Lines r=(2+2t)i+(3-3t)j+(4-2t)k & v=(4+2s)i+(1-6s)j+(5-3s)k are skew. Find the perpendicular distance between?

Question

them at the point(s) where they "cross" over each other. So like imagine if the lines where in 2D and find their intersection then find how far each line is from that intersection. It sounds like I get it except I have no clue how to even solve the problem or even begin it.

Answer 1

Find the minimum (perpendicular) distance between the two skew lines:

r(t) = (2+2t)i + (3-3t)j + (4-2t)k
v(s) = (4+2s)i + (1-6s)j + (5-3s)k

First get the directional vectors for each line.

r(t) = <2, 3, 4> + t<2, -3, -2>
v(s) = <4, 1, 5> + s<2, -6, -3>

For r its u = <2, -3, -2>
For v its w = <2, -6, -3>

Take the cross product of u and w to get the vector n that is normal to both of them.

n = u X w = <-3, 2, -6>

The magnitude of n is calulated below.

|| n || = √[(-3)² + 2² + (-6)²] = √(9 + 4 + 36) = √49 = 7

Pick and a point on each line by setting t and s to zero.

From line r we have P(2, 3, 4).
From line v we have Q(4, 1, 5).

Create a vector between the two points.

m = PQ = Q - P = <4-2, 1-3, 5-4> = <2, -2, 1>

The minimum distance d, between the skew lines is:

d = || m • n || / || n || = | <2, -2, 1> • <-3, 2, -6> | / 7

d = | - 6 - 4 - 6 | / 7 = 16/7