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Math Help - 3d distance problem

  1. #1
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    3d distance problem

    The question is what is the equation that includes the points T so that the distance from T to O = (4,0,1) is twice the distance from T to C = (1, -2, 4)?
    I set the equation up like this sqrt((x-4)^2 + (y-0)^2 + (z-4)^2) = 2sqrt((x-1)^2 + (y+2)^2 + (z-4)^2)

    I get it down to this:
    x^2 + (y + 16/6)^2 + (z - 5)^2 = -137/9
    The problem is the equation doesn't work because the second side is negative
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  2. #2
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    Quote Originally Posted by collegestudent123 View Post
    The question is what is the equation that includes the points T so that the distance from T to O = (4,0,1) is twice the distance from T to C = (1, -2, 4)?
    I set the equation up like this sqrt((x-4)^2 + (y-0)^2 + (z-{\color{red}4})^2) = 2sqrt((x-1)^2 + (y+2)^2 + (z-4)^2)

    I get it down to this:
    x^2 + (y + 16/6)^2 + (z - 5)^2 = -137/9
    The problem is the equation doesn't work because the second side is negative
    See your error in red above.
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  3. #3
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    Sorry the 4 was actually a typo, in the work I did it was a 1
    I tried it again and still come up with that answer

    Edit: I made a algebra error, I figured it out
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