相对于原点 $O$，点 $A$、$B$ 和 $C$ 的位置向量分别为 $\mathbf{a}$、$\mathbf{b}$ 和 $3\mathbf{a}+\mathbf{b}$。点 $D$ 在 $AB$ 上，满足 $AD = k\,AB$，其中 $0 < k < 1$。

1. Find the position vector of $D$ using vector addition or the ratio theorem.

$\vec{AD} = k\vec{AB} = k(\mathbf{b} - \mathbf{a})$
$\vec{OD} = \vec{OA} + \vec{AD} = \mathbf{a} + k(\mathbf{b} - \mathbf{a})$
$\vec{OD} = (1-k)\mathbf{a} + k\mathbf{b}$

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

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

1. Find the position vector of $D$ (as required from the dependent part (a)).

$\vec{OD} = \vec{OA} + \vec{AD}$
Since $AD = k\,AB$,
$\vec{OD} = \mathbf{a} + k(\vec{OB} - \vec{OA})$
$\vec{OD} = \mathbf{a} + k(\mathbf{b} - \mathbf{a}) = (1-k)\mathbf{a} + k\mathbf{b}$

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

Since $E$ is the midpoint of $BC$,
$\vec{OE} = \frac{\vec{OB} + \vec{OC}}{2}$
$\vec{OE} = \frac{\mathbf{b} + (3\mathbf{a} + \mathbf{b})}{2} = \frac{3\mathbf{a} + 2\mathbf{b}}{2}$
$\vec{OE} = \frac{3}{2}\mathbf{a} + \mathbf{b}$

3. Set up an equation using the condition that $O$, $D$, and $E$ are collinear.

Since $O$, $D$, and $E$ are collinear, $\vec{OD} = \mu \vec{OE}$ for some scalar $\mu$.
$(1-k)\mathbf{a} + k\mathbf{b} = \mu \left( \frac{3}{2}\mathbf{a} + \mathbf{b} \right)$
$(1-k)\mathbf{a} + k\mathbf{b} = \frac{3}{2}\mu\mathbf{a} + \mu\mathbf{b}$

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

Comparing the coefficients of $\mathbf{b}$:
$k = \mu$

Comparing the coefficients of $\mathbf{a}$:
$1 - k = \frac{3}{2}\mu$

Substitute $\mu = k$ into the second equation:
$1 - k = \frac{3}{2}k$
$1 = \frac{5}{2}k$
$k = \frac{2}{5}$

1. Expand the scalar product expression.

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

2. Substitute the known magnitudes to find the numerical value of $\mathbf{a}\cdot\mathbf{b}$.

Since $(3\mathbf{a}-2\mathbf{b})\cdot(3\mathbf{a}-2\mathbf{b}) = |3\mathbf{a}-2\mathbf{b}|^2$, we substitute the given magnitudes:\n$(\sqrt{31})^2 = 9(1)^2 - 12\mathbf{a}\cdot\mathbf{b} + 4(2)^2$\n$31 = 9 - 12\mathbf{a}\cdot\mathbf{b} + 16$\n$31 = 25 - 12\mathbf{a}\cdot\mathbf{b}$\n$12\mathbf{a}\cdot\mathbf{b} = -6$\n$\mathbf{a}\cdot\mathbf{b} = -0.5$

3. Calculate the angle between $\mathbf{a}$ and $\mathbf{b}$.

Let the angle between the vectors be $\theta$.\n$\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$\n$\cos\theta = \frac{-0.5}{(1)(2)} = -0.25$\n$\theta = \cos^{-1}(-0.25) \approx 104.4775^\circ$\n$\theta \approx 104.5^\circ \text{ (to 1 d.p.)}$

1. Identify the geometrical meaning using the unit magnitude of vector a

From part (c), we are given that $|\mathbf{a}| = 1$. This means that $\mathbf{a}$ is a unit vector (i.e. $\mathbf{a} = \hat{\mathbf{a}}$).
The expression $|\mathbf{a} \cdot \mathbf{b}|$ is equivalent to $|\mathbf{b} \cdot \hat{\mathbf{a}}|$, which by definition is the length of the projection of vector $\mathbf{b}$ onto the direction of vector $\mathbf{a}$.