Drone Corridors and the 0.8 km Exclusion Bubble
Two survey drones fly along straight paths over a construction site in Greater Noida, with positions in kilometres from a fixed origin and parameters t and s in minutes after take-off. Drone 1 is at r₁ = (1 + 2t)i + (3 − t)j + (2 + t)k. Drone 2 is at r₂ = (3 + s)i + (1 + s)j + (5 − s)k. The safety officer must certify that the paths never come closer than 0.8 km, the radius of the dust-and-debris exclusion bubble carried around each drone. The two operators filed separate plans and each claims 'we solved for the crossing point from the first two coordinates, so the paths meet and we have timed them to avoid each other'. The officer has one shift in which to decide whether to ground a drone, and wants the true minimum separation between the two flight paths, the points at which it occurs, and the minutes at which each drone reaches its closest point.
1. Decide whether the two paths intersect, are parallel, or are skew. Address the operators' reasoning directly.
Read the worked answer
Directions: b₁ = (2, −1, 1), b₂ = (1, 1, −1).
b₁ × b₂ = i[(−1)(−1) − (1)(1)] − j[(2)(−1) − (1)(1)] + k[(2)(1) − (−1)(1)] = (0, 3, 3) ≠ 0, so the lines are not parallel.
Try for an intersection. x and y give 1 + 2t = 3 + s ⇒ 2t − s = 2, 3 − t = 1 + s ⇒ t + s = 2. Adding: 3t = 4, t = 4/3, s = 2/3. Now test the third coordinate: z₁ = 2 + 4/3 = 10/3, z₂ = 5 − 2/3 = 13/3 - not equal.
So there is no common point: the paths are skew. The operators' method is exactly the trap - two equations in two unknowns always produce a 'solution', but a genuine intersection needs all three coordinates to agree. A second confirmation: with a₁ = (1,3,2) and a₂ = (3,1,5), (a₂ − a₁)·(b₁ × b₂) = (2, −2, 3)·(0, 3, 3) = −6 + 9 = 3 ≠ 0, which is the standard test for skew lines.
2. Find the minimum distance between the two paths.
Read the worked answer
For skew lines, the shortest distance is the projection of the join of any two points on the common normal:
d = |(a₂ − a₁)·(b₁ × b₂)| / |b₁ × b₂| = 3/|(0, 3, 3)| = 3/(3√2) = 1/√2 = 0.7071 km.
This is the perpendicular distance between the two infinite flight paths. Since 0.7071 < 0.8, the two 0.8 km bubbles overlap at the closest approach, so the plans as filed cannot be certified - the shortfall is 0.8 − 0.707 = 0.093 km. Note what the number does not mean: it is not the distance between the drones at an arbitrary instant, but the best separation the geometry ever permits.
3. Find the closest points on the two paths and the times at which they occur. Is there a cheap scheduling fix?
Read the worked answer
Let P = (1 + 2t, 3 − t, 2 + t) and Q = (3 + s, 1 + s, 5 − s). Then Q − P = (2 + s − 2t, −2 + s + t, 3 − s − t). Here b₁·b₂ = 2(1) + (−1)(1) + (1)(−1) = 0, so the directions are perpendicular and the minimum occurs where Q − P is perpendicular to both: (Q − P)·b₁ = (4 + 2s − 4t) + (2 − s − t) + (3 − s − t) = 9 − 6t = 0 ⇒ t = 1.5 (Q − P)·b₂ = (2 + s − 2t) − (−2 + s + t) − (3 − s − t) = −3 + 3s = 0 ⇒ s = 1
At t = 1.5 min, drone 1 is at (4, 1.5, 3.5); at s = 1 min, drone 2 is at (4, 2, 4). The gap is |(0, −0.5, 0.5)| = √0.5 = 0.707 km, matching part (b).
So the two drones reach their closest points at different clock times - drone 1 at 1.5 min, drone 2 at 1 min. They are never simultaneously at the pinch point, but the exclusion bubble is a sphere around each drone, so the 0.707 km separation still breaches the 0.8 km rule. Grounding is not automatically required: the officer should ask for either a small lateral shift of one path (about 0.1 km is enough to clear 0.8 km) or a schedule that keeps the drones away from the neighbourhood of the pinch point at the relevant minutes. The operators' claim of a 'crossing point' is rejected outright, since the paths are skew.