Drone path, restricted plane, and tower clearance
An inspection drone follows the straight-line path r(λ)=(1+2λ)i+(2−λ)j+(1+3λ)k, λ≥0, distances in km. A sensitive installation is protected by the plane 2x+y−z=4; the restricted side is 2x+y−z>4. The control tower is at the origin O. The drone's manufacturer warns that the drone must stay at least 2.5 km from the tower, or it loses line-of-sight. The flight controller has to decide whether the planned path enters the restricted airspace, find the closest point on the path to the tower, and determine the angle between the path and the boundary plane. The drone starts at λ=0 and travels in the direction of increasing λ; it does not turn. The planned path is fixed. Use exact surd forms where possible and give distances to three significant figures.
1. Determine whether the path intersects the plane, find the distance from the path to the plane, and decide whether the drone enters the restricted side.
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Substitute the line into the plane expression: 2x + y − z = 2(1 + 2λ) + (2 − λ) − (1 + 3λ) = 2 + 4λ + 2 − λ − 1 − 3λ = 3. This is independent of λ, so the path never satisfies 2x + y − z = 4; it is parallel to the plane. Since 3 < 4, every point lies on the side 2x + y − z < 4, not the restricted side > 4. The distance from the line to the plane equals the distance from any point on the line to the plane: |3 − 4|/√(2² + 1² + (−1)²) = 1/√6 ≈ 0.408 km. The drone does not enter the restricted airspace.
2. Find the point on the allowed path closest to the tower and assess the 2.5 km warning.
Read the worked answer
The squared distance from the origin to the point at parameter λ is D² = (1 + 2λ)² + (2 − λ)² + (1 + 3λ)² = 14λ² + 6λ + 6. Differentiating gives d(D²)/dλ = 28λ + 6. The stationary point λ = −3/14 is negative and is not on the allowed path λ ≥ 0. On λ ≥ 0 the derivative at λ = 0 is 6 > 0 and increases, so D² increases from λ = 0. Hence the closest allowed point is λ = 0, i.e. (1, 2, 1), and the distance is √6 ≈ 2.449 km. Since 2.449 < 2.5, the drone already violates the 2.5 km line-of-sight warning at the start; it does not become safe later along the path.
3. Find the angle between the path and the boundary plane, and interpret the result.
Read the worked answer
The direction vector is d = (2, −1, 3) and the plane normal is n = (2, 1, −1). Their dot product is d·n = 4 − 1 − 3 = 0, so d is perpendicular to n and therefore parallel to the plane. The angle between the path and the plane is 0°. This agrees with the substitution in the first task: a parallel path can never cross the boundary. If the controller changes the path, a nonzero d·n would be needed to intersect the plane, and then the sign of the plane expression on both sides would determine entry.