Answer:
L1 and L2 are skew
Step-by-step explanation:
Since the equation of the line is
L1:x=9+6t,y=12-3t,z=3+9t
L2:x=4+16s, y=12-8s, z=16+20s
then if they intersect each other , they will have both in that point P=(xp , yp ,zp) then
1)9+6t = 4+16s
2) 12-3t =2-8s
3) 3+9t = 16+20s
adding 2*2) to 1)
9+6*t + 24-6t = 4+16*s + 4-16*s
33 = 8
since this is not possible , the error comes from our assumption that the lines intersect each other
then they are skew or parallel. They are parallel if their corresponding vectors are parallel , that is
L1 (x,y,z) = (9,12,3) + (6,-3,9)*t
L1 (x,y,z) = (4,2,16) + (16,-8,20)*t
then if they are parallel
(16,-8,20)= k*(6,-3,9)
16=6*k
-8 = -3*k
20= 9*k
since there is no k that satisfy for x , y and z simultaneously then L1 and L2 are not parallel
therefore L1 and L2 are skew
