Drone Path, Tower Shadow and Skew Flight Lines
A survey drone follows a straight path modelled by L1: r=(1,2,3)+t(2,-1,2), t∈R, with distances in tens of metres. A communication tower stands at P(5,0,1). The flight engineer needs the shortest distance from the tower to the drone path and the point on the path closest to the tower, because a signal shadow is suspected within 5 units. A second drone is assigned path L2: r=(6,0,0)+s(1,1,-1), s∈R. The operations manual says two paths are conflicting only if they intersect; if they do not intersect and are not parallel, they are classified as skew. The engineer has no graph paper and must decide from the vector equations whether the two paths conflict. A previous intern claimed that because the direction vectors are not parallel, the lines must intersect somewhere.
1. Compute the shortest distance from P to L1 and find the foot of the perpendicular on L1.
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Take A=(1,2,3) on L1 and d=(2,-1,2). Then AP=P-A=(4,-2,-2). AP×d = (-6,-12,0), whose magnitude is √(36+144)=6√5. Also |d|=√9=3. Distance = |AP×d|/|d| = 6√5/3=2√5≈4.47 units. The foot is A+td with t=(AP·d)/|d|^2=(8+2-4)/9=6/9=2/3. So the closest point is (1+4/3, 2-2/3, 3+4/3)=(7/3,4/3,13/3). Since 4.47<5, the suspected shadow zone is real.
2. Determine whether L1 and L2 intersect. If not, classify their relationship and correct the intern's claim.
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Equate coordinates: 1+2t=6+s, 2-t=s, 3+2t=-s. From the first two equations, s=2t-5 and s=2-t, so 2t-5=2-t, giving 3t=7 and t=7/3. Then s=2-7/3=-1/3. Check the third coordinate: L1 gives 3+2(7/3)=23/3, while L2 gives -s=1/3. These are not equal, so no triple (t,s) satisfies all three equations. The lines do not intersect. Their direction vectors (2,-1,2) and (1,1,-1) are not scalar multiples, so the lines are not parallel. Hence they are skew. Non-parallel directions in 3D do not guarantee intersection; the intern's claim is false.