a point common to both planes.
     
  15. Oct 11, 2011 #14
    How can I do that? 3 variables with only two equations
     
  16. Oct 11, 2011 #15

    SammyS

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    You get to pick the value of one of them. Pick x = -1, for instance. Then z is determined by new plane2, with those two, y is determined.
     
  17. Oct 11, 2011 #16
    so my final answer is now, I chose x=0

    x^2 + (y-2.75)^2+(z-13/4)^2.

    right?
     
  18. Oct 11, 2011 #17

    SammyS

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    x^2 + (y-2.75)^2+(z-13/4)^2 = 9 , I hope.
     
  19. Oct 11, 2011 #18
    lol are you beck's calc 3 i have the same problem
     
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