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Determine whether the lines L1 and L2 are parallel, skew, or intersecting.
L1: x = 6+4t, y = 8-2t, z = 2+6t
L2: x = 4+16s, y = 12-8s, z = 16+20s
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Consider the guiding vectors of these lines.
The guiding vectors are comprised of the coefficients of parametric equations.
For line L1 the guiding vector is ( 4,-2,6);
For line L2 the guiding vector is (16,-8,20).
It is clearly seen that the guiding vectors are not proportional - hence, lines L1 and L2 are not parallel.
To find out, if the lines L1 and L2 do intersect, you should consider this system of 3 equations in 2 unknowns
6 + 4t = 4 + 16s, (1)
8 - 2t = 12 - 8s, (2)
2 + 6t = 16 + 20t. (3)
Multiply equation (2) by 2 (both sides). Keep equation (1) as is. You will get
6 + 4t = 4 + 16s, (1')
16 - 4t = 24 - 16s. (2')
Add equations (1) and (2'). The terms with "t" will cancel each other; the terms with "s" will cancel each other, too.
Thus, you will get a self-contradictory equality
22 = 28.
It means that the system of equations (1), (2) has no solutions (is inconsistent).
From it, we conclude that the system of equations (1), (2), (3) has no solutions.
So, lines L1 and L2 are skew lines in 3D : they are not parallel and are not intersecting.
Solved.