Question

Question #2eed1

Answer 1

They do not intersect

Explanation:

There are 3 kinds lines in 3D space . Lines which are parallel , lines which intersect and lines that are skewed (neither intersecting nor parallel ).

L1 is of the form [2,2,3] + t [-3,6,1]
L2 is of the form [3,5,-2] + s[2,1,4] .

The lines are intersecting if there are real values of t and s , such that [2,2,3] + t[-3,6,1] = [3,5,-2] + s[2,1,4].
Lets check that ,
2 - 3t = 3 + 2s -> equation 1
2 + 6t = -5 + s -> equation 2
3 + t = -2 + s -> equation 3

2 * (equation 1) + equation 2 => 6 = 1 + 5s
=> 5s = 5
=> `s = 1`
substitute value of s in equation 2 to get value of t ,
2 + 6t = -5 + 1

=> 6t = -5+1-2 = -6
=>`t = -1`.
With value of t and s , we can check if LHS of equation 3 = RHS.
lhs = 3 + (-1) = 2
rhs = -2 + 1 = -1.
well , lhs `cancel=` rhs . So the lines are not intersecting.

Next lets check if they are parallel . Two lines are parallel if their direction ratios/ direction cosines are equally proportional

`(-3t)/(2s) cancel= (6t)/s cancel= t/(4s)` . They are not proportional . So they must be skew lines.

L1 is of the form `vec a_1 + t vecb_1` and L2 is of the form `veca_2 + s vecb_2`.

Distance between skew lines is given by `abs((veca_2 - veca_1) . (vecb_1 X vecb_2))/abs(vecb_1X vecb_2)`.

hope this helps !