点 $A$ 和 $B$ 相对于原点 $O$ 的位置向量分别为 $\mathbf{a}$ 和 $\mathbf{b}$。点 $C$ 使得 $OACB$ 为平行四边形。$AC$ 上的点 $P$ 满足 $AP : PC = 1 : 2$，$BC$ 上的点 $Q$ 满足 $BQ : QC = 1 : 2$。

1. Find the position vectors of points $P$ and $Q$.

Since $OACB$ is a parallelogram, we have $\overrightarrow{AC} = \overrightarrow{OB} = \mathbf{b}$ and $\overrightarrow{BC} = \overrightarrow{OA} = \mathbf{a}$.

Using the given ratios $AP : PC = 1 : 2$ and $BQ : QC = 1 : 2$:
$\overrightarrow{AP} = \frac{1}{3}\overrightarrow{AC} = \frac{1}{3}\mathbf{b}$
$\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \frac{1}{3}\mathbf{b}$

$\overrightarrow{BQ} = \frac{1}{3}\overrightarrow{BC} = \frac{1}{3}\mathbf{a}$
$\overrightarrow{OQ} = \overrightarrow{OB} + \overrightarrow{BQ} = \mathbf{b} + \frac{1}{3}\mathbf{a}$

2. Calculate $\overrightarrow{QP}$ by vector subtraction.

$\overrightarrow{QP} = \overrightarrow{OP} - \overrightarrow{OQ}$
$\overrightarrow{QP} = \left(\mathbf{a} + \frac{1}{3}\mathbf{b}\right) - \left(\mathbf{b} + \frac{1}{3}\mathbf{a}\right)$
$\overrightarrow{QP} = \frac{2}{3}\mathbf{a} - \frac{2}{3}\mathbf{b}$

1. Find the vector $\overrightarrow{QP}$ using the results from part (i).

From part (i), using the ratio theorem or vector addition:
$\overrightarrow{AP} = \frac{1}{3}\overrightarrow{AC} = \frac{1}{3}\mathbf{b} \implies \overrightarrow{OP} = \mathbf{a} + \frac{1}{3}\mathbf{b}$
$\overrightarrow{BQ} = \frac{1}{3}\overrightarrow{BC} = \frac{1}{3}\mathbf{a} \implies \overrightarrow{OQ} = \mathbf{b} + \frac{1}{3}\mathbf{a}$
$\overrightarrow{QP} = \overrightarrow{OP} - \overrightarrow{OQ} = \left(\mathbf{a} + \frac{1}{3}\mathbf{b}\right) - \left(\mathbf{b} + \frac{1}{3}\mathbf{a}\right) = \frac{2}{3}\mathbf{a} - \frac{2}{3}\mathbf{b}$

2. Express $\overrightarrow{OC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ and set up the scalar product.

Since $OACB$ is a parallelogram, $\overrightarrow{OB}$ is parallel to $\overrightarrow{AC}$ and $\overrightarrow{OA}$ is parallel to $\overrightarrow{BC}$.
$\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{OB} = \mathbf{a} + \mathbf{b}$

$\overrightarrow{OC}\cdot\overrightarrow{QP} = (\mathbf{a} + \mathbf{b}) \cdot \left(\frac{2}{3}\mathbf{a} - \frac{2}{3}\mathbf{b}\right)$

3. Expand the scalar product and use dot product properties to find $\lambda$ and $\mu$.

$\begin{aligned} \overrightarrow{OC}\cdot\overrightarrow{QP} &= \frac{2}{3}(\mathbf{a}\cdot\mathbf{a}) - \frac{2}{3}(\mathbf{a}\cdot\mathbf{b}) + \frac{2}{3}(\mathbf{b}\cdot\mathbf{a}) - \frac{2}{3}(\mathbf{b}\cdot\mathbf{b}) \end{aligned}$
Since the scalar product is commutative ($\mathbf{a}\cdot\mathbf{b} = \mathbf{b}\cdot\mathbf{a}$) and $\mathbf{v}\cdot\mathbf{v} = |\mathbf{v}|^2$:
$= \frac{2}{3}|\mathbf{a}|^2 - \frac{2}{3}|\mathbf{b}|^2$
Comparing this with $\lambda |\mathbf{a}|^2 + \mu |\mathbf{b}|^2$ gives $\lambda = \frac{2}{3}$ and $\mu = -\frac{2}{3}$.

1. Use the result from part (ii) to find the relationship between the magnitudes of $\mathbf{a}$ and $\mathbf{b}$.

From part (ii), we established that $\overrightarrow{OC}\cdot\overrightarrow{QP} = \frac{2}{3}|\mathbf{a}|^2 - \frac{2}{3}|\mathbf{b}|^2$.
Given $\overrightarrow{OC}\cdot\overrightarrow{QP} = 0$, we have:
$\frac{2}{3}|\mathbf{a}|^2 - \frac{2}{3}|\mathbf{b}|^2 = 0$
$|\mathbf{a}|^2 = |\mathbf{b}|^2$
Since magnitudes are non-negative,
$|\mathbf{a}| = |\mathbf{b}|$

2. Identify the shape of the parallelogram and justify the answer.

The magnitudes $|\mathbf{a}|$ and $|\mathbf{b}|$ represent the lengths of the adjacent sides $OA$ and $OB$ of the parallelogram $OACB$.
Since the adjacent sides are equal in length ($OA = OB$), the parallelogram $OACB$ is a rhombus.