Autonomous Drone Flight Path Separation (Aerospace & Computer Engineering) Solution

Problem Overview

We evaluate the 3D spatial relationship between two drone trajectories $L_1$ and $L_2$ and compute the perpendicular clearance distance to verify collision avoidance envelopes.

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Step-by-Step Mathematical Solution

Step 1: Formulate Vector and Parametric Equations

State the Parametric Equations of a Line:

(Defined in Equations of Lines in 3D.)

• Drone 1 ($L_1$): Through $P(2, 0, 1)$ with $\vec{v}_1 = \langle 1, 2, -1 \rangle$:
  $$x_1(t) = 2 + t, \quad y_1(t) = 2t, \quad z_1(t) = 1 - t$$
• Drone 2 ($L_2$): Through $Q(1, 3, 2)$ with $\vec{v}_2 = \langle 2, -1, 1 \rangle$:
  $$x_2(s) = 1 + 2s, \quad y_2(s) = 3 - s, \quad z_2(s) = 2 + s$$

Step 2: Classify the Spatial Line Relationship

State the Line Classification Protocol: (Defined in Relationships Between Lines in Space.)

1. Parallel Test: Check direction vector proportionality:
   $$\frac{2}{1} \ne \frac{-1}{2} \implies \vec{v}_1 \nparallel \vec{v}_2 \quad (\text{Not Parallel})$$
2. Intersection Test: Set $\vec{r}_1(t) = \vec{r}_2(s)$:
   $$\begin{cases} 2 + t = 1 + 2s & (1) \\ 2t = 3 - s & (2) \\ 1 - t = 2 + s & (3) \end{cases}$$
   From (1): $t = 2s - 1$. Substitute into (2):
   $$2(2s - 1) = 3 - s \implies 4s - 2 = 3 - s \implies 5s = 5 \implies s = 1$$
   Then $t = 2(1) - 1 = 1$. Test $(t=1, s=1)$ in equation (3):
   $$1 - (1) \stackrel{?}{=} 2 + (1) \implies 0 \ne 3 \quad (\text{Inconsistent})$$
   Because the system is inconsistent and the direction vectors are non-parallel, $L_1$ and $L_2$ are Skew.

Step 3: Compute Perpendicular Clearance from $P(2, 0, 1)$ to Line $L_2$

State the Point-to-Line Distance Formula:

(Defined in Distance Between a Point and a Line in Space.)

1. Displacement vector from reference point $Q(1, 3, 2)$ to $P(2, 0, 1)$:
   $$\vec{QP} = \langle 2 - 1, \; 0 - 3, \; 1 - 2 \rangle = \langle 1, -3, -1 \rangle$$
2. Cross product $\vec{QP} \times \vec{v}_2$:
   $$\vec{QP} \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & -1 \\ 2 & -1 & 1 \end{vmatrix} = \langle (-3)(1) - (-1)(-1), \; (-1)(2) - (1)(1), \; (1)(-1) - (-3)(2) \rangle$$
   $$\vec{QP} \times \vec{v}_2 = \langle -3 - 1, \; -2 - 1, \; -1 + 6 \rangle = \langle -4, -3, 5 \rangle$$
3. Compute magnitudes:
   $$\|\vec{QP} \times \vec{v}_2\| = \sqrt{(-4)^2 + (-3)^2 + 5^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2}$$
   $$\|\vec{v}_2\| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$$
4. Calculate distance $d$:
   $$d = \frac{5\sqrt{2}}{\sqrt{6}} = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \approx 2.89\text{ meters}$$

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Engineering Conclusion

The perpendicular clearance distance between waypoint $P$ and flight corridor $L_2$ is $2.89\text{ m}$. Because $d = 2.89\text{ m} \ge 2.0\text{ m}$, the flight plan fulfills FAA safety clearance criteria and no emergency avoidance maneuver is triggered.