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，使得 $\overrightarrow{OC}=p\overrightarrow{OB}$，其中 $p$ 为常数。$D$ 在 $AC$ 上且满足 $AD:DC=2:3$，$E$ 在 $AB$ 上且满足 $AE:EB=1:3$。

1. Find $\overrightarrow{OD}$ using the ratio theorem

Given that $C$ lies on $OB$ such that $\overrightarrow{OC} = p\overrightarrow{OB}$, we have $\overrightarrow{OC} = p\mathbf{b}$.
Using the ratio theorem for $D$ on $AC$ with $AD:DC = 2:3$,
$\overrightarrow{OD} = \frac{3\overrightarrow{OA} + 2\overrightarrow{OC}}{2 + 3}$
$\overrightarrow{OD} = \frac{3\mathbf{a} + 2p\mathbf{b}}{5} = \frac{3}{5}\mathbf{a} + \frac{2p}{5}\mathbf{b}$

2. Find $\overrightarrow{OE}$ using the ratio theorem

Using the ratio theorem for $E$ on $AB$ with $AE:EB = 1:3$,
$\overrightarrow{OE} = \frac{3\overrightarrow{OA} + 1\overrightarrow{OB}}{1 + 3}$
$\overrightarrow{OE} = \frac{3\mathbf{a} + \mathbf{b}}{4} = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$

1. Calculate the position vectors $\overrightarrow{OD}$ and $\overrightarrow{OE}$ from part (i).

From part (i), applying the ratio theorem gives:
$\overrightarrow{OD} = \frac{3\overrightarrow{OA} + 2\overrightarrow{OC}}{2+3} = \frac{3}{5}\mathbf{a} + \frac{2p}{5}\mathbf{b}$
$\overrightarrow{OE} = \frac{3\overrightarrow{OA} + 1\overrightarrow{OB}}{1+3} = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$

2. Set up the collinearity condition for points $O$, $D$, and $E$.

Since $O$, $D$, and $E$ are collinear, $\overrightarrow{OD}$ must be a scalar multiple of $\overrightarrow{OE}$. Let this scalar be $k$.
$\overrightarrow{OD} = k\overrightarrow{OE}$
$\frac{3}{5}\mathbf{a} + \frac{2p}{5}\mathbf{b} = k\left(\frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}\right)$

3. Compare the coefficients of $\mathbf{a}$ and $\mathbf{b}$ to solve for $p$.

Assuming $\mathbf{a}$ and $\mathbf{b}$ are non-parallel vectors, we can equate their coefficients.

Comparing coefficients of $\mathbf{a}$:
$\frac{3}{5} = \frac{3}{4}k \implies k = \frac{4}{5}$

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

1. State the position vector $\overrightarrow{OE}$ from part (i) and set up the shortest distance formula.

From part (i), using the ratio theorem for $AE:EB = 1:3$:
$\overrightarrow{OE} = \frac{3\overrightarrow{OA} + 1\overrightarrow{OB}}{1 + 3} = \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$

The shortest distance $d$ from point $E$ to the line $OB$ can be found using the cross product with the line's unit direction vector:
$d = \frac{|\overrightarrow{OE} \times \overrightarrow{OB}|}{|\overrightarrow{OB}|}$

2. Evaluate the vector product $\overrightarrow{OE} \times \overrightarrow{OB}$.

$\overrightarrow{OE} \times \overrightarrow{OB} = \left( \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b} \right) \times \mathbf{b}$
By expanding using the distributive property:
$= \frac{3}{4}(\mathbf{a} \times \mathbf{b}) + \frac{1}{4}(\mathbf{b} \times \mathbf{b})$
Since the cross product of any vector with itself is the zero vector ($\mathbf{b} \times \mathbf{b} = \mathbf{0}$):
$\overrightarrow{OE} \times \overrightarrow{OB} = \frac{3}{4}(\mathbf{a} \times \mathbf{b})$

3. Substitute the magnitude $|\overrightarrow{OB}| = 5$ to deduce the constant $k$.

Given that $|\overrightarrow{OB}| = 5$:
$d = \frac{\left| \frac{3}{4}(\mathbf{a} \times \mathbf{b}) \right|}{5}$
$d = \frac{\frac{3}{4}|\mathbf{a} \times \mathbf{b}|}{5} = \frac{3}{20}|\mathbf{a} \times \mathbf{b}|$
Comparing this to $k\,|\mathbf{a} \times \mathbf{b}|$, we have:
$k = \frac{3}{20}$

1. Relate the expression to projection using the definition of the scalar product

$|\mathbf{a}\cdot\mathbf{b}| = |\mathbf{a}| \,|\mathbf{b}| \,|\cos\theta|$, where $\theta$ is the angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$.
This can be written as $|\mathbf{a}| \,(|\mathbf{b}| \,|\cos\theta|)$ or $|\mathbf{b}| \,(|\mathbf{a}| \,|\cos\theta|)$.
Recall that $|\mathbf{b}| \,|\cos\theta|$ is the length of the projection of $\overrightarrow{OB}$ onto $\overrightarrow{OA}$.

2. State the geometric interpretation

Therefore, $|\mathbf{a}\cdot\mathbf{b}|$ is the product of the length of $OA$ and the length of the projection of $\overrightarrow{OB}$ onto the line $OA$ (or the product of the length of $OB$ and the length of the projection of $\overrightarrow{OA}$ onto the line $OB$).