Finding two points on the vector joining two lines

[aeromat]

Homework Statement

You are given the two lines with the following parametric equations:

L1:
x = 4 + 2t
y = 4 + t
z = -3 - t

L2:
x = -2 + 3s
y = -7 + 2s
z = 2 - 3s

Both s, and t are elements of a real number.

Determine the coordinates of P1 that lies on L1, and P2 that lies on L2, if vector L1L2 is perpendicular to each of the two lines.

Homework Equations

Cross Product
Dot Product
Parametric,vector, symmetric equations

The Attempt at a Solution

Alright, so I converted the L1 and L2 into their vector equations:
L1{ r = <4,4,-3> + t<2,1,-1> }
L2{ r = <-2,-7,2> + t<3,2,-3> }

The direction vectors are:
m1 = <2,1,-1>
m2 = <3,2,-3>

I took the cross product of both vectors to find the P1P2 vector that is also perpendicular to both the direction vectors of the line; if the line segment is perpendicular to the line, then it is bound to be perpendicular to its direction vectors as well.

m1 X m2 = P1P2
P1P2 = <-1,3,1> after doing all the cross product work.

I know <x - x0, y - y0, z - z0> = P1P2

I assign A(4,4,-3) from L1
I assign B(-2,-7,2) from L2
I know that the two lines share the same normal vector, P1P2, which is <-1,3,1>.

I am stuck here because I don't know how to go on with finding P1 and P2...
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I tried:

Making a line segment AP1 and another line segment BP2 to ultimately solve for the two points, by dot producting each line segment individually with P1P2.

(1) AP1 dot P1P2 = 0
(2) BP2 dot P1P2 = 0

Stopped at (1) because I realized all I accomplished was getting the Cartesian equation -_-
(1) <x - 4, y -4, z - (-3)> dot <-1,3,1> = 0

-1(x-4) + 3(y-4) + 1(z+3) = 0
-1x +4 + 3y -12 +1z + 3 = 0
-1x +3y + 1z -5 = 0

 

[LCKurtz]

Using different letters for the two lines:

$$\vec R = \langle 4, 4, -3\rangle + t\langle 2,1,-1\rangle$$

$$\vec S = \langle -2,-7,2\rangle + s\langle 3,2,-3\rangle$$

and you have a normal direction

$$\vec N = \langle -1,3,1\rangle$$

which is all good so far. What you are looking for is a point on R and a point on S such that the vector between those points is parallel to N:

$$\vec R(t) - \vec S(s) =\lambda \vec N$$

Write that out and you should get three equations in the 3 unknowns t, s, λ. Once you know t and s you have your two points.