Skew Drone Paths
Two drones move inside a rectangular hangar; coordinates are in metres and time t is in seconds. Drone A has position r_A(t)=(1+2t, 4−t, 3+t), and drone B has position r_B(t)=(5−t, 1+2t, 2+2t), for t≥0. A collision alarm sounds if the straight-line distance between the drones is less than 1 m. A trainee computes the direction vectors u=(2,−1,1) and v=(−1,2,2), finds u·v=−2, and concludes that because the paths are not perpendicular they cannot intersect; the trainee then says no alarm is possible. Another trainee solves the line equations, finds no intersection, and wants to dismiss the alarm for that reason. The controllers need a correct decision: do the flight paths intersect, what is the minimum distance between the two path lines, and what is the minimum actual separation when both drones are at the same time t? Also identify whether the alarm sounds and at what time the closest simultaneous approach occurs.
1. Do the two path lines intersect? Evaluate the trainee's claim that non-perpendicular paths cannot intersect.
Read the worked answer
Use separate parameters: r_A(t)=P+t u and r_B(s)=Q+s v. Equating coordinates gives
1+2t=5−s, 4−t=1+2s, 3+t=2+2s.
The last two equations give t=1 and s=1, but then the first gives 3=4, which is impossible. Hence the lines are skew; they do not intersect. The dot product u·v=−2 only says the directions are not perpendicular. Non-perpendicular lines can intersect, so the trainee's inference is invalid.
2. Find the minimum distance between the two path lines.
Read the worked answer
Take P=(1,4,3), Q=(5,1,2), u=(2,−1,1), and v=(−1,2,2). Then Q−P=(4,−3,−1) and
u×v=(−4,−5,3).
The distance between the skew lines is
|(Q−P)·(u×v)|/|u×v| = |−16+15−3|/√50 = 4/(5√2)=2√2/5≈0.566 m.
This is the shortest gap between the infinite lines, but the closest points occur at different parameters s and t.
3. Find the minimum actual separation when both drones are at the same time t, decide whether the alarm sounds, and give the time.
Read the worked answer
Subtract positions:
r_A(t)−r_B(t)=(3t−4, 3−3t, 1−t).
Then
D(t)^2=(3t−4)^2+(3−3t)^2+(1−t)^2 =19t^2−44t+26 =19(t−22/19)^2+10/19.
Thus the minimum occurs at t=22/19≈1.16 s, with
D_min=√(10/19)≈0.725 m.
Since 0.725<1, the alarm sounds. This is larger than the line-to-line distance 0.566 m because that figure uses different times on the two paths.