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Math Help - Vector Problem

  1. #1
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    Vector Problem

    Given 3 points A(0,2,7) , B(5,-3,2) and C(1,1,1). (1)Find the position vector of the point R on AB such that CR is perpendicular to AB. Hence (2)find the perpendicular distance from C to AB and the (3)position vector of the reflection of C in AB.

    Here's my solution, i just have trouble finding the "position vector of the reflection of C in AB."

    Vector AB = OB - OA
    = (5-0,-3-2,2-7)
    =(5,-5,-5)

    vector equation of line AB l: (0,2,7) + λ(5,-5,-5)

    vector OR = (5λ, 2-5λ, 7-5λ)

    vector CR= OR - OC = (5λ-1, 1-5λ, 6-5λ)

    vector CR.AB=0 , since cos(90) = 0, given by scalar product

    (5λ-1, 1-5λ, 6-5λ).(5,-5,-5) = 0

    λ=8/15

    vector OR= (8/3, -2/3, 13/3)= 1/3(8,-2,13)


    vector CR=(5/3,-5/3,10/3)
    Magnitude of CR = (50/3)^0.5

    now i'm having trouble with the reflection part
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  2. #2
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    Consider  OC (the position of point  C)

     = OR + RC

    Now the position vector of the reflection of  C in  AB ,

    it should be  OR - RC

     = \frac{1}{3} [(8,-2,13) - (5,-5,10) ]

     = \frac{1}{3} (3 , 3 , 3 ) = (1,1,1)
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  3. #3
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    the answer is 1/3(13,-7,23) but thanks for helping
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  4. #4
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    Quote Originally Posted by simplependulum View Post
    Consider  OC (the position of point  C)

     = OR + RC

    Now the position vector of the reflection of  C in  AB ,

    it should be  OR - RC

     = \frac{1}{3} [(8,-2,13) - (5,-5,10) ]

     = \frac{1}{3} (3 , 3 , 3 ) = (1,1,1)
    Dear simplependulum,

    You have made a slight mistake, instead of \frac{1}{3} [(8,-2,13) - (5,-5,10)] it should be \frac{1}{3} [(8,-2,13) - (-5,5,-10)] since RC=(-5,5,-10). (You have accidently taken CR instead of RC).

    Dear mephisto50,

    The above correction will give you the correct answer.

    Thank you.
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