prove that the lines through a(0,-1,-1) and b(4,5,1) intersects the line through c (3,9,4) and d (-4,4,4). also find their point of intersection

L1:

x = 0 + 4t
y = -1 + 6t
z = -1 + 2t

L2:
x = 3 - 7s
y = 9 - 5s
z = 4 + 0s

L1 intersects L2 if

4t = 3-7s
-1+6t = 9-5s
-1+2t = 4

(t,s) = (5/2, -1) is the solution, so
(x,y,z) = (10,14,4)

direction vector of 1st line = (4,6,2) or reduced to (2,3,1)

parametric equation of 1st:
x = 0 + 2t
y = -1 + 3t
z = -1 + t

direction vector of 2nd line = (7, 5, 0)
parametric equation of 2nd:
x = 3 + 7k
y = 9 + 5k
z = 4

solving for x's
0+2t = 3 + 7k
2t = 3+ 7k (#1)

solving for y's
3t-1 = 5k+9
3t - 5k = 10 (#2)

using the z's :
t-1 = 4
t = 5

sub that into #1:
10 = 3 + 7k
k = 1

so t=5 and k = 1
check for consistency in #2 (the equation not used to solve)
3t - 5k = 10
LS = 15 - 5 = 10 = RS

so the point of intersection is: (Using the 1st, could use either one)
x = 0+2t = 10
y = -1+3t = 14
z = -1 + t = 4

They intersect at (10,14,4)

L(ab): (0,-1,-1)+(4u,6u,2u)

L(cd): (3,9,4)+(-7v,-5v,0)

If they intersect at P, then corresponding x,y,z values match, hence
2u-1 = 0 => u=2.5
where
P=(0+4*2.5, -1+6*2.5, -1+2*2.5)=(10,14,4)
For line L(cd),
we have P(3-7v, 9-5v, 4+0v)
=>
3-7v=10 => v=-1
or
9-5v=14 => v=-1
or
4+0v=4 => v=-1 (works).
Therefore they intersect at P, since P(10,14,4) lies on both lines.

To prove that two lines intersect, we need to show that the lines are not parallel. Two lines are parallel if their direction vectors are proportional.

Let's find the direction vectors of the two lines:

Line AB passing through points A(0, -1, -1) and B(4, 5, 1):
The direction vector of AB, let's call it vector v1, is calculated as:
v1 = B - A = (4, 5, 1) - (0, -1, -1) = (4, 6, 2)

Line CD passing through points C(3, 9, 4) and D(-4, 4, 4):
The direction vector of CD, let's call it vector v2, is calculated as:
v2 = D - C = (-4, 4, 4) - (3, 9, 4) = (-7, -5, 0)

Now, let's check if the direction vectors are proportional or not:
If v1 and v2 are proportional, then v1 = k * v2 for some scalar k.

v1 = (4, 6, 2)
v2 = (-7, -5, 0)

Equating the coordinates, we get:
4 = -7k
6 = -5k
2 = 0k

The second and third equations already show that k is not a real number because -5k cannot be equal to 6, and 0k cannot be equal to 2. Therefore, the lines are not parallel, and they intersect.

To find the point of intersection, we need to find the parametric equations of the two lines and solve them simultaneously.

The parametric equation of line AB passing through points A and B is:
x = 0 + 4t
y = -1 + 6t
z = -1 + 2t

The parametric equation of line CD passing through points C and D is:
x = 3 - 7s
y = 9 - 5s
z = 4

To find the point of intersection, we equate the x, y, and z coordinates of both lines and solve the system of equations:

0 + 4t = 3 - 7s
-1 + 6t = 9 - 5s
-1 + 2t = 4

From the third equation, we get:
2t = 5 --> t = 5/2

Substituting t = 5/2 into the first equation:
4 * (5/2) = 3 - 7s
10 = 3 - 7s
7s = -7
s = -1

Substituting s = -1 into the second equation:
-1 + 6t = 9 - 5*(-1)
-1 + 6t = 9 + 5
6t = 15
t = 15/6 --> t = 5/2

So the point of intersection of the two lines is when t = 5/2 and s = -1:

x = 0 + 4(5/2) = 10
y = -1 + 6(5/2) = 14
z = -1 + 2(5/2) = 4

Therefore, the lines intersect at the point (10, 14, 4).