The position vectors of the points $A$, $B$ and $C$ with respect to the origin $O$ are $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively. Point $C$ is on $AB$ produced such that $5AB = 3AC$. Given that $\mathbf{a} \cdot \mathbf{c} = 2$, $\mathbf{a}$ is a unit vector and the angle $AOB$ is $\frac{\pi}{3}$, show that $|\mathbf{b}| = \frac{16}{5}$.

1. Express vector $\mathbf{c}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ using the collinearity and length ratios.

Since point $C$ is on $AB$ produced, the points $A$, $B$, and $C$ are collinear in that order. This means vectors $\vec{AB}$ and $\vec{AC}$ have the same direction.
Given $5AB = 3AC$, we can write this as a vector equation:
$5\vec{AB} = 3\vec{AC}$
$5(\mathbf{b} - \mathbf{a}) = 3(\mathbf{c} - \mathbf{a})$
$5\mathbf{b} - 5\mathbf{a} = 3\mathbf{c} - 3\mathbf{a}$
$3\mathbf{c} = 5\mathbf{b} - 2\mathbf{a}$

2. Substitute the expression for $3\mathbf{c}$ into the given dot product condition.

We are given that $\mathbf{a} \cdot \mathbf{c} = 2$. Multiplying both sides by 3 gives:
$3(\mathbf{a} \cdot \mathbf{c}) = 6$
$\mathbf{a} \cdot (3\mathbf{c}) = 6$
Substitute $3\mathbf{c} = 5\mathbf{b} - 2\mathbf{a}$ into the equation:
$\mathbf{a} \cdot (5\mathbf{b} - 2\mathbf{a}) = 6$
$5(\mathbf{a} \cdot \mathbf{b}) - 2(\mathbf{a} \cdot \mathbf{a}) = 6$

3. Use the property of the unit vector $\mathbf{a}$ to find the value of $\mathbf{a} \cdot \mathbf{b}$.

Since $\mathbf{a}$ is a unit vector, its magnitude $|\mathbf{a}| = 1$, which implies $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2 = 1$.
$5(\mathbf{a} \cdot \mathbf{b}) - 2(1) = 6$
$5(\mathbf{a} \cdot \mathbf{b}) = 8$
$\mathbf{a} \cdot \mathbf{b} = \frac{8}{5}$

4. Use the scalar product definition with the given angle to find $|\mathbf{b}|$.

By the definition of the scalar product:
$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(\angle AOB)$
Substitute $|\mathbf{a}| = 1$, $\mathbf{a} \cdot \mathbf{b} = \frac{8}{5}$, and $\angle AOB = \frac{\pi}{3}$:
$\frac{8}{5} = (1)|\mathbf{b}|\cos\left(\frac{\pi}{3}\right)$
$\frac{8}{5} = |\mathbf{b}|\left(\frac{1}{2}\right)$
$|\mathbf{b}| = \frac{16}{5} \text{ (shown)}$