By reduction to the absurd, the two lines do not intercept each other because of the following absurd: 5 = - 4.
From statement we have two lines in space, whose vector forms are described below:
(x, y, z) = (1, - 9, 1) + t · (4, 1, 2) (1)
(x, y, z) = (1, 0, - 9) + s · (5, - 1, 5) (2)
The point of intersection satisfies the following condition:
(1, - 9, 1) + t · (4, 1, 2) = (1, 0, - 9) + s · (5, - 1, 5)
(1, - 9, 1) - (1, 0, - 9) = s · (5, - 1, 5) + t · (- 4, - 1, - 2)
s · (5, - 1, 5) + t · (- 4, - 1, - 2) = (0, - 9, - 8)
There is a point of intersection if the resulting system of linear equations has at least a solution:
5 · s - 4 · t = 0 (2)
- s - t = - 9 (3)
5 · s - 2 · t = - 8 (4)
By (2):
s = 4 · t/5
(2) in (3):
- 4 · t/5 - t = - 9
- 9 · t/5 = - 9
t/5 = 1
t = 5
(2) in (4):
5 · (4 · t/5) - 2 · t = - 8
4 · t - 2 · t = - 8
2 · t = - 8
t = - 4
By reduction to the absurd, the two lines do not intercept each other because of the following absurd: 5 = - 4.
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