You are given the two lines L1: x= 4 + 2t, y = 4 + t, z = -3 - t, and

Question

L2: x= -2 + 3s, y= -7 + 2s, z = 2 - 3s, If the point P1 lies on L1
and the point P2 lies on L2, determine the coordinates of these two points
if P1P2 is perpendicular to each of the two lines.

Don't really know what to do, I got the direction vector of each point and got their dot product (-1,3,1)

Answer 1

cross product*

Answer 2

One of the classic problems in vector geometry.
In effect you are given two skew lines, (they don't intersect in 3D)
and you are finding the line of shortest distance between them.

We can label P1(4+2t, 4+t, -3-t) and P2(-2+3s, -7+2s, 2-3s)
then vector P2P1 = (-2+3s-4-2t, -7+2s-4-t, 2-3s+3+t)
= (3s-2t-6 , 2s-t-11 , -3s+t+5)

We then get two dot product equations, since this vector is perpendicular
with each of the lines:
(3s-2t-6 , 2s-t-11 , -3s+t+5) dot (2,1,-1) = 0
6s -4t -12 + 2s - t - 11 + 3s - t - 5 = 0
11s -6t = 28

(3s-2t-6 , 2s-t-11 , -3s+t+5) dot (3,2,-3) = 0
9s -6t -18 + 4s - 2t -22 + 9s - 3t - 15 = 0
22s -5t = 55

solve for s and t, sub into the corresponding definitions for P1 and P2
and you are done.

(I got s = 190/77 and t = -1/7
was expecting "nicer" results, so I might have made an arithmetic error.
Better check my calculations)

Answer 3

You are right with your steps, but you have indeed made an arithmetic error on the step when simplifying 9s -6t -18 + 4s - 2t -22 + 9s - 3t - 15 = 0. The correct answer should be s=3 and t=1

Answer 4

Should be s=2 and t=-1