Question

Determine whether the lines L1 and L2 are parallel, skew, or intersecting. L1:x=9+6t,y=12-3t,z=3+9t L2:x=4+16s, y=12-8s, z=16+20s parallel skew intersecting If they intersect, find the point of intersection. (If an answer does not exist, enter DNE.)

Answer 1

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