一个位于水平平坦地面上的机场必须 management 飞机从其跑道起飞的飞行航线。这些飞行航线可以假设为直线，飞机可以假设为质点。 在下方图中，标记的点是相对于机场上坐标为 $(0,0,0)$ 的原点 O 定义的。$x$ 轴、$y$ 轴和 $z$ 轴的方向分别为正东、正北和垂直向上，单位为米。方向向量 $\mathbf{i}$、$\mathbf{j}$ 和 $\mathbf{k}$ 分别对应正东、正北和垂直向上的方向。点 A 的坐标为 $(2000,500,0)$，点 B 的坐标为 $(1400,2900,0)$。 阿尔法（Alpha）号飞机从点 A 出发，沿 $4\mathbf{i}+3\mathbf{j}+\mathbf{k}$ 方向起飞，如图所示。 一座雄伟山峰的顶峰坐标为 $(18000,4500,2000)$。

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.