# Vectors of two skew lines

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• Jul 1st 2010, 09:53 PM
arze
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!
• Jul 1st 2010, 11:13 PM
earboth
Quote:

Originally Posted by arze
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:

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