图中显示了一个棱锥 DOABC。取如图所示的单位向量 i, j, k，A, B, C, D 的位置向量分别为 $\overrightarrow{OA} = 4i,\quad \overrightarrow{OB} = 4i + 2j,\quad \overrightarrow{OC} = 2j,\quad \overrightarrow{OD} = 6k。$ AD, BD 和 AB 的中点分别为 L, M 和 N。

1. Find the position vectors of the mid-points $M$ and $N$.

Since $M$ is the mid-point of $BD$:
$\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OB} + \overrightarrow{OD}) = \frac{1}{2}(4\mathbf{i} + 2\mathbf{j} + 6\mathbf{k}) = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k}$
Since $N$ is the mid-point of $AB$:
$\overrightarrow{ON} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB}) = \frac{1}{2}(4\mathbf{i} + (4\mathbf{i} + 2\mathbf{j})) = 4\mathbf{i} + \mathbf{j}$

2. Find the displacement vector $\overrightarrow{MN}$.

$\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM}$
$\overrightarrow{MN} = (4\mathbf{i} + \mathbf{j}) - (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) = 2\mathbf{i} - 3\mathbf{k}$

3. Calculate the scalar product and the magnitudes of $\overrightarrow{MN}$ and $\overrightarrow{OB}$.

$\overrightarrow{MN} \cdot \overrightarrow{OB} = (2\mathbf{i} - 3\mathbf{k}) \cdot (4\mathbf{i} + 2\mathbf{j}) = (2)(4) + (0)(2) + (-3)(0) = 8$
$|\overrightarrow{MN}| = \sqrt{2^2 + 0^2 + (-3)^2} = \sqrt{13}$
$|\overrightarrow{OB}| = \sqrt{4^2 + 2^2 + 0^2} = \sqrt{20}$

4. Determine the angle between the two vectors using the scalar product formula.

Let $\theta$ be the required angle.
$\cos \theta = \frac{\overrightarrow{MN} \cdot \overrightarrow{OB}}{|\overrightarrow{MN}| |\overrightarrow{OB}|}$
$\cos \theta = \frac{8}{\sqrt{13}\sqrt{20}} = \frac{8}{\sqrt{260}} = \frac{4}{\sqrt{65}}$
$\theta = \cos^{-1}\left(\frac{4}{\sqrt{65}}\right) \approx 60.255^\circ$
$\theta \approx 60.3^\circ \text{ (to 1 d.p.)}$

1. Find the position vector of the point $L$.

$L$ is the mid-point of $AD$. Using the ratio theorem:
$\overrightarrow{OL} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OD})$
$\overrightarrow{OL} = \frac{1}{2}(4\mathbf{i} + 6\mathbf{k}) = 2\mathbf{i} + 3\mathbf{k}$

2. Determine the vector equation of the line passing through $C$ and $L$.

The direction vector of line $CL$ is:
$\overrightarrow{CL} = \overrightarrow{OL} - \overrightarrow{OC} = (2\mathbf{i} + 3\mathbf{k}) - 2\mathbf{j} = 2\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}$
Let $\mu$ be a scalar parameter. The vector equation of line $CL$ is:
$\mathbf{r} = 2\mathbf{j} + \mu(2\mathbf{i} - 2\mathbf{j} + 3\mathbf{k})$

3. Determine the vector equation of the line passing through $P$ and $B$.

The direction vector of line $PB$ is:
$\overrightarrow{PB} = \overrightarrow{OB} - \overrightarrow{OP} = (4\mathbf{i} + 2\mathbf{j}) - p\mathbf{k} = 4\mathbf{i} + 2\mathbf{j} - p\mathbf{k}$
Let $\lambda$ be a scalar parameter. The vector equation of line $PB$ is:
$\mathbf{r} = p\mathbf{k} + \lambda(4\mathbf{i} + 2\mathbf{j} - p\mathbf{k})$

4. Form a system of equations by equating the components of the two lines at their point of intersection.

At the point of intersection, the position vectors from both lines must be identical:
$2\mu\mathbf{i} + (2 - 2\mu)\mathbf{j} + 3\mu\mathbf{k} = 4\lambda\mathbf{i} + 2\lambda\mathbf{j} + (p - p\lambda)\mathbf{k}$
Comparing the $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ coefficients respectively:
$2\mu = 4\lambda \implies \mu = 2\lambda \quad \text{--- (1)}$
$2 - 2\mu = 2\lambda \quad \text{--- (2)}$
$3\mu = p(1 - \lambda) \quad \text{--- (3)}$

5. Solve for the scalar parameters $\lambda$ and $\mu$.

Substitute equation (1) into equation (2):
$2 - 2(2\lambda) = 2\lambda$
$2 - 4\lambda = 2\lambda$
$6\lambda = 2 \implies \lambda = \frac{1}{3}$
Using equation (1) to find $\mu$:
$\mu = 2\left(\frac{1}{3}\right) = \frac{2}{3}$

6. Find the value of $p$ using the $\mathbf{k}$ component equation.

Substitute $\lambda = \frac{1}{3}$ and $\mu = \frac{2}{3}$ into equation (3):
$3\left(\frac{2}{3}\right) = p\left(1 - \frac{1}{3}\right)$
$2 = p\left(\frac{2}{3}\right)$
$p = 3$