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Math Help - Point to a line 3-d

  1. #1
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    Point to a line 3-d

    alright so I am having doubt in my professors answer to this question or maybe I am doing something wrong so before i go speak with him, i wanna ask some other professionals

    Find the distance of the point P = (1,2,-3) to the line: (1,-2,1) + t(2,3,4)

    The vector from (1,2,-3) to (1,-2,1) is (0,4, -4) = A
    and the direction vector is (2,3,4) = B

    so sinθ = |AxB|/ [|A|][|B|]

    where |AxB| = 12
    |A| = √32
    |B| = √29

    so |A|sinθ = 12/√29

    however my profs answer is √(32 - 16/29)

    sorry i din't latex it i was in a rush
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  2. #2
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    I get the same answer as your prof, (by a different method from the one that you have used).
    I think that your value for |AxB| is wrong.
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  3. #3
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    Quote Originally Posted by treetheta View Post
    alright so I am having doubt in my professors answer to this question or maybe I am doing something wrong so before i go speak with him, i wanna ask some other professionals

    Find the distance of the point P = (1,2,-3) to the line: (1,-2,1) + t(2,3,4)

    The vector from (1,2,-3) to (1,-2,1) is (0,4, -4) = A
    and the direction vector is (2,3,4) = B

    so sinθ = |AxB|/ [|A|][|B|]

    where |AxB| = 12
    |A| = √32
    |B| = √29

    so |A|sinθ = 12/√29

    however my profs answer is √(32 - 16/29)
    You should find that A\times B = (28,-8,-8) = 4(7,-2,-2), from which |A\times B| = 4\sqrt{57}, giving the answer to the problem as 4\sqrt{\tfrac{57}{29}}.

    The prof's method was probably to minimise the distance from the point (1+2t,–2+3t,1+4t) to the point (1,2,–3) by completing the square. That distance is \sqrt{4t^2+(3t-4)^2+(4t+4)^2} = \sqrt{29(t+\tfrac4{29})^2 + 32-\tfrac{16}{29}}, which also comes out as 4\sqrt{\tfrac{57}{29}} when t has the minimising value –4/29.

    Personally, I prefer the cross product method to the minimisation method (provided that it's done correctly).
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