相对于原点 $O$，点 $A$ 和 $B$ 的位置向量分别为 $4\mathbf{i} + 2\mathbf{j}$ 和 $3\mathbf{i} + \mathbf{k}$。点 $C$ 使得 $OACB$ 为平行四边形，点 $E$ 满足 $\overrightarrow{BE} = \frac{1}{4}\overrightarrow{BC}$。点 $M$ 是 $AB$ 的中点。

1. Find the direction vector of the line $AB$.

2. State the vector equation of the line $AB$.

$\text{Line } AB: \mathbf{r} = 4\mathbf{i} + 2\mathbf{j} + \lambda(-\mathbf{i} - 2\mathbf{j} + \mathbf{k}), \text{ where } \lambda \in \mathbb{R}$

3. Find the position vector of $E$.

Since $OACB$ is a parallelogram, $\overrightarrow{BC} = \overrightarrow{OA} = 4\mathbf{i} + 2\mathbf{j}$.
$\overrightarrow{BE} = \frac{1}{4}\overrightarrow{BC} = \frac{1}{4}(4\mathbf{i} + 2\mathbf{j}) = \mathbf{i} + \frac{1}{2}\mathbf{j}$
$\overrightarrow{OE} = \overrightarrow{OB} + \overrightarrow{BE} = (3\mathbf{i} + \mathbf{k}) + \left(\mathbf{i} + \frac{1}{2}\mathbf{j}\right) = 4\mathbf{i} + \frac{1}{2}\mathbf{j} + \mathbf{k}$

4. State the vector equation of the line $OE$.

$\text{Line } OE: \mathbf{r} = \mu\left(4\mathbf{i} + \frac{1}{2}\mathbf{j} + \mathbf{k}\right), \text{ where } \mu \in \mathbb{R}$
By scaling the direction vector by a factor of 2, we can also write this as:
$\mathbf{r} = \mu(8\mathbf{i} + \mathbf{j} + 2\mathbf{k}), \text{ where } \mu \in \mathbb{R}$

1. Formulate the vector equations of lines $AB$ and $OE$ (required from part i).

From part (i), we establish the vector equations for lines $AB$ and $OE$.
Since $OACB$ is a parallelogram, $\overrightarrow{BC} = \overrightarrow{OA} = \begin{pmatrix} 4 \\ 2 \\ 0 \end{pmatrix}$.
$\overrightarrow{BE} = \frac{1}{4} \overrightarrow{BC} = \begin{pmatrix} 1 \\ 0.5 \\ 0 \end{pmatrix} \implies \overrightarrow{OE} = \overrightarrow{OB} + \overrightarrow{BE} = \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 0.5 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 0.5 \\ 1 \end{pmatrix}$
Equation of line $OE$: $\mathbf{r} = \mu \begin{pmatrix} 4 \\ 0.5 \\ 1 \end{pmatrix}$

$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} - \begin{pmatrix} 4 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix}$
Equation of line $AB$: $\mathbf{r} = \begin{pmatrix} 4 \\ 2 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix}$

2. Equate the equations of lines $AB$ and $OE$ to set up a system of equations for the point of intersection.

At the point of intersection $F$, the position vectors from both line equations must be equal:
$\begin{pmatrix} 4 \\ 2 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix} = \mu \begin{pmatrix} 4 \\ 0.5 \\ 1 \end{pmatrix}$
Equating the components gives a system of linear equations:
$4 - \lambda = 4\mu \quad \text{--- (1)}$
$2 - 2\lambda = 0.5\mu \quad \text{--- (2)}$
$\lambda = \mu \quad \text{--- (3)}$

3. Solve the system of equations to find the value of the parameter.

Substitute equation (3) into equation (1):
$4 - \mu = 4\mu$
$5\mu = 4 \implies \mu = \frac{4}{5}$
(Checking consistency with equation (2): $2 - 2(\frac{4}{5}) = \frac{2}{5}$ and $0.5(\frac{4}{5}) = \frac{2}{5}$, which matches.)
Thus, the lines intersect when $\mu = \frac{4}{5}$ (and $\lambda = \frac{4}{5}$).

