# Thread: Find the shortest distance between the lines:

1. ## Find the shortest distance between the lines:

Sorry again if this is a noobie post.

In R3, find the shortest distance between the lines:

l1 = x (2, -1, 3) + t(1, -1, 4)
l2 = x (1, 0, 1) + t(1, -1, 4)

Unless I am mistaken I would take vector 1 to be (1, -1, 4) and vector 2 to be (1, -1, 4) as well.
Using the equation distance = |P1P2 dot n| / ||n||
with n = cross product of vector 1 and vector 2

But when I do the cross product, I receive a value of 0.

Can anyone help steer me in the right direction.

I'm sorry if this is a trivial question, I have not taken math courses in a few years.

2. ## Re: Find the shortest distance between the lines:

Originally Posted by biggerleaffan
In R3, find the shortest distance between the lines:
l1 = x (2, -1, 3) + t(1, -1, 4)
l2 = x (1, 0, 1) + t(1, -1, 4)
Those x's have no place in the question.

These two lines are parallel. Thus take any point on $\displaystyle \ell_1$ and find its distance to $\displaystyle \ell_2$.

If $\displaystyle P\notin \ell=Q+tD$ then $\displaystyle D(\ell;P)=\frac{|\overrightarrow {QP}\times D| }{\|D\|}$

3. ## Re: Find the shortest distance between the lines:

Yes, the cross product is 0 because the "direction vectors" are the same, (1, -1, 4). And that means that the two lines are parallel. Take any point on one of the lines, say (1, 0, 1) and find the equation of the plane containing that point perpendicular to the direction vector. The distance between the two line is the distance from (1, 0, 1) to the point at which the other line crosses that plane.