相对于原点 $O$，点 $A$ 和 $B$ 的位置向量分别为 $\mathbf{a}$ 和 $\mathbf{b}$，其中 $\mathbf{a}$ 与 $\mathbf{b}$ 不平行。已知 $B$ 落在线段 $AC$ 上，使得 $\overrightarrow{BC} = 3\mathbf{b} - \mu\mathbf{a}$。

1. Express $\overrightarrow{AB}$ and use collinearity to relate it to $\overrightarrow{BC}$.

$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$

Since $B$ lies on the line segment $AC$, the points $A$, $B$, and $C$ are collinear. Thus, $\overrightarrow{BC}$ is a scalar multiple of $\overrightarrow{AB}$:
$\overrightarrow{BC} = k\overrightarrow{AB}$
$3\mathbf{b} - \mu\mathbf{a} = k(\mathbf{b} - \mathbf{a}) = k\mathbf{b} - k\mathbf{a}$

2. Compare coefficients of the non-parallel vectors $\mathbf{a}$ and $\mathbf{b}$ to find $\mu$.

Since $\mathbf{a}$ and $\mathbf{b}$ are non-parallel, we can equate their respective coefficients:
For $\mathbf{b}$: $k = 3$
For $\mathbf{a}$: $-\mu = -k$

Therefore, $\mu = 3$.

3. Find $\overrightarrow{OC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

Using vector addition:
$\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC}$

Substitute $\mu = 3$ into $\overrightarrow{BC}$:
$\overrightarrow{OC} = \mathbf{b} + (3\mathbf{b} - 3\mathbf{a})$
$\overrightarrow{OC} = 4\mathbf{b} - 3\mathbf{a}$

1. State $\overrightarrow{OQ}$ in terms of $\mathbf{a}$, $\mathbf{b}$ and $t$ using the result from part (a).

From part (a), we have $\overrightarrow{OC} = 4\mathbf{b} - 3\mathbf{a}$.
Given $\overrightarrow{OQ} = t\,\overrightarrow{OC}$,
$\overrightarrow{OQ} = t(4\mathbf{b} - 3\mathbf{a}) = 4t\mathbf{b} - 3t\mathbf{a}$

2. Apply the ratio theorem to find an expression for $\overrightarrow{OP}$.

Since $P$ divides the line segment $AQ$ in the ratio $\lambda : 1$, by the ratio theorem:
$\overrightarrow{OP} = \frac{1\cdot \overrightarrow{OA} + \lambda\cdot \overrightarrow{OQ}}{\lambda + 1}$

3. Express $\overrightarrow{OP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

$\overrightarrow{OP} = \frac{\mathbf{a} + \lambda(4t\mathbf{b} - 3t\mathbf{a})}{\lambda + 1}$
$\overrightarrow{OP} = \left(\frac{1 - 3\lambda t}{\lambda + 1}\right)\mathbf{a} + \left(\frac{4\lambda t}{\lambda + 1}\right)\mathbf{b}$

4. Use the collinearity of $O, P, B$ to set the coefficient of $\mathbf{a}$ to zero and solve for $\lambda t$.

Since $P$ lies on the line segment $OB$, the points $O$, $P$, and $B$ are collinear. Thus, $\overrightarrow{OP}$ must be a scalar multiple of $\mathbf{b}$.
Because $\mathbf{a}$ and $\mathbf{b}$ are non-parallel, the coefficient of $\mathbf{a}$ in $\overrightarrow{OP}$ must be $0$.
$\frac{1 - 3\lambda t}{\lambda + 1} = 0$
$1 - 3\lambda t = 0$
$3\lambda t = 1$
$\lambda t = \frac{1}{3}$

1. Use the result from part (b) and the given value of $\lambda$ to express $\overrightarrow{OP}$ in terms of $\mathbf{b}$.

From part (b), applying the Ratio Theorem gave $\overrightarrow{OP} = \frac{1 \cdot \mathbf{a} + \lambda \cdot (t\overrightarrow{OC})}{\lambda + 1}$.
Since $P$ lies on $OB$, the $\mathbf{a}$ component evaluates to zero, leaving:
$\overrightarrow{OP} = \frac{4\lambda t}{\lambda + 1}\mathbf{b}$
We also deduced that $\lambda t = \frac{1}{3}$.
Given $\lambda = 5$, we substitute these values into the expression for $\overrightarrow{OP}$:
$\overrightarrow{OP} = \frac{4\left(\frac{1}{3}\right)}{5 + 1}\mathbf{b}$
$\overrightarrow{OP} = \frac{\frac{4}{3}}{6}\mathbf{b} = \frac{2}{9}\mathbf{b}$

2. Interpret the vector equation geometrically to find the required ratio of lengths.

The equation $\overrightarrow{OP} = \frac{2}{9}\mathbf{b}$ implies that $\overrightarrow{OP} = \frac{2}{9}\overrightarrow{OB}$.
This means that the length $OP$ is $\frac{2}{9}$ of the total length $OB$.
Consequently, the length $PB$ is $1 - \frac{2}{9} = \frac{7}{9}$ of $OB$.
The ratio of the lengths is therefore:
$OP : PB = 2 : 7$