# SOLVEDVectors of two skew lines

#### 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!

Last edited:

#### earboth

MHF Hall of Honor
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:

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

#### Attachments

• arze