# Thread: To find the points of intersection of a line that is perpendicular to two skew lines.

1. ## To find the points of intersection of a line that is perpendicular to two skew lines.

So given two SKEW lines (in Euclidian space) finding the shortest distance between them is fine, but how would you find the points of intersection of the two lines with the line of shortest distance (i.e. the line that is mutually perpendicular to both skew lines)?

E.g. given L1: r= (2+m)i + (3+2m)j + (4+2m)k
and L2: r= (3+f)i + (3+ f)j + (4+3f)k

(where f and m are parameters)
I believe they are skew, now how to find the line that passes through L1 and L2 and is mutually perpendicular to both? I can find the direction vector (by crossing the two other vectors) but I need a known point...

Help very appreciated!!! thanks!!!

2. Originally Posted by Yehia
So given two SKEW lines (in Euclidian space) finding the shortest distance between them is fine, but how would you find the points of intersection of the two lines with the line of shortest distance (i.e. the line that is mutually perpendicular to both skew lines)?

E.g. given L1: r= (2+m)i + (3+2m)j + (4+2m)k
and L2: r= (3+f)i + (3+ f)j + (4+3f)k

(where f and m are parameters)
I believe they are skew, now how to find the line that passes through L1 and L2 and is mutually perpendicular to both? I can find the direction vector (by crossing the two other vectors) but I need a known point...

Help very appreciated!!! thanks!!!
1. I'll re-write the equations of the lines to:

$\displaystyle L_1:\vec r = \langle 2,3,4 \rangle+m \cdot \langle 1,2,2 \rangle$

$\displaystyle L_2:\vec r = \langle 3,3,4 \rangle+f \cdot \langle 1,1,3 \rangle$

2. The line $\displaystyle L_3$ is perpendicular to both lines $\displaystyle L_1$ and $\displaystyle L_2$. Therefore the direction of $\displaystyle L_3$ is:

$\displaystyle \langle 1,2,2 \rangle \times \langle 1,1,3 \rangle = \langle 4,-1,-1 \rangle$

3. Now P is a point on $\displaystyle L_1: \vec p = \langle 2,3,4 \rangle+p \cdot \langle 1,2,2 \rangle$. $\displaystyle L_3$ passes through P and intersects with $\displaystyle L_2$:

$\displaystyle \underbrace{\left(\langle 2,3,4 \rangle+p \cdot \langle 1,2,2 \rangle\right)}_{\text{staionary vector of P}} + s \cdot \langle 4,-1,-1 \rangle = \langle 3,3,4 \rangle+f \cdot \langle 1,1,3 \rangle$

4. Solve the system of simultaneous equations for (p, s, f).

3. Suppose that $\displaystyle \ell _1 (t) = P + tD\;\& \,\ell _2 (t) = Q + tE$ are two skew lines.
The distance between the two lines is $\displaystyle \frac{{\left\| {\overrightarrow {PQ} \cdot \left( {D \times E} \right)} \right\|}}{{\left\| {D \times E} \right\|}}$.
That is rather simple formula to calculate.

However to find the points on each line that are endpoints of the shortest line segment between the lines is complicated.
The vector $\displaystyle U=D\times(D\times E)$ is perpendicular to both vectors $\displaystyle D~\&~(D\times E)$.
Now write the equation of the plane that contains the line $\displaystyle \ell_1$ with normal $\displaystyle U$.
That plane is $\displaystyle U\cdot(<x,y,z>-P)=0$.
The point of intersection of $\displaystyle \ell_2$ an that plane is one of the endpoints we need.
That point is $\displaystyle \,\ell _2 \left( {\frac{{U \cdot \overrightarrow {PQ} }}{{U \cdot E}}} \right)$.

To find the other point we use the vector $\displaystyle W=E\times(D\times E)$ we get $\displaystyle \ell_1\left( {\frac{{W \cdot \overrightarrow {QP} }}{{W \cdot D}}} \right)$.

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