以原点 $O$ 为基准，三个点 $A$、$B$ 和 $P$ 的位置向量分别为 $\mathbf{a}$、$\mathbf{b}$ 和 $\mathbf{a} + 5\mathbf{b}$。已知 $\mathbf{b}$ 是单位向量，角 $AOB$ 为 $60^\circ$ 且 $AB$ 垂直于 $OP$，求 $|\mathbf{a}|$。 在以 $O$ 为原点的三角形 $OAB$ 中，点 $A$ 和 $B$ 的位置向量分别为 $\mathbf{a}$ 和 $\mathbf{b}$。点 $C$ 是 $OA$ 的中点， $BC$ 上的点 $E$ 满足 $CE : EB = 3 : 4$，且直线 $OE$ 与 $AB$ 交于点 $D$。求比值 $AD : AB$。

1. Find the vector $\vec{AB}$ and set up the dot product equation for perpendicularity.

$\vec{AB} = \mathbf{b} - \mathbf{a}$

Since $AB$ is perpendicular to $OP$, their dot product is zero:
$\vec{AB} \cdot \vec{OP} = 0$
$(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{a} + 5\mathbf{b}) = 0$

2. Expand the dot product expression.

$\mathbf{b} \cdot \mathbf{a} + 5(\mathbf{b} \cdot \mathbf{b}) - (\mathbf{a} \cdot \mathbf{a}) - 5(\mathbf{a} \cdot \mathbf{b}) = 0$
$5|\mathbf{b}|^2 - |\mathbf{a}|^2 - 4(\mathbf{a} \cdot \mathbf{b}) = 0$

3. Substitute the known magnitude and angle into the equation.

We are given that $\mathbf{b}$ is a unit vector ($|\mathbf{b}| = 1$) and $\angle AOB = 60^\circ$.

The dot product $\mathbf{a} \cdot \mathbf{b}$ can be expressed as:
$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos 60^\circ = |\mathbf{a}|(1)\left(\frac{1}{2}\right) = \frac{1}{2}|\mathbf{a}|$

Substituting this into the expanded equation:
$5(1)^2 - |\mathbf{a}|^2 - 4\left(\frac{1}{2}|\mathbf{a}|\right) = 0$
$5 - |\mathbf{a}|^2 - 2|\mathbf{a}| = 0$
$|\mathbf{a}|^2 + 2|\mathbf{a}| - 5 = 0$

4. Solve the resulting quadratic equation for $|\mathbf{a}|$.

Using the quadratic formula for $|\mathbf{a}|^2 + 2|\mathbf{a}| - 5 = 0$:
$|\mathbf{a}| = \frac{-2 \pm \sqrt{2^2 - 4(1)(-5)}}{2}$
$|\mathbf{a}| = \frac{-2 \pm \sqrt{24}}{2}$
$|\mathbf{a}| = \frac{-2 \pm 2\sqrt{6}}{2} = -1 \pm \sqrt{6}$

Since the magnitude of a vector must be strictly positive ($|\mathbf{a}| > 0$), we reject the negative root.
$|\mathbf{a}| = \sqrt{6} - 1$

1. Express the position vector of $E$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

Given $C$ is the midpoint of $OA$, its position vector is:
$\vec{OC} = \frac{1}{2}\mathbf{a}$
By the ratio theorem, since $E$ lies on $BC$ and $CE : EB = 3 : 4$, we cross-multiply the ratios with the position vectors of $B$ and $C$:
$\vec{OE} = \frac{4\vec{OC} + 3\vec{OB}}{3 + 4}$
$\vec{OE} = \frac{4(\frac{1}{2}\mathbf{a}) + 3\mathbf{b}}{7} = \frac{2}{7}\mathbf{a} + \frac{3}{7}\mathbf{b}$

2. Set up an expression for $\vec{OD}$ using the collinearity of $O, E$, and $D$.

Since $O, E$, and $D$ lie on the same straight line, $\vec{OD}$ is a scalar multiple of $\vec{OE}$. Let $\vec{OD} = k\vec{OE}$:
$\vec{OD} = k\left(\frac{2}{7}\mathbf{a} + \frac{3}{7}\mathbf{b}\right) = \frac{2k}{7}\mathbf{a} + \frac{3k}{7}\mathbf{b}$

3. Use the collinearity of $A, D$, and $B$ to solve for $k$ and the parameter $\lambda$.

Since $D$ lies on the line segment $AB$, we can also express $\vec{OD}$ as:
$\vec{OD} = (1 - \lambda)\mathbf{a} + \lambda\mathbf{b}$
where $\lambda = \frac{AD}{AB}$.
Equating the coefficients of $\mathbf{a}$ and $\mathbf{b}$ from both expressions for $\vec{OD}$:
$1 - \lambda = \frac{2k}{7} \quad \text{and} \quad \lambda = \frac{3k}{7}$
Adding these two equations together yields:
$1 = \frac{5k}{7} \implies k = \frac{7}{5}$
Substitute $k$ back to find $\lambda$:
$\lambda = \frac{3}{7} \left(\frac{7}{5}\right) = \frac{3}{5}$

4. Determine the required ratio $AD : AB$.

Since $\lambda$ represents the ratio of the length of $AD$ to the total length $AB$:
$\lambda = \frac{AD}{AB} = \frac{3}{5}$
Therefore, the ratio is:
$AD : AB = 3 : 5$