4. Substitute the parameter back into the line equation to find the position vector of $F$.

Substitute $\mu = \frac{4}{5}$ into the vector equation of line $OE$:
$\overrightarrow{OF} = \frac{4}{5} \begin{pmatrix} 4 \\ 0.5 \\ 1 \end{pmatrix} = \begin{pmatrix} 16/5 \\ 2/5 \\ 4/5 \end{pmatrix}$
So the position vector of $F$ is $\frac{16}{5}\mathbf{i} + \frac{2}{5}\mathbf{j} + \frac{4}{5}\mathbf{k}$.

5. Deduce the ratio $\frac{OF}{OE}$ using the scalar multiple found for the parameter.

Since $\overrightarrow{OF} = \frac{4}{5}\overrightarrow{OE}$ and the points $O$, $F$, and $E$ are collinear, the ratio of their lengths is equal to the scalar parameter:
$\frac{OF}{OE} = \frac{4}{5}$

1. Find the position vector of midpoint $M$ and the direction vectors $\overrightarrow{MO}$ and $\overrightarrow{MA}$.

$\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB}) = \frac{1}{2} \left[ \begin{pmatrix} 4 \\ 2 \\ 0 \end{pmatrix} + \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} \right] = \begin{pmatrix} 3.5 \\ 1 \\ 0.5 \end{pmatrix}$

$\overrightarrow{MO} = -\overrightarrow{OM} = \begin{pmatrix} -3.5 \\ -1 \\ -0.5 \end{pmatrix}$

$\overrightarrow{MA} = \overrightarrow{OA} - \overrightarrow{OM} = \begin{pmatrix} 4 \\ 2 \\ 0 \end{pmatrix} - \begin{pmatrix} 3.5 \\ 1 \\ 0.5 \end{pmatrix} = \begin{pmatrix} 0.5 \\ 1 \\ -0.5 \end{pmatrix}$

2. Calculate the scalar product $\overrightarrow{MO} \cdot \overrightarrow{MA}$ and their respective magnitudes.

$\overrightarrow{MO} \cdot \overrightarrow{MA} = (-3.5)(0.5) + (-1)(1) + (-0.5)(-0.5) = -1.75 - 1 + 0.25 = -2.5$

$|\overrightarrow{MO}| = \sqrt{(-3.5)^2 + (-1)^2 + (-0.5)^2} = \sqrt{12.25 + 1 + 0.25} = \sqrt{13.5}$

$|\overrightarrow{MA}| = \sqrt{0.5^2 + 1^2 + (-0.5)^2} = \sqrt{0.25 + 1 + 0.25} = \sqrt{1.5}$

3. Apply the scalar product formula to find angle $OMA$.

$\cos(\angle OMA) = \frac{\overrightarrow{MO} \cdot \overrightarrow{MA}}{|\overrightarrow{MO}| |\overrightarrow{MA}|} = \frac{-2.5}{\sqrt{13.5}\sqrt{1.5}}$

$\cos(\angle OMA) = \frac{-2.5}{\sqrt{20.25}} = \frac{-2.5}{4.5} = -\frac{5}{9}$

$\angle OMA = \cos^{-1}\left(-\frac{5}{9}\right) \approx 123.748^\circ \approx 123.7^\circ \text{ (to 1 d.p.)}$

1. Find the position vector $\overrightarrow{OM}$ and the direction vector $\overrightarrow{AB}$.

$\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB}) = \frac{1}{2} \begin{pmatrix} 4+3 \\ 2+0 \\ 0+1 \end{pmatrix} = \begin{pmatrix} 7/2 \\ 1 \\ 1/2 \end{pmatrix}$
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 3-4 \\ 0-2 \\ 1-0 \end{pmatrix} = \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix}$

2. Evaluate the scalar product $\overrightarrow{OM} \cdot \overrightarrow{AB}$ and the magnitude $|\overrightarrow{AB}|$.

$\overrightarrow{OM} \cdot \overrightarrow{AB} = \left(\frac{7}{2}\right)(-1) + (1)(-2) + \left(\frac{1}{2}\right)(1) = -\frac{7}{2} - 2 + \frac{1}{2} = -5$
$|\overrightarrow{AB}| = \sqrt{(-1)^2 + (-2)^2 + 1^2} = \sqrt{6}$

3. Calculate the exact length of the projection using the projection formula.

$\text{Length of projection} = \frac{|\overrightarrow{OM} \cdot \overrightarrow{AB}|}{|\overrightarrow{AB}|}$
$= \frac{|-5|}{\sqrt{6}} = \frac{5}{\sqrt{6}} = \frac{5\sqrt{6}}{6}$