An airfield on a horizontal flat piece of ground must manage the take-off flight paths of aircraft from its runways. These flight paths may be assumed to be straight lines, and the aircraft may be assumed to be point objects. In the diagram below, the labelled points are defined relative to an origin O with coordinates $(0,0,0)$ on the airfield. The x-, y- and z- axes are in the directions east, north and vertically upwards respectively, with units in metres. The direction vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ correspond to the east, north and vertically upwards directions respectively. The point A has coordinates $(2000,500,0)$ and the point B has coordinates $(1400,2900,0)$. Aircraft Alpha takes off from Point A in the direction $4\mathbf{i}+3\mathbf{j}+\mathbf{k}$ as shown. The peak of a majestic mountain has coordinates $(18000,4500,2000)$.

1. Identify the direction vector of the flight path and the normal vector to the horizontal ground.

The direction vector of Aircraft Alpha's flight path is $\mathbf{d} = \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix}$.
The horizontal ground corresponds to the $xy$-plane, which has a normal vector $\mathbf{n} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

2. Apply the scalar product formula to find the angle between the line and the plane.

Let $\theta$ be the angle between the flight path and the horizontal ground. Using the formula $\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$:
$\sin \theta = \frac{\left| \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right|}{\sqrt{4^2 + 3^2 + 1^2} \sqrt{0^2 + 0^2 + 1^2}}$
$\sin \theta = \frac{1}{\sqrt{26} (1)}$
$\theta = \sin^{-1}\left(\frac{1}{\sqrt{26}}\right) \approx 11.3099^\circ$
$\theta \approx 11.3^\circ \text{ (to 1 d.p.)}$

1. Set up the position vector of a general point on the flight path of Aircraft Alpha.

Let $F$ be the closest point on the flight path. The line passes through $A(2000, 500, 0)$ with direction vector $\mathbf{d} = \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix}$.
$\overrightarrow{OF} = \begin{pmatrix} 2000 \\ 500 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 2000 + 4\lambda \\ 500 + 3\lambda \\ \lambda \end{pmatrix}$

2. Express the displacement vector from the peak to the general point $F$.

Let the peak be $P(18000, 4500, 2000)$.
$\overrightarrow{PF} = \overrightarrow{OF} - \overrightarrow{OP} = \begin{pmatrix} 2000 + 4\lambda \\ 500 + 3\lambda \\ \lambda \end{pmatrix} - \begin{pmatrix} 18000 \\ 4500 \\ 2000 \end{pmatrix} = \begin{pmatrix} 4\lambda - 16000 \\ 3\lambda - 4000 \\ \lambda - 2000 \end{pmatrix}$

3. Apply the condition that the shortest distance occurs when $\overrightarrow{PF}$ is perpendicular to the flight path direction.

For $F$ to be the closest point, $\overrightarrow{PF}$ must be perpendicular to $\mathbf{d}$. Hence, $\overrightarrow{PF} \cdot \mathbf{d} = 0$.
$\begin{pmatrix} 4\lambda - 16000 \\ 3\lambda - 4000 \\ \lambda - 2000 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} = 0$
$4(4\lambda - 16000) + 3(3\lambda - 4000) + 1(\lambda - 2000) = 0$
$16\lambda - 64000 + 9\lambda - 12000 + \lambda - 2000 = 0$
$26\lambda - 78000 = 0 \implies 26\lambda = 78000 \implies \lambda = 3000$

4. Find the coordinates of the closest point $F$.

Substitute $\lambda = 3000$ back into $\overrightarrow{OF}$:
$\overrightarrow{OF} = \begin{pmatrix} 2000 + 4(3000) \\ 500 + 3(3000) \\ 3000 \end{pmatrix} = \begin{pmatrix} 14000 \\ 9500 \\ 3000 \end{pmatrix}$
Therefore, the coordinates of the closest point are $(14000, 9500, 3000)$.

5. Calculate the magnitude of $\overrightarrow{PF}$ to find the closest distance.

Substitute $\lambda = 3000$ into $\overrightarrow{PF}$:
$\overrightarrow{PF} = \begin{pmatrix} 4(3000) - 16000 \\ 3(3000) - 4000 \\ 3000 - 2000 \end{pmatrix} = \begin{pmatrix} -4000 \\ 5000 \\ 1000 \end{pmatrix}$
Distance $= |\overrightarrow{PF}| = \sqrt{(-4000)^2 + 5000^2 + 1000^2}$
$= \sqrt{16000000 + 25000000 + 1000000} = \sqrt{42000000} \approx 6480.74$
The closest distance is $6481\text{ m}$ (to the nearest metre).

