Some vectors help needed pt2?

Question

If the lines L1 and L2 have equations as follows:

L1 =(3,1,0) + t(1,2,4) and L2 = (1,-1,1) + s(2,1,-1)
where t and s are parameters.

How can I:

Show that L1 passes through the point (2, -1, -4) but L2 does not pass through this point, as well as

Find the acute angle between L2 and the line joining the points (1, -1, 1) and (2, -1, -4).

Answer 1

If the lines L1 and L2 have equations as follows:

L1 = (3,1,0) + t(1,2,4) and L2 = (1,-1,1) + s(2,1,-1)
where t and s are parameters.

1) Show that L1 passes through the point (2, -1, -4) but L2 does not pass through this point.

Solve for t and s. If you get an inconsistency then the point is not on the line.

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

z: 4t = -4
t = -1
y: 1 + 2*(-1) = -1
x: 3 - 1 = 2

This is consistent so the point (2, -1, -4) is on L1 at t = -1.

L2:
x = 1 + 2s = 2
y = -1 + s = -1
z = 1 - s = -4

y: -1 + s = -1
s = 0
x: 1 + 0 = 2 an inconsistency

This is inconsistent so the point (2, -1, -4) not is on L2.
_________

2) Find the acute angle between L2 and the line joining the points P(1, -1, 1) and Q(2, -1, -4).

The directional vector of L2 is u = <2, 1, -1>.

The directional vector of the line PQ is

v = <2-1, -1--1, -4-1> = <1, 0, -5>

Use the dot product to find the angle between the lines.

cosθ = (u • v) / || u || || v ||

cosθ = (<2, 1, -1> • <1, 0, -5>) / || u || || v ||

cosθ = (2 + 0 + 5) / {√[2² + 1² + (-1)²] * √[1² + 0² + (-5)²]}

cosθ = 7 / {√6 * √26} = 7 / √156

θ = arccos(7/√156) ≈ 55.9°

Answer 2

Set L1 equal to that point and solve for t. If a t exists, then it passes through the point. Do same for line L2, if there isn't a value that u can get for s that doesn't give u that point, then it doesn't pass through that point.
For part b,
find the vector between the two points, then dot product the two vectors, then divide by the product of the two magnitudes. This is cos(theta) So take arccose of that value.