Distance Between Nonparallel Lines: Linear Algebra Solution

[dangish]

I already posted this once, have a hard time believing nobody knows how to do it, and it's due tomorrow, so here it is again:

Find the distance between the nonparallel lines,

L1:

|x| ...|4| ...|1|
|y| = |-1| + t|0|
|x| ...|3| ...|2|

and L2:

|x| ...|1| ...|-3|
|y| = |2| + s|1|
|x| ...|2| ...|-1|

picture the above as matricies, sorry i don't know how to properly make them :...(

Attempt at a solution:

The hint told me to find a vector orthogonal to both lines. so i took the normal vector of each line ( the three numbers after s and t ) and got a vector parallel to each.

The vector I used was v = |6,15,-3| <-- again picture this as a matricie.

The next part in the hint said take a plane with this normal containing one of the lines.

so I did and got, 6x +15y - 3z = d , then I subbed in a point from L2 to get,

6(1)+15(2)-3(2)=d ==> d=30

the equation of the plane is now 6x+15y-3z=30

The next part of the hint says "use a projection", and this is where I am stuck.

how will I project the line onto the plane? any help would be much appreciated, thanks.

 

[Dick]

You don't really need the plane equation. Take the normal vector v=[6,15,-3] and find a vector of unit length pointing in the same direction. If you take the dot product of that with the difference between any point in L1 and any point in L2, the result will be the distance. That's 'projecting' a difference vector between the two lines onto the normal direction.