SOLVED Vectors of two skew lines

Jul 2009
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Singapore
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!
 
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earboth

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Jan 2006
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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:

\(\displaystyle \overrightarrow{AB} \cdot \overrightarrow{CR} = \overrightarrow{AB} \cdot \frac12(\overrightarrow{AP} + \overrightarrow{BQ}) = \frac12(\overrightarrow{AB} \cdot \overrightarrow{AP} + \overrightarrow{AB} \cdot \overrightarrow{BQ}) = \frac12(0+0)=0\)
 

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