1. Set up the parametric vector equations for the flight paths of Aircraft Alpha and Aircraft Bravo.

Let the flight path of Aircraft Alpha be $l_1$ with parameter $\lambda$, and the flight path of Aircraft Bravo be $l_2$ with parameter $\mu$:
$l_1: \mathbf{r} = \begin{pmatrix} 2000 \\ 500 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix}$
$l_2: \mathbf{r} = \begin{pmatrix} 1400 \\ 2900 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 2 \\ h \end{pmatrix}$

2. Equate the $x$ and $y$ components of the two lines to form a system of equations.

For the flight paths to cross, the lines must intersect, so there exist $\lambda$ and $\mu$ such that the coordinates are equal.
Equating $x$-components:
$2000 + 4\lambda = 1400 + 4\mu \implies 4\mu - 4\lambda = 600 \implies \mu = \lambda + 150 \quad \text{--- (1)}$
Equating $y$-components:
$500 + 3\lambda = 2900 + 2\mu \implies 3\lambda - 2\mu = 2400 \quad \text{--- (2)}$

3. Solve the system of equations to find the values of $\lambda$ and $\mu$.

Substitute (1) into (2):
$3\lambda - 2(\lambda + 150) = 2400$
$3\lambda - 2\lambda - 300 = 2400$
$\lambda = 2700$
Substituting $\lambda = 2700$ back into (1):
$\mu = 2700 + 150 = 2850$

4. Equate the $z$ components and substitute the values of $\lambda$ and $\mu$ to find $h$.

Equating $z$-components:
$0 + 1\lambda = 0 + h\mu \implies \lambda = h\mu$
Substitute $\lambda = 2700$ and $\mu = 2850$:
$2700 = h(2850)$
$h = \frac{2700}{2850} = \frac{18}{19}$
Rounding to 3 decimal places:
$h \approx 0.947$

1. Find the normal vector to the plane containing point $B$ and the flight path of Aircraft Alpha.

Let the plane be $\Pi$ and its normal vector be $\mathbf{n}$.
The plane contains point $A(2000, 500, 0)$ and point $B(1400, 2900, 0)$.
The vector connecting them is $\vec{AB} = \begin{pmatrix} 1400 \\ 2900 \\ 0 \end{pmatrix} - \begin{pmatrix} 2000 \\ 500 \\ 0 \end{pmatrix} = \begin{pmatrix} -600 \\ 2400 \\ 0 \end{pmatrix} = 600 \begin{pmatrix} -1 \\ 4 \\ 0 \end{pmatrix}$.

The flight path of Alpha is a line in $\Pi$ with direction vector $\mathbf{d}_A = \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix}$.
The normal $\mathbf{n}$ must be perpendicular to both $\vec{AB}$ and $\mathbf{d}_A$:
$\mathbf{n} = \begin{pmatrix} -1 \\ 4 \\ 0 \end{pmatrix} \times \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} (4)(1) - (0)(3) \\ (0)(4) - (-1)(1) \\ (-1)(3) - (4)(4) \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ -19 \end{pmatrix}$

2. Find the cartesian equation of the plane.

The equation of the plane is $\mathbf{r} \cdot \mathbf{n} = \vec{OA} \cdot \mathbf{n}$.
$\mathbf{r} \cdot \begin{pmatrix} 4 \\ 1 \\ -19 \end{pmatrix} = \begin{pmatrix} 2000 \\ 500 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 1 \\ -19 \end{pmatrix} = 4(2000) + 1(500) - 19(0) = 8500$.

The cartesian equation of the plane is $4x + y - 19z = 8500$.

3. Verify if Aircraft Bravo's flight path lies on the plane by taking the dot product of its direction with the normal.

Aircraft Bravo takes off from $B$ in the direction $\mathbf{d}_B = \begin{pmatrix} 4 \\ 2 \\ 1 \end{pmatrix}$.
Check if this line is parallel to the plane by computing $\mathbf{d}_B \cdot \mathbf{n}$:
$\mathbf{d}_B \cdot \mathbf{n} = \begin{pmatrix} 4 \\ 2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 1 \\ -19 \end{pmatrix} = 4(4) + 2(1) + 1(-19) = 16 + 2 - 19 = -1$.

Since $\mathbf{d}_B \cdot \mathbf{n} \neq 0$, Aircraft Bravo's flight path is not parallel to the plane and thus does not lie within the plane.

4. Provide a logical deduction to explain why the two flight paths will not cross.

Because Aircraft Bravo's flight path starts at point $B$ (which is on the plane) and is not parallel to the plane, it intersects the plane at exactly one point: $B$.

Aircraft Alpha's flight path is entirely contained within the plane. If the two flight paths were to cross, they must intersect at a point on the plane, which can only be $B$.

For Alpha's flight path to pass through $B$, there must be a scalar $\lambda$ such that:
$\begin{pmatrix} 2000 \\ 500 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 1400 \\ 2900 \\ 0 \end{pmatrix}$
Equating the $z$-components gives $0 + \lambda(1) = 0 \implies \lambda = 0$.
However, when $\lambda = 0$, Alpha's position is $A$, which is not $B$.
Thus, Aircraft Alpha does not pass through $B$, and therefore the flight paths will not cross.