Find the distance between two skew lines.?

Question

Find the distance between two skew lines:
X=1+t, Y=1+6t, Z=2t
and
X=1+2s, Y=5+15s, Z=-2+6s

Answer 1

Find the distance between two skew lines:
L1: x = 1 + t, y = 1 + 6t, z = 2t
and
L2: x = 1 + 2s, y = 5 + 15s, z = -2 + 6s

The directional vector of L1 is v1 = <1, 6, 2>.
The directional vector of L2 is v2 = <2, 15, 6>.

The directional vector of the line connecting the closest points on each line is perpendicular to both of the directional vectors. Take the cross product.

n = v1 X v2 = <1, 6, 2> X <2, 15, 6> = <6, -2, 3>

Divide by the magnitude of n to get the unit normal vector.

|| n || = √[6² + (-2)² + 3²] = √(36 + 4 + 9) = √49 = 7

n / || n || = <6, -2, 3> / 7

Pick a point on each line. Set t and s equal to zero to pick.
P1(1, 1, 0) and P2(1, 5, -2).

u = P1P2 = <1-1, 5-1, -2-0> = <0, 4, -2>

Take the dot product to find the minimum distance between the two skew lines.

Distance = | u • n | / || n || = | <0, 4, -2> • <6, -2, 3> | / 7
Distance = | 0 - 8 - 6 | / 7 = 14/7 = 2
_______

Please show the courtesy of picking someone's answer for best answer.

Answer 2

pizzazz

♠ start choosing answers you ∫hίt hεad! Otherwise you never get any help;
♣ thus
♪ X=1+t, Y=1+6t, Z=2t;
♫ X=1+2s, Y=5+15s, Z=-2+6s;
♥ let point A belong to line (♪) and point B belong to line (♫), the distance squared
AB^2 being r^2= [(1+2s)-(1+t)]^2 + [(5+15s)-(1+6t)]^2 + [(-2+6s)-(2t)]^2; hence
d(r^2) = (∂r^2/∂s)*ds +(∂r^2/∂t)*dt; and distance in question means shortest r, that is d(r^2)=0, or dr=0 or
∂r^2/∂s=0 and ∂r^2/∂t=0;
1♦ 0.5*(∂r^2/∂s) =
= [(1+2s)-(1+t)]*2 + [(5+15s)-(1+6t)]*15 + [(-2+6s)-(2t)]*6 =0;
2♦ -0.5*(∂r^2/∂t) =
= [(1+2s)-(1+t)]*1 + [(5+15s)-(1+6t)]*6 + [(-2+6s)-(2t)]*2 =0;
♦ now solving system (1♦-2♦) find s and t, then plug them into r in (♥);
I’m not going to nurse you further on;

Answer 3

How do you define a distance between two skew lines?