# Thread: Distance between lines in R^3

1. ## Distance between lines in R^3

Hi,

I'm trying to determine the distance between the following lines:

To solve this question, I'll be using the following distance formula:

I know that the vector 'm' would be (1,1,3), but in terms of vector 'QP' I'm not sure how to determine this. Can someone help?

- otownsend

2. ## Re: Distance between lines in R^3

You realize, don't you, that because both lines have direction vector <1, 1, 3?> the two lines are parallel?

Take any point on the first line, say take (3, 1, -2) with m= 0. The plane perpendicular to the first line, through that point, is 1(x- 3)+ 1(y- 1)+ 3(z+2)= x+ y+ 3z+ 2= 0. The second line, x= 1+ n, y= n, z= 1+ 3n, intersects the plane where 1+ n+ n+ 3(1+ 3n)+ 2= 11n+ 6= 0 so n= -6/11. So the point is x= 1- 6/11= 5/11, y= -6/11, z= 1- 18/11= -7/11. The distance between the two lines is the distance between the points (3, 1, -2) and (5/11, -6/11, -7/11).

3. ## Re: Distance between lines in R^3

Originally Posted by otownsend
I'm trying to determine the distance between the following lines:

To solve this question, I'll be using the following distance formula:
[ATTACH=CONFIG]38505[/ATTACH
I know that the vector 'm' would be (1,1,3), but in terms of vector 'QP' I'm not sure how to determine this. Can someone help?
Just let $P=(3,1,-2)~\&~Q=(1,0,1)$ and as you said $m=(1,1,3)$ then apply the distance formula.
$\overrightarrow {QP} = \left\langle {2,1, - 3} \right\rangle$