Linear Algebra (distance from two nonparallel lines)

[dangish]

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.

 

[mighty2000]

To find the distance between two nonparallel lines, you can use the vector projection formula. Let's call the two lines L1 and L2, with equations:

L1: x = 4 + t, y = -1, z = 3t
L2: x = 1 + s, y = 2 + s, z = -1 + 2s

First, we need to find a vector that is orthogonal to both lines. This can be done by taking the cross product of the direction vectors of the two lines. So we have:

v = (1,0,3) x (1,1,2) = (-3, -1, 1)

Next, we need to choose a point on one of the lines to use as our base point. Let's use the point (4, -1, 0) on L1.

Now, we can use the vector projection formula to find the distance between the two lines:

distance = |(p - q) * v| / |v|

where p is a point on L1 (in this case, (4, -1, 0)), q is a point on L2 (in this case, (1, 2, -1)), and v is the vector we found earlier.

Substituting in the values, we get:

distance = |(4-1, -1-2, 0-(-1)) * (-3, -1, 1)| / |(-3, -1, 1)|
= |(3, -3, 1) * (-3, -1, 1)| / |(-3, -1, 1)|
= |-12 + 3 + 1| / |-3 + 1 + 1|
= 4 / 1
= 4

So the distance between L1 and L2 is 4 units.