# Homework Help: Finding the Distance Between 2 Lines and One Orthogonal Line

1. Oct 16, 2011

### TranscendArcu

1. The problem statement, all variables and given/known data
Let L1 be the line (0,4,5) + (1,2,-1)t. Let L2 be the line (-10,9,17) + (-11,3,1)t.

Find the line L passing through and orthogonal to L1 and L2.

What is the distance between L1 and L2?

2. Relevant equations
Vector Projection Equation: V • W/|W|

3. The attempt at a solution
I think finding the equation is the more difficult part of the question, so I'll begin by finding the distance. First, I find a vector orthogonal to both lines by cross product. This vector is <5,10,25>, which I reduce to <1,2,5>.

Using the points given in the definitions of lines, I subtract to get another vector, <10,-5,22>.

By the equation for vector projections, <10,-5,22> • <1,2,5> = 110, which I divide by the magnitude of <1,2,5>. So, the distance equals 110/sqrt(30).

Now, I want the equation for this orthogonal line to be in the form (a,b,c) + <1,2,5>*s, where (a,b,c) is a point on the line. By multiplying out,

xs: a + s
ys: b +2s
zs: c + 5s

I suppose I want to solve for s in terms of t. I know

xt1: t
yt1: 4 + 2t
zt1: 5 - t

So,

t = a + s, then....

I think this whole system of equations will turn into a large mess. Is there a simpler way to approach this problem?

2. Oct 16, 2011

### verty

The way I thought of this was, you have a line L3 that intersects L1 and L2, and L3 has the direction vector of L1 x L2.

So there is a point P1 on L1 and a point P2 on L2 such that P2 - P1 = (L1 x L2)t. This is a 3-vector equation in 3 variables which should be easy to solve using linear algebra.

Learning things like this, I always like to find the shortest way of describing the structure of the problem.

3. Oct 16, 2011

### TranscendArcu

I haven't taken any linear algebra classes, so I don't know the method that you're referring to. Could you please demonstrate?