CHSE Odisha Elective Mathematics study materials
CHSE Odisha Elective Mathematics study resources

Drone Flight Paths and a Safety Bubble

CHSE Odisha Elective Mathematics · Put your knowledge to work in context.

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Drone Flight Paths and a Safety Bubble

Two survey drones fly straight lines over a proposed mining block. Using coordinates in kilometres, drone A’s path is r1(t) = (1, 2, 0) + t(3, −1, 2), and drone B’s path is r2(s) = (4, 0, 1) + s(−1, 2, 3), where t and s are independent path parameters. The operations manual requires a 0.50 km separation between the two flight paths at all points, not just at the same clock time, because scheduling can change. The survey manager has been told the paths look close near the middle of the block. A junior mapper solves the equations for a collision by setting t = s and finds no solution, and concludes the paths are safe. The manager wants a proper check: whether the paths intersect, the shortest distance between the two lines, and whether the 0.50 km rule is met. If not, the team must shift drone B’s path by a vector parallel to the x-axis and report the smallest shift that works.

1. Check properly whether the two paths intersect. Explain the error in setting t=s.

Read the worked answer

Set r1(t) = r2(s): 1 + 3t = 4 − s, 2 − t = 2s, 2t = 1 + 3s.

The first equation gives s = 3 − 3t. Substituting in the second: 2 − t = 2(3 − 3t) = 6 − 6t, so 5t = 4, t = 0.8 and s = 0.6. Check the third: 2t = 1.6, while 1 + 3s = 2.8. The three equations are inconsistent, so the lines do not intersect.

The junior mapper’s error was setting t = s. The two drones can pass the same point at different parameter values only if the lines intersect, and here they do not.

2. Find the shortest distance between the two flight paths.

Read the worked answer

Direction vectors d1 = (3, −1, 2) and d2 = (−1, 2, 3). A point on line 1 is a = (1, 2, 0), and on line 2 b = (4, 0, 1), so b − a = (3, −2, 1).

The cross product d1 × d2 = (−7, −11, 5), with norm sqrt(49 + 121 + 25) = sqrt(195) ≈ 13.96.

The shortest distance between the lines is |(b − a)·(d1 × d2)|/|d1 × d2| = |3(−7) + (−2)(−11) + 1(5)|/sqrt(195) = |−21 + 22 + 5|/sqrt(195) = 6/sqrt(195) ≈ 0.43 km.

Because the lines are skew (no intersection), this perpendicular distance is the minimum separation between the two flight paths.

3. Is the 0.50 km rule met? If not, find the smallest shift of drone B’s path parallel to the x-axis that satisfies the rule.

Read the worked answer

The required separation is 0.50 km, but the shortest distance is about 0.43 km, so the rule is not met.

Shift B’s path by (d, 0, 0): its point becomes b′ = (4 + d, 0, 1), and b′ − a = (3 + d, −2, 1). The new distance is |(b′ − a)·n|/|n| with n = (−7, −11, 5): |−7(3 + d) + 22 + 5|/sqrt(195) = |6 − 7d|/sqrt(195).

Require |6 − 7d| ≥ 0.50 sqrt(195) ≈ 6.98. This gives d ≤ −0.14 or d ≥ 1.85. The smallest shift is about 0.14 km in the negative x-direction. The manager should also check that this shifted path remains inside the survey block; otherwise the 1.85 km shift in the positive x-direction is the next option.