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Math Help - Intersection of common perpendicular and lines

  1. #1
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    Intersection of common perpendicular and lines

    Hello everyone,

    15. The common perpendicular of two skew lines with direction vectors  \vec{d_1} and  \vec{d_2} is the line that intersects both the skew lines and has direction vector  \vec{n} = \vec{d_1} + \vec{d_2} . Find the points of intersection of the common perpendicular with each of the lines  [x, y, z] = [0, -1, 0] + s[1, 2, 1] and  [d, e, f] = [-2, 2, 0] + t[2, -1, 2].

    Thank you for your help.

    ---

     \vec{n} = \vec{d_1} \times \vec{d_2} ,

    so:  \vec{n} = [1, 2, 1] \times [2, -1, 2]

     \vec{n} = [5, 0, -5]

    However, I am not sure how to proceed from here.

    I called the common perpendicular  \vec{p} , whose vector equation is  \vec{p} = [a, b, c] + p[5, 0, -5].

    I then tried to find the points of intersection with the first given line ([x, y, z]) but I did not get useful solutions.
    Last edited by scherz0; April 6th 2009 at 03:22 PM.
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  2. #2
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    Quote Originally Posted by scherz0 View Post
    15. The common perpendicular of two skew lines with direction vectors  \vec{d_1} and  \vec{d_2} is the line that intersects both the skew lines and has direction vector  \vec{n} = \vec{d_1} + \vec{d_2} . Find the points of intersection of the common perpendicular with each of the lines  [x, y, z] = [0, -1, 0] + s[1, 2, 1] and  [d, e, f] = [-2, 2, 0] + t[2, -1, 2].  \vec{n} = \vec{d_1} + \vec{d_2} ,
    You have a very basic mistake. The direction of the common perpendicular is:
     \vec{n} = \vec{d_1} \times \vec{d_2}
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  3. #3
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    Quote Originally Posted by scherz0 View Post
    Hello everyone,

    15. The common perpendicular of two skew lines with direction vectors  \vec{d_1} and  \vec{d_2} is the line that intersects both the skew lines and has direction vector  \vec{n} = \vec{d_1} + \vec{d_2} . Find the points of intersection of the common perpendicular with each of the lines  [x, y, z] = [0, -1, 0] + s[1, 2, 1] and  [d, e, f] = [-2, 2, 0] + t[2, -1, 2].

    Thank you for your help.

    ...
    Have a look here: http://www.mathhelpforum.com/math-he...kew-lines.html
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  4. #4
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    Thanks for your reply.

    According to to other post, I would have:

     [-2, 2, 0] - [0, -1, 0] + s \bullet [1, 2, 1] - t \bullet [2, -1, 2] = k \bullet [5, 0, -5]

     [-2, 3, 0] + s \bullet [1, 2, 1] - t \bullet [2, -1, 2] = k \bullet [5, 0, -5] .

    Could you please tell me how I would construct a system of equations based on this equation?
    Last edited by scherz0; April 6th 2009 at 04:19 PM.
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  5. #5
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    Quote Originally Posted by scherz0 View Post
    Thanks for your reply.

    According to to other post, I would have:

     [-2, 2, 0] - [0, -1, 0] + s \bullet [1, 2, 1] - t \bullet [2, -1, 2] = k \bullet [5, 0, -5]

     [-2, 3, 0] + s \bullet [1, 2, 1] - t \bullet [2, -1, 2] = k \bullet [5, 0, -5] .

    Could you please tell me how I would construct a system of equations based on this equation?
     [-2, 3, 0] + s \cdot [1, 2, 1] - t \cdot [2, -1, 2] = k \cdot [5, 0, -5] ~\implies~\left\{\begin{array}{rcl}-2+s-2t&=&5k\\3+2s+t&=&0 \\ s-2t&=&-5k\end{array}\right.

    This system has the solution (s,t,k)=\left(-1, -1, -\frac15 \right)
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