Drone Inspection Paths and Safety Envelope
Two inspection drones move in a warehouse. Their positions in metres at time parameters t and s seconds are modelled by r_A(t) = (1, 2, 3) + t(2, −1, 1) and r_B(s) = (4, 0, 5) + s(−1, 2, 2), with 0 ≤ t ≤ 10 and 0 ≤ s ≤ 8. A safety engineer must decide whether the straight-line flight paths intersect, how close the drones actually come, and whether the 1.2 m separation rule is met. Path A also passes near a fixed crate at C = (3, 1, 5), which has a spherical no-fly zone of radius 1 m. Drone B can be shifted vertically by changing its altitude by d metres, so r_B becomes (4, 0, 5 + d) + s(−1, 2, 2), without changing its direction. The engineer wants to know whether a vertical shift is necessary and, under a stated approximation, how large it must be.
1. Show that the two infinite flight lines are skew, and find the shortest distance between them. Using the given time intervals, state whether the 1.2 m separation rule is violated.
Read the worked answer
The direction vectors are d₁ = (2, −1, 1) and d₂ = (−1, 2, 2). They are not scalar multiples, so the lines are not parallel. A vector between the starting points is b − a = (3, −2, 2). Now d₁ × d₂ = (−4, −5, 3), whose magnitude is √50 = 5√2.
The shortest distance between the infinite lines is |(b − a) · (d₁ × d₂)| / |d₁ × d₂| = |3(−4) + (−2)(−5) + 2(3)| / (5√2) = |−12 + 10 + 6| / (5√2) = 4/(5√2) ≈ 0.566 m.
However, the closest points on the infinite lines occur at s = −0.04, outside the allowed interval. On the actual path, the closest approach is at s = 0 and t = 5/3, giving distance √3/3 ≈ 0.577 m. Both values are below 1.2 m, so the separation rule is violated.
2. Determine whether path A enters the crate's spherical no-fly zone of radius 1 m centred at C = (3, 1, 5). Give the time parameter values where the boundary is crossed.
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A point on path A is (1 + 2t, 2 − t, 3 + t). Its displacement from C is (2t − 2, 1 − t, t − 2). The squared distance is
(2t − 2)^2 + (1 − t)^2 + (t − 2)^2 = 6t^2 − 14t + 9.
The minimum occurs at t = 14/(12) = 7/6, and the minimum squared distance is 5/6, so the minimum distance is √(5/6) ≈ 0.913 m, which is less than 1 m. Therefore path A enters the no-fly zone.
The boundary is crossed when 6t^2 − 14t + 9 = 1, i.e. 3t^2 − 7t + 4 = 0, so (3t − 4)(t − 1) = 0. Hence t = 1 and t = 4/3. Both lie in [0, 10], so the drone enters at t = 1 s and leaves at t = 4/3 s.
3. Drone B is shifted vertically by d metres. Assuming the closest-approach pair remains at s = 0 and t = 5/3, find the smallest d > 0 that makes the separation at least 1.2 m. Comment on whether this guarantees safety over the whole flight.
Read the worked answer
At s = 0 and t = 5/3, the unshifted separation vector is (1/3, 1/3, −1/3). After shifting B upward by d, the vertical component becomes −1/3 − d. The squared separation is
D² = (1/3)^2 + (1/3)^2 + (1/3 + d)^2 = 2/9 + (1/3 + d)^2.
We need D ≥ 1.2, so D² ≥ 1.44. Hence (1/3 + d)^2 ≥ 1.44 − 2/9 = 1.2178, giving 1/3 + d ≥ 1.1035 and d ≥ 0.770 m.
So the smallest shift under this assumption is about 0.77 m. This does not by itself guarantee safety over the whole flight: shifting B changes the closest-approach parameters, and the engineer should re-minimise the distance over the full intervals 0 ≤ t ≤ 10 and 0 ≤ s ≤ 8 before certifying the route.