一群男孩想搭建一个露营帐篷。他们在水平地面上铺了一块矩形防水布 $OABC$，其中 $OA = 3\text{ m}$ 且 $AB = 1.5\text{ m}$，并将点 $D$ 和 $E$ 分别固定在 $O$ 和 $B$ 的正上方，使得 $OD = BE = 2\text{ m}$。 假设帐篷的形状如上图所示，具有 6 个三角形面和一个矩形底面。设点 $O$ 为原点，单位向量 $\mathbf{i}$、$\mathbf{j}$ 和 $\mathbf{k}$ 分别取在 $\overrightarrow{OA}$、$\overrightarrow{OC}$ 和 $\overrightarrow{OD}$ 的方向上。

1. Determine the position vectors of points D and E from the given coordinates and lengths.

Given $O$ is the origin, $OA$ is along the $x$-axis ($\mathbf{i}$), $OC$ is along the $y$-axis ($\mathbf{j}$), and $OD$ is along the $z$-axis ($\mathbf{k}$).
Since $OA = 3$ and $AB = 1.5$ for the rectangular base, the position vector of $B$ is $3\mathbf{i} + 1.5\mathbf{j}$.
Since $D$ and $E$ are vertically above $O$ and $B$ respectively by $2\text{ m}$, we have:
$\overrightarrow{OD} = 2\mathbf{k}$
$\overrightarrow{OE} = 3\mathbf{i} + 1.5\mathbf{j} + 2\mathbf{k}$

2. Find the direction vector of the line passing through D and E.

The displacement vector $\overrightarrow{DE}$ is:
$\overrightarrow{DE} = \overrightarrow{OE} - \overrightarrow{OD}$
$\overrightarrow{DE} = (3\mathbf{i} + 1.5\mathbf{j} + 2\mathbf{k}) - 2\mathbf{k}$
$\overrightarrow{DE} = 3\mathbf{i} + 1.5\mathbf{j}$

3. Form the vector equation of the line DE.

The direction vector can be scaled by taking out a factor of $1.5$ to match the required form:
$\overrightarrow{DE} = 1.5(2\mathbf{i} + \mathbf{j})$
Taking $2\mathbf{i} + \mathbf{j}$ as the direction vector and the point $D$ (with position vector $2\mathbf{k}$) as the starting point, the vector equation of the line is:
$\mathbf{r} = \overrightarrow{OD} + \lambda(2\mathbf{i} + \mathbf{j})$
$\mathbf{r} = 2\mathbf{k} + \lambda(2\mathbf{i} + \mathbf{j}), \lambda \in \mathbb{R}\text{ (Shown)}$

1. Find two non-parallel vectors parallel to the plane $ADE$.

From the given dimensions and coordinate system, the coordinates of $A$ and $D$ are $A(3, 0, 0)$ and $D(0, 0, 2)$.
$\overrightarrow{DA} = \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ -2 \end{pmatrix}$
From part (i), the direction vector of line $DE$ is:
$\mathbf{d}_{DE} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$

2. Find a normal vector to the plane $ADE$ using the cross product.

A normal vector $\mathbf{n}$ can be found by evaluating the cross product of $\overrightarrow{DA}$ and $\mathbf{d}_{DE}$:
$\mathbf{n} = \begin{pmatrix} 3 \\ 0 \\ -2 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} (0)(0) - (-2)(1) \\ (-2)(2) - (3)(0) \\ (3)(1) - (0)(2) \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ 3 \end{pmatrix}$

3. Determine the Cartesian equation of the plane $ADE$.

Using point $D(0, 0, 2)$ and the normal vector $\mathbf{n}$:
$\mathbf{r} \cdot \mathbf{n} = \mathbf{d} \cdot \mathbf{n}$
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ -4 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -4 \\ 3 \end{pmatrix} = 6$
Thus, replacing $\mathbf{r}$ with $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$, the Cartesian equation of the plane is:
$2x - 4y + 3z = 6$

1. Determine the normal vector to the plane $ADE$ and calculate the acute angle it makes with the plane $OABC$.

Let $\mathbf{n}_1$ be the normal to the plane $ADE$. Deriving from the coordinates $A(3, 0, 0)$, $D(0, 0, 2)$ and $E(3, 1.5, 2)$ (as expected in part (ii)):
$\overrightarrow{AD} = \begin{pmatrix} -3 \\ 0 \\ 2 \end{pmatrix}, \quad \overrightarrow{AE} = \begin{pmatrix} 0 \\ 1.5 \\ 2 \end{pmatrix}$
$\mathbf{n}_1 = \overrightarrow{AD} \times \overrightarrow{AE} = \begin{pmatrix} -3 \\ 0 \\ 2 \end{pmatrix} \times \begin{pmatrix} 0 \\ 1.5 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 6 \\ -4.5 \end{pmatrix}$
Taking a simpler scalar multiple, let $\mathbf{n}_1 = \begin{pmatrix} 2 \\ -4 \\ 3 \end{pmatrix}$.
The plane $OABC$ lies flat on the $xy$-plane, so its normal is $\mathbf{n}_2 = \mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.
Let $\theta$ be the acute angle between the planes $ADE$ and $OABC$:
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{\left| \begin{pmatrix} 2 \\ -4 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right|}{\sqrt{2^2 + (-4)^2 + 3^2}\sqrt{1^2}} = \frac{3}{\sqrt{29}}$
$\theta = \cos^{-1}\left(\frac{3}{\sqrt{29}}\right) \approx 56.145^\circ \approx 56.1^\circ$

2. Find the normal vector to the plane $CDE$ and use the dot product to find the acute angle between $ADE$ and $CDE$.

To find the angle between planes $ADE$ and $CDE$, we first find the normal to the plane $CDE$, $\mathbf{n}_3$. Using $C(0, 1.5, 0)$:
$\overrightarrow{CD} = \begin{pmatrix} 0 \\ -1.5 \\ 2 \end{pmatrix}, \quad \overrightarrow{CE} = \begin{pmatrix} 3 \\ 0 \\ 2 \end{pmatrix}$
$\mathbf{n}_3 = \overrightarrow{CD} \times \overrightarrow{CE} = \begin{pmatrix} 0 \\ -1.5 \\ 2 \end{pmatrix} \times \begin{pmatrix} 3 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 6 \\ 4.5 \end{pmatrix}$
Taking a scalar multiple, we use $\mathbf{n}_3 = \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix}$.
Let $\phi$ be the acute angle between the planes $ADE$ and $CDE$:
$\cos \phi = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_3|}{|\mathbf{n}_1||\mathbf{n}_3|} = \frac{\left| \begin{pmatrix} 2 \\ -4 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix} \right|}{\sqrt{29}\sqrt{29}} = \frac{|4 + 16 - 9|}{29} = \frac{11}{29}$
$\phi = \cos^{-1}\left(\frac{11}{29}\right) \approx 67.706^\circ \approx 67.7^\circ$
(Note for "Hence" method: Since $\mathbf{n}_3$ is a reflection of $\mathbf{n}_1$ across the $xy$-plane, the angle between them is identically $180^\circ - 2\theta = 180^\circ - 2(56.145^\circ) = 67.7^\circ$.)