以原点 $O$ 为基准，点 $A$ 和点 $B$ 满足 $\overrightarrow{OA}=\mathbf{a}$ 且 $\overrightarrow{OB}=\mathbf{b}$。点 $C$ 在 $OB$ 上，满足 $\overrightarrow{OC} = p\,\overrightarrow{OB}$，其中 $p$ 为常数。点 $D$ 在 $AC$ 上，满足 $AD:DC = 2:3$，点 $E$ 在 $AB$ 上，满足 $AE:EB = 1:3$。

1. Express $\overrightarrow{OC}$ in terms of $\mathbf{b}$ and use the ratio theorem to find $\overrightarrow{OD}$.

$\overrightarrow{OC} = p\,\overrightarrow{OB} = p\mathbf{b}$
Since $D$ divides $AC$ in the ratio $2:3$, by the ratio theorem:
$\overrightarrow{OD} = \frac{3\overrightarrow{OA} + 2\overrightarrow{OC}}{2 + 3}$
$\overrightarrow{OD} = \frac{3\mathbf{a} + 2p\mathbf{b}}{5} = \frac{3}{5}\mathbf{a} + \frac{2p}{5}\mathbf{b}$

2. Use the ratio theorem to find $\overrightarrow{OE}$.

Since $E$ divides $AB$ in the ratio $1:3$, by the ratio theorem:
$\overrightarrow{OE} = \frac{3\overrightarrow{OA} + 1\overrightarrow{OB}}{1 + 3}$
$\overrightarrow{OE} = \frac{3\mathbf{a} + \mathbf{b}}{4} = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$

1. Determine the position vectors $\overrightarrow{OD}$ and $\overrightarrow{OE}$ (from part i) and set up the collinearity equation.

Using the ratio theorem to calculate the vectors from part (i):
$\overrightarrow{OD} = \frac{3\overrightarrow{OA} + 2\overrightarrow{OC}}{2+3} = \frac{3}{5}\mathbf{a} + \frac{2}{5}p\mathbf{b}$
$\overrightarrow{OE} = \frac{3\overrightarrow{OA} + 1\overrightarrow{OB}}{1+3} = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$
Since $O$, $D$ and $E$ are collinear, $\overrightarrow{OD}$ must be a scalar multiple of $\overrightarrow{OE}$. Let $\overrightarrow{OD} = \lambda \overrightarrow{OE}$:
$\frac{3}{5}\mathbf{a} + \frac{2}{5}p\mathbf{b} = \lambda \left( \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b} \right)$

2. Compare the coefficients of $\mathbf{a}$ to find the scalar $\lambda$.

Comparing the coefficients of $\mathbf{a}$:
$\frac{3}{5} = \frac{3}{4}\lambda$
$\lambda = \frac{3}{5} \times \frac{4}{3} = \frac{4}{5}$

3. Compare the coefficients of $\mathbf{b}$ to evaluate $p$.

Comparing the coefficients of $\mathbf{b}$:
$\frac{2}{5}p = \frac{1}{4}\lambda$
Substitute $\lambda = \frac{4}{5}$ into the equation:
$\frac{2}{5}p = \frac{1}{4} \left(\frac{4}{5}\right)$
$\frac{2}{5}p = \frac{1}{5}$
$2p = 1$
$p = \frac{1}{2}$

1. Express the shortest distance from $E$ to the line $OB$ using the vector product.

The line $OB$ passes through the origin $O$ and has direction vector $\mathbf{b}$. The shortest perpendicular distance $d$ from point $E$ to the line $OB$ is given by the magnitude of the cross product of the position vector of $E$ and the unit direction vector of the line:
$d = \frac{|\overrightarrow{OE} \times \mathbf{b}|}{|\mathbf{b}|}$

2. Substitute the position vector $\overrightarrow{OE}$ from part (i) and the given magnitude $|\mathbf{b}|$.

From part (i), we know $\overrightarrow{OE} = \frac{3\mathbf{a} + \mathbf{b}}{4}$. We are given that the length $OB = 5$, so $|\mathbf{b}| = 5$.
$d = \frac{1}{5} \left| \left( \frac{3\mathbf{a} + \mathbf{b}}{4} \right) \times \mathbf{b} \right|$
$d = \frac{1}{20} \left| (3\mathbf{a} + \mathbf{b}) \times \mathbf{b} \right|$

3. Expand the vector product and simplify to find $k$.

Using the distributive property of the cross product:
$d = \frac{1}{20} |3\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{b}|$
Since the cross product of any vector with itself is the zero vector ($\mathbf{b} \times \mathbf{b} = \mathbf{0}$):
$d = \frac{1}{20} |3(\mathbf{a} \times \mathbf{b})|$
$d = \frac{3}{20} |\mathbf{a} \times \mathbf{b}|$
Comparing this with $k\,|\mathbf{a} \times \mathbf{b}|$, we get:
$k = \frac{3}{20}$

1. State the geometric interpretation of the magnitude of the dot product

We know that $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta$, where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$.
$|\mathbf{a} \cdot \mathbf{b}| = |\mathbf{a}| \big| |\mathbf{b}| \cos \theta \big| = |\mathbf{a}| \times (\text{length of projection of } \mathbf{b} \text{ onto } \mathbf{a})$
Hence, it represents the product of the length of $OA$ and the length of the projection of $OB$ onto $OA$.