(716, #15) Determine whether the lines and are parallel, skew, or intersecting. If they intersect, find the point of intersection.

Solution:

The parametric equations for are:

1) : x = 4 + 2t, y = -5 + 4t, z = 1 + 3t

2) : x = 2 + s, y = -1 + 3s, z = 2s

The direction vectors for and are < 2, 4, 3 > and < 1, 3, 0 >. The lines are not parallel because their vectors are not proportional.

If the lines were to intersect we would have three equations in two unknowns s and t:

4 + 2t = 2 + s

-5 + 4t = -1 + 3s

1 + 3t = 2s

Using the first two equations to solve for s and t we get s = -8 and t = -5:

```
2 = s - 2t  -> -6 = -3s + 6t   ->  2t = -10   -> t = -5  and s = -8
-4 = 3s - 4t     -4 = 3s - 4t

```
Using the last two equations to solve for s and t we get s = -16 and t = -11:

```
-4 = 3s - 4t -> -8 = 6s - 8t ->  t = -11 and s = -16
1 = 2s - 3t        3 = -6s + 9t

```

Thus the lines do not intersect because different values of t and s are obtained for solving the first and second, and then the second and third equations.

Since the lines are not parallel and they do not intersect, then they must be skew.

by: nl