Answer:
Step-by-step explanation:
Given are two lines equation in parametric form as
[tex]l1: x=3-2t, y=7+4t, z=-3+8t\\l2: x=-1-u, y=18+3u, z=7+2u[/tex]
The direction ratios of the I line are -2, 4 and 8
Ii line are -1, 3, 2
These two are not proportional
Hence l1 and l2 are not parallel
Either they are skew or intersect
If intersect common point P would have coordinates as
[tex]3-2t =- 1-u : 2t-u =4 \\7+4t = 18+3u: 4t-3u = 11\\-3+8t = 7+2u\\8t-2u = 10[/tex]
Let us solve first two equatins for u and t and check whether 3rd equation is satisfied by this.
2*(i) -ii gives u = -3
t =0.5
Substitute in III equation
8(0.5)-2(-3) = 10 so satisifed
These two are not skew lines
The point of intersection is got by substituting either t or u
Point of intersection (2, 9, 1)
