Let $T T (0, 0, 0.1)$ denote the position of the control tower, $A (4, 3, 1)$ denote a point on the path of the aircraft and $F$ denote the point when the aircraft is at its closest point to the air traffic controllers.

\begin{aligned} \vec{A T} & = \vec{O T} - \vec{O A} \\ = (\begin{array}{c} 0 \\ 0 \\ 0.1 \end{array}) - (\begin{array}{c} 4 \\ 3 \\ 1 \end{array}) \\ = (\begin{array}{c} - 4 \\ - 3 \\ - 0.9 \end{array}) \end{aligned}

\begin{aligned} \vec{A F} & = (\vec{A T} \cdot \hat{𝐝}) \hat{𝐝} \\ = ((\begin{array}{c} - 4 \\ - 3 \\ - 0.9 \end{array}) \cdot \frac{(\begin{array}{c} 1 \\ - 1 \\ 2 \end{array})}{| (\begin{array}{c} 1 \\ - 1 \\ 2 \end{array}) |}) \frac{(\begin{array}{c} 1 \\ - 1 \\ 2 \end{array})}{| (\begin{array}{c} 1 \\ - 1 \\ 2 \end{array}) |} \\ = \frac{(- 4) (1) + (- 3) (- 1) + (- 0.9) (2)}{6} (\begin{array}{c} 1 \\ - 1 \\ 2 \end{array}) \\ = - \frac{7}{15} (\begin{array}{c} 1 \\ - 1 \\ 2 \end{array}) \end{aligned}

\begin{aligned} \vec{A F} & = \vec{O F} - \vec{O A} \\ - \frac{7}{15} (\begin{array}{c} 1 \\ - 1 \\ 2 \end{array}) & = \vec{O F} - (\begin{array}{c} 4 \\ 3 \\ 1 \end{array}) \\ \vec{O F} & = - \frac{7}{15} (\begin{array}{c} 1 \\ - 1 \\ 2 \end{array}) + (\begin{array}{c} 4 \\ 3 \\ 1 \end{array}) \\ = \frac{1}{15} (\begin{array}{c} 53 \\ 52 \\ 1 \end{array}) \end{aligned}

Coordinates of $F = (\frac{53}{15}, \frac{52}{15}, \frac{1}{15}) ∎$

(ii)

\begin{aligned} \vec{T F} & = \vec{O T} - \vec{O F} \\ = (\begin{array}{c} 0 \\ 0 \\ 0.1 \end{array}) - (\begin{array}{c} \frac{53}{15} \\ \frac{52}{15} \\ \frac{1}{15} \end{array}) \\ = \frac{1}{30} (\begin{array}{c} 106 \\ 104 \\ - 1 \end{array}) \end{aligned}

\begin{aligned} | \vec{T F} | & = | \frac{1}{30} (\begin{array}{c} 106 \\ 104 \\ - 1 \end{array}) | \\ = \frac{1}{30} \sqrt{22053} \\ = 4.9501 > 4 \end{aligned}

Hence the air traffic controllers will not see this aircraft.

(b)

Let $l_{1}$ and $l_{2}$ represent the paths of the aircraft and drone respectively.

l_{2} : 𝐫 = (\begin{array}{c} \frac{3}{2} \\ - 1 \\ k \end{array}) + μ (\begin{array}{c} \frac{3}{2} \\ 1 \\ - 1 \end{array})

We observe that $l_{1}$ and $l_{2}$ are not parallel since their direction vectors are not scalar multiples of each other.

Now assuming that the paths intersect,

(\begin{array}{c} 4 + λ \\ 3 - λ \\ 1 + 2 λ \end{array}) = (\begin{array}{c} \frac{3}{2} + \frac{3}{2} μ \\ - 1 + μ \\ k - μ \end{array})

\begin{array}{lrlr} 4 + λ & = \frac{3}{2} + \frac{3}{2} μ \\ 3 - λ & = - 1 + μ \\ 1 + 2 λ & = k - μ \end{array}

From $(1),$

\begin{array}{lrr} λ - \frac{3}{2} μ = - \frac{5}{2} \end{array}

From $(2),$

\begin{array}{lrr} λ + μ = 4 \end{array}

Solving $(4)$ and $(5)$ with a GC,

λ = \frac{7}{5}, μ = \frac{13}{5}

Since the paths do not intersect, considering equation $(3),$

\begin{aligned} 1 + 2 (\frac{7}{5}) & \neq k - (\frac{13}{5}) \\ k & \neq \frac{32}{5} \end{aligned}

Hence the possible values of $k$ are: $k \in ℝ^{+},$ $k \neq \frac{32}{5} ∎$

(c)

Let $θ$ denote the acute angle between the paths of the aircraft and the drone

\begin{matrix} | (\begin{array}{c} 1 \\ - 1 \\ 2 \end{array}) \cdot (\begin{array}{c} \frac{3}{2} \\ 1 \\ - 1 \end{array}) | = | (\begin{array}{c} 1 \\ - 1 \\ 2 \end{array}) | | (\begin{array}{c} \frac{3}{2} \\ 1 \\ - 1 \end{array}) | \cos θ \\ | (1) (\frac{3}{2}) + (- 1) (1) + (2) (- 1) | = \frac{1}{2} \sqrt{6} \sqrt{17} \cos θ \\ \cos θ = \frac{3}{\sqrt{6} \sqrt{17}} \\ θ = 72 . 7^{\circ} ∎ \end{matrix}

(d)

(i)

Let $B$ and $C$ denote the points $(4, 3, 1)$ and $(4.2, 0.8, 0.2)$ respectively

\begin{aligned} \vec{B C} & = \vec{O B} - \vec{O C} \\ = (\begin{array}{c} 4 \\ 3 \\ 1 \end{array}) - (\begin{array}{c} 4.2 \\ 0.8 \\ 0.2 \end{array}) \\ = (\begin{array}{c} - 0.2 \\ 2.2 \\ 0.8 \end{array}) \end{aligned}

\begin{aligned} Distance & = | \vec{B C} | \\ = | (\begin{array}{c} - 0.2 \\ 2.2 \\ 0.8 \end{array}) | \\ = 2.3495 \\ = 2.35 km (3 s.f.) ∎ \end{aligned}

(ii)

Let $𝐝_{1}$ and $𝐝_{2}$ denote the direction vectors of the paths of the aircraft and the drone respectively

\begin{aligned} \vec{B C} \cdot 𝐝_{1} \\ = (\begin{array}{c} - 0.2 \\ 2.2 \\ 0.8 \end{array}) \cdot (\begin{array}{c} 1 \\ - 1 \\ 2 \end{array}) \\ = (- 0.2) (1) + (2.2) (- 1) + (0.8) (2) \\ = - 0.8 \\ \neq 0 \end{aligned}

\begin{aligned} \vec{B C} \cdot 𝐝_{2} \\ = (\begin{array}{c} - 0.2 \\ 2.2 \\ 0.8 \end{array}) \cdot (\begin{array}{c} \frac{3}{2} \\ 1 \\ - 1 \end{array}) \\ = (- 0.2) (\frac{3}{2}) + (2.2) (1) + (0.8) (- 1) \\ = 1.1 \\ \neq 0 \end{aligned}

Hence $B C$ is not perpendicular to either $l_{1}$ or $l_{2},$ so the distance found in (d)(i) is not the shortest distance between the two paths $∎$