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Thread: Vectors of two skew lines

  1. #1
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    Vectors of two skew lines

    The line AB is a common perpendicular to to two skew lines AP and BQ, and C and R are the midpoints of AB and PQ respectively. Prove by vector methods, that CR and AB are perpendicular.

    I tried the vectors, AP=s, AB=b, BQ=c, and PQ=d
    a.b=0, and b.c=0
    CR=$\displaystyle \frac{1}{2}b+c-\frac{1}{2}d$
    then AB.CR=$\displaystyle b.(\frac{1}{2}b+c-\frac{1}{2}d)$
    this would equal 0 if b.d is equal to one but I don't see how that relationship comes about.
    Thanks!
    Last edited by arze; Jul 1st 2010 at 09:58 PM. Reason: Typo
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  2. #2
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    Quote Originally Posted by arze View Post
    The line AB is a common perpendicular to to two skew lines AP and BQ, and C and R are the midpoints of AB and PQ respectively. Prove by vector methods, that CR and AB are perpendicular.

    I tried the vectors, AP=s, AB=b, BQ=c, and PQ=d
    a.b=0, and b.c=0
    CR=$\displaystyle \frac{1}{2}b+c-\frac{1}{2}d$
    then AB.CR=$\displaystyle b.(\frac{1}{2}b+c-\frac{1}{2}d)$
    this would equal 0 if b.d is equal to one but I don't see how that relationship comes about.
    Thanks!
    1. Let $\displaystyle \vec a$ denote the stationary vector of point A, $\displaystyle \vec b$ the stationary vector of point B, etc..

    2. According to the text you know:

    $\displaystyle \left|\begin{array}{l}\overrightarrow{AB} \cdot \overrightarrow{AP} = 0 \\ \overrightarrow{AB} \cdot \overrightarrow{BQ} = 0\end{array}\right.$

    3. The stationary vector of point C is calculated by:

    $\displaystyle \vec c = \frac12 (\vec a+\vec b)$

    The stationary vector of point R is calculated by:

    $\displaystyle \vec r = \frac12 (\vec p+\vec q)$

    Consequently the direction vector

    $\displaystyle \overrightarrow{CR} = \vec r - \vec c = \frac12 (\vec p+\vec q) - \frac12 (\vec a+\vec b) = \frac12((\vec p - \vec a) + (\vec q - \vec b)) = \frac12(\overrightarrow{AP} + \overrightarrow{BQ})$

    4. Now you have to calculate:

    $\displaystyle \overrightarrow{AB} \cdot \overrightarrow{CR} = \overrightarrow{AB} \cdot \frac12(\overrightarrow{AP} + \overrightarrow{BQ}) = \frac12(\overrightarrow{AB} \cdot \overrightarrow{AP} + \overrightarrow{AB} \cdot \overrightarrow{BQ}) = \frac12(0+0)=0$
    Attached Thumbnails Attached Thumbnails Vectors of two skew lines-dopplt_orthogonal.png  
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