Drone Paths, Skew Lines and a No-Fly Sphere
Two survey drones fly straight-line paths in a hangar. Path 1 passes through A(1, 2, 3) with direction d1 = (2, -1, 1); path 2 passes through B(4, 0, 5) with direction d2 = (1, 1, 2). Distances are in metres and the paths are treated as infinite lines for planning. The safety officer wants three checks before authorising simultaneous flight: whether the two centre-lines intersect; the shortest distance between the centre-lines, compared with the required 1 m separation; and whether path 1 enters a spherical no-fly zone centred at C(3, 3, 3) with radius 1 m. A junior planner argues that because the two lines do not meet, they are automatically safe. The officer points out that skew lines can still pass close together, and that a line can miss a sphere even when its closest approach is not at one of the given base points. The team must produce the three numerical checks and a clear recommendation.
1. Show that the two paths are skew, not intersecting.
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Write the lines as
L1: (x, y, z) = (1, 2, 3) + t(2, -1, 1) L2: (x, y, z) = (4, 0, 5) + u(1, 1, 2).
Equate coordinates: 1 + 2t = 4 + u ⇒ 2t - u = 3 2 - t = u ⇒ t + u = 2 3 + t = 5 + 2u ⇒ t - 2u = 2.
From the second equation, u = 2 - t. Substitute into the first: 2t - (2 - t) = 3 ⇒ 3t = 5 ⇒ t = 5/3, u = 1/3.
Check the third equation: t - 2u = 5/3 - 2/3 = 1, but the required value is 2. The contradiction means no pair (t, u) satisfies all three equations. The lines are not parallel (d1 and d2 are not scalar multiples), so they are skew.
2. Find the shortest distance between the two centre-lines and decide whether the 1 m separation requirement is met.
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For skew lines, the shortest distance is
d = |(B - A) · (d1 × d2)| / |d1 × d2|.
Here B - A = (3, -2, 2), and d1 × d2 = (-3, -3, 3).
Dot product: (3, -2, 2) · (-3, -3, 3) = -9 + 6 + 6 = 3.
Magnitude: |d1 × d2| = sqrt(9 + 9 + 9) = 3sqrt(3).
Thus d = 3/(3sqrt(3)) = 1/sqrt(3) ≈ 0.577 m.
This is less than the required 1 m separation, so the infinite centre-lines come closer than the safety margin. However, this is a geometric distance, not proof of a timed collision: the drones may occupy the closest points at different times. The recommendation is not to authorise simultaneous flight without a time-separation plan or rerouting; the junior planner's 'they do not meet' argument is insufficient.
3. Determine whether path 1 enters the no-fly sphere centred at C(3, 3, 3) with radius 1 m.
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The distance from C to L1 is
D = |(C - A) × d1| / |d1|.
C - A = (2, 1, 0). Then (2, 1, 0) × (2, -1, 1) = (1, -2, -4).
Its magnitude is sqrt(1 + 4 + 16) = sqrt(21), and |d1| = sqrt(6). Therefore
D = sqrt(21)/sqrt(6) = sqrt(7/2) ≈ 1.871 m.
Since 1.871 m > 1 m, the line stays outside the sphere; path 1 does not enter the no-fly zone. The closest point is not one of the stated base points, which is why the perpendicular-distance formula is needed. The check is independent of the skew-line separation check: missing the sphere does not fix the close approach between the two paths.