Drone Flight Paths: Skew Lines and Coplanarity
A drone control team models two straight flight paths in metres relative to a tower at the origin. Drone 1 follows r1=(1,2,3)+t(3,4,0), 0≤t≤1, from A(1,2,3) to B(4,6,3). Drone 2 follows r2=(2,0,5)+s(1,1,−1), s∈R, through C(2,0,5) with direction d2=(1,1,−1). The controller wants to know whether the paths intersect, whether they are coplanar, and the minimum separation if they are skew. A safety rule requires at least 2 m separation between the two paths. A third point D(5,6,4) is suggested as a waypoint; a trainee claims A, B, C and D must be coplanar because all four points are in the same block of airspace. The controller is unsure whether checking intersection is enough and whether the closest approach might occur beyond the segment flown by Drone 1.
1. Determine whether the two lines intersect. What does a negative result imply about coplanarity of the two lines?
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Equate the coordinates: 1+3t=2+s, 2+4t=s, and 3=5−s. The third equation gives s=2. Substituting into the second gives 2+4t=2, so t=0. The first equation then gives 1+0=2+2, i.e. 1=4, which is false. Hence the lines do not intersect. Their direction vectors (3,4,0) and (1,1,−1) are not scalar multiples, so the lines are not parallel. Non-parallel, non-intersecting lines in space are skew and therefore not coplanar.
2. Compute the shortest distance between the two lines and decide whether the 2 m rule is met. Verify whether the closest point on Drone 1's line lies on its segment.
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Use d=|(a2−a1)·(d1×d2)|/|d1×d2|, with a1=(1,2,3), a2=(2,0,5), d1=(3,4,0), d2=(1,1,−1). Now a2−a1=(1,−2,2) and d1×d2=(−4,3,−1). The dot product is (1)(−4)+(−2)(3)+(2)(−1)=−12, and |d1×d2|=√(16+9+1)=√26. Thus d=12/√26≈2.35 m, which is at least 2 m, so the rule is met. For the closest points, write w=(1+s−3t, s−2−4t, 2−s). The conditions w·d1=0 and w·d2=0 give −5+7s−25t=0 and −3+3s−7t=0. Solving gives t=3/13≈0.231 and s=20/13≈1.538. Since t lies in [0,1], the closest approach occurs on Drone 1's actual segment, so the separation figure applies to the flight.
3. Test the trainee's claim that A, B, C and D are coplanar using a scalar triple product.
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Take AB=(3,4,0), AC=(1,−2,2) and AD=(4,4,1). First compute AC×AD: (−2·1−2·4, 2·4−1·1, 1·4−(−2)·4) = (−10,7,12). Then AB·(AC×AD)=3(−10)+4(7)+0(12)=−30+28=−2. Since the scalar triple product is not zero, the four points are not coplanar. Being in the same block of airspace is not a geometric coplanarity condition; the trainee's reasoning is invalid.