为了即将举行的一场赛车比赛，将在赛道附近设立一个观众看台。作为准备工作的一部分，制作了一个棱柱形的看台模型。 上图展示了该看台的模型，以 $O$ 为原点，单位向量 $\mathbf{i}$、$\mathbf{j}$ 和 $\mathbf{k}$ 分别平行于 $OA$、$OC$ 和 $OE$。点 $(x, y, z)$ 的坐标是相对于原点 $O$ 定义的，单位为米。 已知 $OA = CB = 1\text{ m}$，$OC = AB = ED = 2\text{ m}$，且 $OE = CD = 0.5\text{ m}$。 计划在平面 $ABDE$ 上方建造一个遮阳篷。在模型中，该遮阳篷是一个矩形平面，它与平面 $ABDE$ 相交于直线 $ED$。

1. Determine the position vectors of three points on the plane, such as $A$, $B$, and $E$.

Given the origin $O(0,0,0)$ and unit vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ along $OA$, $OC$, $OE$ respectively:
$\vec{OA} = 1\mathbf{i} \implies A(1, 0, 0)$
$\vec{OC} = 2\mathbf{j}$ and $CB$ is parallel to $OA$ with $CB = 1 \implies B(1, 2, 0)$
$\vec{OE} = 0.5\mathbf{k} \implies E(0, 0, 0.5)$

2. Form two directional vectors on the plane and compute their cross product to find the normal vector.

$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix}$

$\vec{AE} = \vec{OE} - \vec{OA} = \begin{pmatrix} 0 \\ 0 \\ 0.5 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 0.5 \end{pmatrix}$

Let the normal vector to the plane be $\mathbf{n}$.
$\mathbf{n} = \vec{AB} \times \vec{AE} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} \times \begin{pmatrix} -1 \\ 0 \\ 0.5 \end{pmatrix} = \begin{pmatrix} (2)(0.5) - (0)(0) \\ (0)(-1) - (0)(0.5) \\ (0)(0) - (2)(-1) \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$

3. Form the cartesian equation of the plane.

The equation of the plane is given by $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$, where $\mathbf{a}$ is a known position vector on the plane (e.g., $A$).
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = (1)(1) + (0)(0) + (0)(2)$
$x + 2z = 1$

1. Determine the coordinates of a point on the line $ED$, such as point $E$.

From the given information, the origin is $O(0, 0, 0)$ and the unit vector $\mathbf{k}$ is parallel to $OE$.
Since $OE = 0.5\text{ m}$, the coordinates of $E$ are $(0, 0, 0.5)$.

2. Substitute the coordinates of $E$ into the equation of the shelter to show the value of $h$.

The shelter intersects the plane $ABDE$ in the line $ED$, which means the point $E$ lies on the shelter.
Substitute $x = 0$ and $z = 0.5$ into the shelter's equation $-0.5x + z = h$:
$-0.5(0) + 0.5 = h$
$h = 0.5$ (shown)

1. Find the normal vectors for the plane $ABDE$ and the shelter.

To answer this, we need the normal vector of plane $ABDE$ (which would have been found in part (a)).
From the coordinates $A(1,0,0)$, $B(1,2,0)$, and $E(0,0,0.5)$, we have $\vec{AB} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix}$ and $\vec{AE} = \begin{pmatrix} -1 \\ 0 \\ 0.5 \end{pmatrix}$.
A normal vector to the plane $ABDE$ is $\mathbf{n}_1 = \vec{AB} \times \vec{AE} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$.

The equation of the shelter (from part b) is given as $-0.5x + z = 0.5$, which can be equivalently written as $-x + 2z = 1$. A normal vector for the shelter is therefore $\mathbf{n}_2 = \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}$.

