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Math Help - lines in space (vectors)

  1. #1
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    lines in space (vectors)

    I'm having a problem finding the solution to this question from my precal class. Its pretty confusing. Right now we're learning about vectors.

    Find the distance between the given point and the given line.

    Point: (4,1,2)
    Line: r*= (1 + 6/7d)i* + (3 - 2/7)j* + (-2 + 3/7d)k*

    i put a (*) to indicate the vector. dont know how to put the vector sign.

    thanks to anyone who can help. =)
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  2. #2
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    Quote Originally Posted by illustriousLOVE View Post
    I'm having a problem finding the solution to this question from my precal class. Its pretty confusing. Right now we're learning about vectors.

    Find the distance between the given point and the given line.

    Point: (4,1,2)
    Line: r*= (1 + 6/7d)i* + (3 - 2/7)j* + (-2 + 3/7d)k*

    i put a (*) to indicate the vector. dont know how to put the vector sign.

    thanks to anyone who can help. =)
    There are two completely different ways to solve this problem:

    1. Analytic: The distance between P and any point on the line must be a minimum to determine the distance between P and the line:

    Re-write the equation of the line:

    \vec r = (1,3,-2)+d(6,-2,3)

    \overrightarrow{PR}= (1,3,-2)+d(6,-2,3) - (4,1,2) =  (-3,2,-4)+d(6,-2,3)

    |(\overrightarrow{PR}|=\sqrt{\left( (-3,2,-4)+d(6,-2,3) \right)^2} =\sqrt{ 49d^2-68d+29}

    (|(\overrightarrow{PR}|)^2=D(d)=49d^2-68d+29

    D'(d)=98d-68 Now D'(d)=0~\implies~d = \dfrac{34}{49}

    Plug in this value into the equation of D(d). You'll get D = \dfrac{265}{49}\approx 5.41


    2. Geometrically: Construct a plane which contains the point P(4,1,2) and is perpendicular to the given line:

    (6,-2,3)\cdot (\vec q - (4,1,2))=0~\implies~(6,-2,3)\cdot \vec q -28=0

    Let F denote the point of intersection between this plane and the given line: Plug in the term of the line instead of \vec q and solve for d:

    (6,-2,3)\cdot ((1,3,-2)+d(6,-2,3)) -28=0

    49d-6-28=0~\implies~d = \dfrac{34}{49}

    Now calculate the coordinates of F. Afterwards calculate the distance between the given point P and F: That's the distance between the line and the given point.
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