Relative to the origin $O$, the points $A$, $B$, and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $3\mathbf{a}+\mathbf{b}$ respectively. The point $D$ lies on $AB$ such that $AD = k\,AB$, where $0 < k < 1$.

1. Find the position vector of point $D$.

Since $D$ lies on $AB$ such that $AD = k AB$ and $0 < k < 1$, we can express this as a vector relationship:
$\vec{AD} = k\vec{AB}$
$\vec{OD} - \vec{OA} = k(\vec{OB} - \vec{OA})$
$\vec{OD} - \mathbf{a} = k(\mathbf{b} - \mathbf{a})$
$\vec{OD} = \mathbf{a} + k\mathbf{b} - k\mathbf{a}$
$\vec{OD} = (1-k)\mathbf{a} + k\mathbf{b}$

2. Write the vector equation of the line $OD$.

The line $OD$ passes through the origin $O$ (position vector $\mathbf{0}$) and has the direction of $\vec{OD}$.
Introducing a scalar parameter $\lambda$, the vector equation of the line is $\mathbf{r} = \mathbf{0} + \lambda \vec{OD}$.
$\mathbf{r} = \lambda((1-k)\mathbf{a} + k\mathbf{b})$, where $\lambda \in \mathbb{R}$.

1. Find the position vector of point $D$.

Using the information given in part (a), $D$ lies on $AB$ such that $AD = k\,AB$, meaning it divides $AB$ in the ratio $k:(1-k)$. By the ratio theorem:
$\vec{OD} = (1-k)\mathbf{a} + k\mathbf{b}$

2. Find the position vector of point $E$.

$E$ is the midpoint of $BC$, so using the midpoint formula (ratio theorem 1:1):
$\vec{OE} = \frac{\vec{OB} + \vec{OC}}{2}$
$\vec{OE} = \frac{\mathbf{b} + (3\mathbf{a} + \mathbf{b})}{2} = \frac{3}{2}\mathbf{a} + \mathbf{b}$

3. Apply the condition for the collinearity of $O$, $D$, and $E$.

Since the points $O$, $D$, and $E$ are collinear, the position vectors $\vec{OD}$ and $\vec{OE}$ must be parallel. Thus, $\vec{OD} = \mu\vec{OE}$ for some scalar constant $\mu$.
$(1-k)\mathbf{a} + k\mathbf{b} = \mu \left(\frac{3}{2}\mathbf{a} + \mathbf{b}\right)$

4. Compare coefficients of $\mathbf{a}$ and $\mathbf{b}$ to solve for $k$.

Since $\mathbf{a}$ and $\mathbf{b}$ are non-parallel vectors, we can equate their coefficients:
Comparing coefficients of $\mathbf{b}$:
$k = \mu$
Comparing coefficients of $\mathbf{a}$:
$1 - k = \frac{3}{2}\mu$
Substituting $\mu = k$ into the second equation:
$1 - k = \frac{3}{2}k$
$1 = \frac{5}{2}k$
$k = \frac{2}{5}$

1. Expand the scalar product $(3\mathbf{a} - 2\mathbf{b})\cdot(3\mathbf{a} - 2\mathbf{b})$ using distributive properties.

$\begin{aligned} (3\mathbf{a} - 2\mathbf{b})\cdot(3\mathbf{a} - 2\mathbf{b}) &= 9(\mathbf{a}\cdot\mathbf{a}) - 6(\mathbf{a}\cdot\mathbf{b}) - 6(\mathbf{b}\cdot\mathbf{a}) + 4(\mathbf{b}\cdot\mathbf{b}) \\ &= 9|\mathbf{a}|^2 - 12(\mathbf{a}\cdot\mathbf{b}) + 4|\mathbf{b}|^2 \end{aligned}$

2. Equate the expanded expression to $|3\mathbf{a} - 2\mathbf{b}|^2$ and solve for $\mathbf{a}\cdot\mathbf{b}$.

Since $|3\mathbf{a} - 2\mathbf{b}| = \sqrt{31}$, its square is $31$.

$\begin{aligned} 9|\mathbf{a}|^2 - 12(\mathbf{a}\cdot\mathbf{b}) + 4|\mathbf{b}|^2 &= 31 \\ 9(1)^2 - 12(\mathbf{a}\cdot\mathbf{b}) + 4(2)^2 &= 31 \\ 9 - 12(\mathbf{a}\cdot\mathbf{b}) + 16 &= 31 \\ 25 - 12(\mathbf{a}\cdot\mathbf{b}) &= 31 \\ -12(\mathbf{a}\cdot\mathbf{b}) &= 6 \\ \mathbf{a}\cdot\mathbf{b} &= -\frac{1}{2} \end{aligned}$

3. Set up the equation for the angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$.

$\begin{aligned} \cos \theta &= \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \\ &= \frac{-0.5}{(1)(2)} \\ &= -0.25 \end{aligned}$

4. Calculate the angle $\theta$.

$\begin{aligned} \theta &= \cos^{-1}(-0.25) \\ &\approx 104.4775^\circ \\ &\approx 104.5^\circ \text{ (to 1 d.p.)} \end{aligned}$

1. State the geometrical interpretation using the magnitude of $\mathbf{a}$.

From part (c), we know that $|\mathbf{a}| = 1$, which means $\mathbf{a}$ is a unit vector ($\mathbf{a} = \hat{\mathbf{a}}$).

Therefore, $|\mathbf{a}\cdot\mathbf{b}| = |\hat{\mathbf{a}}\cdot\mathbf{b}|$.
This represents the length (or magnitude) of the projection of vector $\mathbf{b}$ onto the direction of vector $\mathbf{a}$.