上图显示了一个半径为 $r$ 单位且圆心为 $O$ 的圆。圆上的点 $A$ 和 $B$ 满足 $\overrightarrow{OA} = \mathbf{a}$，$\overrightarrow{OB} = \mathbf{b}$ 且 $\angle AOB = 120^\circ$。点 $N$ 以 $\lambda : 1-\lambda$ 的比例分割 $AB$，且 $ON$ 垂直于 $OB$。

1. Express $\overrightarrow{ON}$ in terms of $\mathbf{a}$, $\mathbf{b}$ and $\lambda$ using the ratio theorem.

Since $N$ divides $AB$ in the ratio $\lambda : 1-\lambda$, by the ratio theorem:
$\overrightarrow{ON} = \frac{(1-\lambda)\overrightarrow{OA} + \lambda\overrightarrow{OB}}{\lambda + (1-\lambda)} = (1-\lambda)\mathbf{a} + \lambda\mathbf{b}$

2. Formulate an equation using the perpendicularity condition.

Since $ON$ is perpendicular to $OB$, their dot product is zero:
$\overrightarrow{ON} \cdot \overrightarrow{OB} = 0$
$((1-\lambda)\mathbf{a} + \lambda\mathbf{b}) \cdot \mathbf{b} = 0$
$(1-\lambda)(\mathbf{a} \cdot \mathbf{b}) + \lambda(\mathbf{b} \cdot \mathbf{b}) = 0$

3. Evaluate the dot products using the given lengths and angle.

Since $A$ and $B$ lie on the circle with radius $r$, $|\mathbf{a}| = |\mathbf{b}| = r$.
$\mathbf{b} \cdot \mathbf{b} = |\mathbf{b}|^2 = r^2$
$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos 120^\circ = (r)(r)\left(-\frac{1}{2}\right) = -\frac{1}{2}r^2$
Substituting these into the equation:
$(1-\lambda)\left(-\frac{1}{2}r^2\right) + \lambda(r^2) = 0$

4. Solve for $\lambda$ to establish the given expression for $\overrightarrow{ON}$.

Dividing both sides by $r^2$ (since $r \neq 0$):
$-\frac{1}{2}(1-\lambda) + \lambda = 0$
$-\frac{1}{2} + \frac{1}{2}\lambda + \lambda = 0$
$\frac{3}{2}\lambda = \frac{1}{2} \implies \lambda = \frac{1}{3}$
Substitute $\lambda = \frac{1}{3}$ back into the expression for $\overrightarrow{ON}$:
$\overrightarrow{ON} = \left(1-\frac{1}{3}\right)\mathbf{a} + \frac{1}{3}\mathbf{b} = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b} = \frac{1}{3}(2\mathbf{a} + \mathbf{b}) \quad \text{(shown)}$

5. Use the cross product to formulate the area of triangle $OAN$.

The area of triangle $OAN$ is given by $\frac{1}{2} |\overrightarrow{OA} \times \overrightarrow{ON}|$.
$\text{Area} = \frac{1}{2} \left| \mathbf{a} \times \frac{1}{3}(2\mathbf{a} + \mathbf{b}) \right|$

6. Expand the cross product to find the constant $k$.

Using the distributive property of the cross product:
$\mathbf{a} \times \frac{1}{3}(2\mathbf{a} + \mathbf{b}) = \frac{2}{3}(\mathbf{a} \times \mathbf{a}) + \frac{1}{3}(\mathbf{a} \times \mathbf{b})$
Since the cross product of a vector with itself is the zero vector ($\mathbf{a} \times \mathbf{a} = \mathbf{0}$), this simplifies to:
$\frac{1}{3}(\mathbf{a} \times \mathbf{b})$
Therefore, the area is:
$\frac{1}{2} \left| \frac{1}{3}(\mathbf{a} \times \mathbf{b}) \right| = \frac{1}{6} |\mathbf{a} \times \mathbf{b}|$
Thus, $k = \frac{1}{6}$.

1. Calculate the magnitude of $\overrightarrow{ON}$ using the scalar product.

Using the expression from part (i), $\overrightarrow{ON} = \frac{1}{3}(2\mathbf{a}+\mathbf{b})$.
We find the square of its magnitude using the scalar product:

Since $A$ and $B$ lie on the circle of radius $r$, $|\mathbf{a}| = |\mathbf{b}| = r$, giving $\mathbf{a}\cdot\mathbf{a} = r^2$ and $\mathbf{b}\cdot\mathbf{b} = r^2$.
The angle between $\mathbf{a}$ and $\mathbf{b}$ is $120^\circ$, so:
$\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}| \cos 120^\circ = r(r)\left(-\frac{1}{2}\right) = -\frac{1}{2}r^2$
Substituting these into the expression:
$|\overrightarrow{ON}|^2 = \frac{1}{9} \left( 4r^2 + 4\left(-\frac{1}{2}r^2\right) + r^2 \right) = \frac{1}{9}(3r^2) = \frac{1}{3}r^2$
Thus, $|\overrightarrow{ON}| = \frac{r}{\sqrt{3}}$.

(Alternatively, one could use simple geometry: in right-angled $\triangle ONB$, $OB = r$ and $\angle OBN = 30^\circ$, directly giving $ON = r \tan 30^\circ = \frac{r}{\sqrt{3}}$).

2. Relate $\overrightarrow{OC}$ to $\overrightarrow{ON}$ using their lengths and collinearity.

Since point $C$ lies on the circle, its distance from the centre $O$ is exactly the radius $r$, so $|\overrightarrow{OC}| = r$.
Because $O$, $N$, and $C$ are collinear, $\overrightarrow{OC}$ is a scalar multiple of $\overrightarrow{ON}$:
$\overrightarrow{OC} = k \overrightarrow{ON}$
Taking magnitudes on both sides:
$|\overrightarrow{OC}| = |k| |\overrightarrow{ON}| \implies r = |k| \left(\frac{r}{\sqrt{3}}\right) \implies |k| = \sqrt{3}$
This gives $k = \pm \sqrt{3}$.
Thus, the possible expressions for $\overrightarrow{OC}$ are:
$\overrightarrow{OC} = \pm \sqrt{3} \left[ \frac{1}{3}(2\mathbf{a}+\mathbf{b}) \right] = \pm \frac{\sqrt{3}}{3}(2\mathbf{a}+\mathbf{b})$

(Note: Analytically, both directions satisfy the condition. If the diagram strictly visualises $O$ as between $C$ and $N$, it narrows the answer to only the negative scalar variation).