设点 $A$ 和点 $B$ 相对于原点 $O$ 的位置向量分别为 $\mathbf{a}$ 和 $\mathbf{b}$。

1. Apply the ratio theorem to express $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

$\overrightarrow{OM} = \frac{2\overrightarrow{OA} + 1\overrightarrow{OB}}{1 + 2}$
$\overrightarrow{OM} = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$

2. Formulate the area of triangle $OBM$ using the vector product.

$\text{Area of } \triangle OBM = \frac{1}{2} |\overrightarrow{OM} \times \overrightarrow{OB}| = 4$

3. Substitute $\overrightarrow{OM}$ and expand the vector product using the distributive property.

$\frac{1}{2} \left| \left( \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b} \right) \times \mathbf{b} \right| = 4$
$\frac{1}{2} \left| \frac{2}{3}(\mathbf{a} \times \mathbf{b}) + \frac{1}{3}(\mathbf{b} \times \mathbf{b}) \right| = 4$

4. Simplify using the property $\mathbf{b} \times \mathbf{b} = \mathbf{0}$ to solve for $|\mathbf{a} \times \mathbf{b}|$.

Since $\mathbf{b} \times \mathbf{b} = \mathbf{0}$,
$\frac{1}{2} \left| \frac{2}{3}(\mathbf{a} \times \mathbf{b}) + \mathbf{0} \right| = 4$
$\frac{1}{3} |\mathbf{a} \times \mathbf{b}| = 4$
$|\mathbf{a} \times \mathbf{b}| = 12$

1. Interpret the vector product equation to deduce the relationship between the vectors.

$(\mathbf{p} - \mathbf{a}) \times (\mathbf{b} - \mathbf{a}) = \mathbf{0}$
Since the position vectors are given relative to the origin $O$, we have $\mathbf{p} - \mathbf{a} = \overrightarrow{AP}$ and $\mathbf{b} - \mathbf{a} = \overrightarrow{AB}$.
Substituting these into the equation gives:
$\overrightarrow{AP} \times \overrightarrow{AB} = \mathbf{0}$
Since the vector product of these two vectors is the zero vector, we deduce that $\overrightarrow{AP}$ is parallel to $\overrightarrow{AB}$ (or $\overrightarrow{AP}$ is a scalar multiple of $\overrightarrow{AB}$).

2. Find the vector equation of line $l$.

Since $\overrightarrow{AP}$ is parallel to $\overrightarrow{AB}$ and they share the common point $A$, the line $l$ that passes through $A$ and $P$ is the same straight line that passes through $A$ and $B$.
Using $A$ as the support point and $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ as the direction vector, the vector equation of the line is:
$\mathbf{r} = \mathbf{a} + \lambda(\mathbf{b} - \mathbf{a}), \quad \lambda \in \mathbb{R}$

1. Identify the relationship between $\mathbf{a}$ and $\mathbf{b}$.

Since the scalar product $\mathbf{a} \cdot \mathbf{b}$ is a scalar, $4(\mathbf{a} \cdot \mathbf{b})$ is also a scalar.
This means $\mathbf{a}$ is a scalar multiple of $\mathbf{b}$, so $\mathbf{a}$ and $\mathbf{b}$ are parallel vectors.

2. Set up an equation by taking the scalar product with $\mathbf{b}$ on both sides.

Given $\mathbf{a} = 4(\mathbf{a} \cdot \mathbf{b})\mathbf{b}$, take the scalar product with $\mathbf{b}$ on both sides:
$\mathbf{a} \cdot \mathbf{b} = [4(\mathbf{a} \cdot \mathbf{b})\mathbf{b}] \cdot \mathbf{b}$
$\mathbf{a} \cdot \mathbf{b} = 4(\mathbf{a} \cdot \mathbf{b})(\mathbf{b} \cdot \mathbf{b})$

3. Solve for $|\mathbf{b}|$.

Using the property $\mathbf{b} \cdot \mathbf{b} = |\mathbf{b}|^2$:
$\mathbf{a} \cdot \mathbf{b} = 4(\mathbf{a} \cdot \mathbf{b})|\mathbf{b}|^2$
Assuming $\mathbf{a} \cdot \mathbf{b} \ne 0$ (non-trivial vectors):
$1 = 4|\mathbf{b}|^2$
$|\mathbf{b}|^2 = \frac{1}{4}$
Since magnitude must be non-negative, $|\mathbf{b}| > 0$:
$|\mathbf{b}| = \frac{1}{2}$