2. Calculate the acute angle between the two planes using their normal vectors.

Let $\theta$ be the acute angle between the planes. We use the scalar product of their normal vectors:
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$
$\cos \theta = \frac{\left| \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} \right|}{\sqrt{1^2 + 0^2 + 2^2} \sqrt{(-1)^2 + 0^2 + 2^2}}$
$\cos \theta = \frac{|(1)(-1) + (0)(0) + (2)(2)|}{\sqrt{5} \sqrt{5}}$
$\cos \theta = \frac{|-1 + 4|}{5}$
$\cos \theta = \frac{3}{5} = 0.6$
$\theta = \cos^{-1}(0.6) \approx 53.1^\circ \text{ (to 1 d.p.)}$

1. Determine the cartesian equation of the plane $ABDE$ (as required from part a).

From the given side lengths and orientations, the coordinates of the vertices are $A(1, 0, 0)$, $B(1, 2, 0)$, and $E(0, 0, 0.5)$.

The direction vectors of the plane are:
$\vec{AB} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix}$ and $\vec{AE} = \begin{pmatrix} -1 \\ 0 \\ 0.5 \end{pmatrix}$.

The normal vector to plane $ABDE$ is:
$\mathbf{n} = \vec{AB} \times \vec{AE} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} \times \begin{pmatrix} -1 \\ 0 \\ 0.5 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$.

The equation of plane $ABDE$ is:
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = 1 \implies x + 2z = 1$.

2. Express the line $l$ in terms of $z$ and substitute it into the plane equation.

The line $l$ is given by $x - a = z \implies x = z + a$, and $y = 1$.

Substitute $x$ into the equation of plane $ABDE$:
$(z + a) + 2z = 1$
$3z = 1 - a \implies z = \frac{1 - a}{3}$

3. Find the $x$-coordinate and state the position vector of point $M$.

$x = z + a = \frac{1 - a}{3} + a = \frac{2a + 1}{3}$.

Given $y = 1$, the position vector of $M$ is:
$\vec{OM} = \begin{pmatrix} \frac{2a + 1}{3} \\ 1 \\ \frac{1 - a}{3} \end{pmatrix}$

4. Use the boundaries of the rectangular surface $ABDE$ to find the range of $a$.

For $M$ to land on the finite rectangular surface $ABDE$, its $x$-coordinate must lie between the $x$-coordinates of the edges $ED$ (where $x=0$) and $AB$ (where $x=1$).
(Note: The $y$-coordinate is $1$, which safely lies within the $y$-boundaries of $0$ to $2$, and the $z$-boundary check is naturally equivalent to the $x$-boundary check due to the plane equation).

$0 \le x \le 1$
$0 \le \frac{2a + 1}{3} \le 1$
$0 \le 2a + 1 \le 3$
$-1 \le 2a \le 2$
$-\frac{1}{2} \le a \le 1$

1. Find the position vector of point $M$ using the result from part (d) with $a = \frac{1}{2}$.

From part (d), the position vector of $M$ is $\begin{pmatrix} \frac{2a+1}{3} \\ 1 \\ \frac{1-a}{3} \end{pmatrix}$.
Substituting $a = \frac{1}{2}$:
$\overrightarrow{OM} = \begin{pmatrix} \frac{2(1/2)+1}{3} \\ 1 \\ \frac{1-1/2}{3} \end{pmatrix} = \begin{pmatrix} 2/3 \\ 1 \\ 1/6 \end{pmatrix}$

2. Set up the equation for the shortest distance from $M$ to the vertical plane.

The screen plane equation is $12x + 5y + 0z = d$.
The shortest distance from $M \left(\frac{2}{3}, 1, \frac{1}{6}\right)$ to the plane is given as $4\text{ m}$. Using the perpendicular distance formula:
$\frac{\left| 12\left(\frac{2}{3}\right) + 5(1) + 0\left(\frac{1}{6}\right) - d \right|}{\sqrt{12^2 + 5^2 + 0^2}} = 4$
$\frac{|8 + 5 - d|}{\sqrt{144 + 25}} = 4$
$\frac{|13 - d|}{13} = 4$

3. Solve the modulus equation for $d$.

$|13 - d| = 52$
$13 - d = 52 \quad \text{or} \quad 13 - d = -52$
$d = -39 \quad \text{or} \quad d = 65$

4. Determine the correct value of $d$ based on the given constraint.

Since we are given that $d > 50$, we reject $d = -39$.
Therefore, $d = 65$.