Drone Clearance Check Over a Munnar Hill Site
Two survey drones fly straight paths over a hill site near Munnar. Coordinates are in metres, with the origin at the base station. Drone P follows $\mathbf{r}=(2\mathbf{i}+3\mathbf{j}+\mathbf{k})+t(3\mathbf{i}-\mathbf{j}+2\mathbf{k})$; drone Q follows $\mathbf{r}=(\mathbf{i}+4\mathbf{j}+2\mathbf{k})+s(2\mathbf{i}+\mathbf{j}-\mathbf{k})$. Flight control computes the shortest distance between the two paths before clearing the flight; the safety rule is a clearance of at least 1.5 m.
An operator argues that the paths must cross, because the direction vectors $(3,-1,2)$ and $(2,1,-1)$ are not parallel. He also proposes raising drone P by 0.5 m in altitude to gain clearance. The controller wants the actual shortest distance between the two paths, a verdict on whether the lines meet at all, and an assessment of the operator's altitude fix before signing the clearance.
1. Compute the shortest distance between the two flight paths and give a verdict against the 1.5 m rule.
Read the worked answer
Take $\mathbf{a}_1=(2,3,1)$, $\mathbf{d}_1=(3,-1,2)$ and $\mathbf{a}_2=(1,4,2)$, $\mathbf{d}_2=(2,1,-1)$. Then $\mathbf{a}_2-\mathbf{a}_1=(-1,1,1)$ and
$$\mathbf{d}_1\times\mathbf{d}_2=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\3&-1&2\2&1&-1\end{vmatrix}=(-1,,7,,5)$$
$$|\mathbf{d}_1\times\mathbf{d}_2|=\sqrt{1+49+25}=\sqrt{75}=5\sqrt{3}\approx 8.660$$
Shortest distance for skew lines:
$$d=\frac{|(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{d}_1\times\mathbf{d}_2)|}{|\mathbf{d}_1\times\mathbf{d}_2|}=\frac{|1+7+5|}{8.660}=\frac{13}{8.660}\approx 1.501\text{ m}$$
The paths pass the 1.5 m rule, but by roughly one millimetre - well inside survey and GPS error. A controller should treat this as marginal rather than as a comfortable pass, and should not clear the flight on the strength of the third decimal place.
2. Is the operator right that non-parallel direction vectors mean the paths cross? Settle the question using the equations of the two lines.
Read the worked answer
No. In three dimensions two non-parallel lines are usually skew, not intersecting; parallelism is only one of three possibilities (parallel, intersecting, skew).
The lines meet only if there exist $t$ and $s$ satisfying all three coordinates simultaneously:
$$2+3t=1+2s,\qquad 3-t=4+s,\qquad 1+2t=2-s$$
From the first, $3t-2s=-1$; from the second, $t=-1-s$. Substituting: $3(-1-s)-2s=-1$, giving $-5s=2$, so $s=-0.4$ and $t=-0.6$.
Check the third coordinate: $1+2(-0.6)=-0.2$, while $2-(-0.4)=2.4$. These are unequal, so no common point exists.
The same conclusion comes from the scalar triple product: $[(\mathbf{a}_2-\mathbf{a}_1),\mathbf{d}_1,\mathbf{d}_2]=13\neq 0$. The lines are skew, and the operator's inference is invalid.
3. Evaluate the proposal to raise drone P by 0.5 m. Does it improve the clearance?
Read the worked answer
Raising P by 0.5 m changes its anchor point to $\mathbf{a}_1'=(2,3,1.5)$, so $\mathbf{a}_2-\mathbf{a}_1'=(-1,1,0.5)$.
The normal stays $\mathbf{n}=(-1,7,5)$, so
$$d'=\frac{|(-1)(-1)+(1)(7)+(0.5)(5)|}{8.660}=\frac{1+7+2.5}{8.660}=\frac{10.5}{8.660}\approx 1.21\text{ m}$$
The clearance falls from about 1.501 m to about 1.21 m, which breaches the 1.5 m rule. The reason is that the separation depends only on the component of the anchor offset along $\mathbf{n}=(-1,7,5)$; the vertical shift contributed $+0.5\times 5=+2.5$ to a dot product that was previously 13, shrinking it rather than enlarging it. Moving P upward moved it toward Q's path, not away.
The fix should be rejected. Clearance must be gained by shifting the offset in the direction of $\mathbf{n}$, for example by altering the lateral position or the timing of the two flights.