Drone Corridor and a Circular No-Fly Zone
A drone flies in a straight line across a coordinate grid where one unit represents 100 m. It starts at point A(1, 2) and is programmed to reach point B(9, 8) at constant speed. A circular restricted zone is centred at C(5, 6) and has radius 2 units. The flight controller represents the path as r(t) = (1 + 0.8t)i + (2 + 0.6t)j, where t is the distance travelled in grid units from A, so 0 ≤ t ≤ 10. A reviewer claims the drone is safe because the straight-line distance from C to A is greater than 2 units and the distance from C to B is also greater than 2 units. The controller must determine whether the path actually enters the zone, the length of the path inside it, and the closest approach distance. The drone's battery allows a detour only if the path inside the zone exceeds 300 m.
1. Find the closest distance from C to the drone's path and decide whether the path enters the restricted zone. Explain why the reviewer's endpoint check is insufficient.
Read the worked answer
Use vector projection. AB = (8, 6) and AC = (4, 4). The closest point on the line to C is P = A + λAB, where λ = (AC·AB)/|AB|² = (4×8 + 4×6)/(8² + 6²) = (32 + 24)/100 = 56/100 = 0.56.
Thus P = (1 + 0.56×8, 2 + 0.56×6) = (5.48, 5.36). The closest distance is CP = √[(5.48 − 5)² + (5.36 − 6)²] = √(0.48² + 0.64²) = √(0.2304 + 0.4096) = √0.64 = 0.8 units.
Since 0.8 < 2, the path enters the zone. Checking only the endpoints is insufficient because both endpoints can lie outside a circle while the middle of the segment passes through it; here CA ≈ 5.66 and CB ≈ 4.47, yet the closest approach is only 0.8.
2. Find the values of t where the path crosses the boundary of the zone and calculate the length of the path inside the zone. Does the battery rule require a detour?
Read the worked answer
Solve |r(t) − C|² = 4: (0.8t − 4)² + (0.6t − 4)² = 4. Expanding gives 0.64t² − 6.4t + 16 + 0.36t² − 4.8t + 16 = 4, so t² − 11.2t + 28 = 0. The discriminant is 11.2² − 4×28 = 125.44 − 112 = 13.44, and √13.44 ≈ 3.666. Hence t = (11.2 ± 3.666)/2, giving t ≈ 3.767 and t ≈ 7.433. Both lie in [0, 10].
The length inside the zone is (7.433 − 3.767) = 3.666 units. Since each unit is 100 m, this is about 366.6 m, which exceeds 300 m. Under the stated battery rule, a detour is required.
3. The controller proposes shifting every point of the path 2 units north. Does this clear the restricted zone? Justify with a distance calculation.
Read the worked answer
The original line through A and B has equation 3x − 4y + 5 = 0. Shifting the path 2 units north replaces y by y − 2, giving 3x − 4(y − 2) + 5 = 0, or 3x − 4y + 13 = 0.
The distance from C(5, 6) to this new line is |3(5) − 4(6) + 13|/√(3² + 4²) = |15 − 24 + 13|/5 = 4/5 = 0.8 units. The line has moved to the other side of C but is still only 0.8 units away, so the path still enters the zone. A shift of d units north changes the distance to |4d − 4|/5; clearing the zone needs |4d − 4|/5 > 2, so d > 3.5. A 2-unit shift is therefore insufficient.