The diagram shows a solid with a horizontal rectangular base $OABC$ in which the lengths of $OA$ and $OC$ are $12\text{ m}$ and $8\text{ m}$ respectively. $M$ and $N$ are the mid-points of $OC$ and $AB$ respectively, and $D$ and $E$ are $7\text{ m}$ and $5\text{ m}$ vertically above $M$ and $N$ respectively. The point $O$ is taken as the origin for position vectors. Perpendicular vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ are such that $\mathbf{i}$ and $\mathbf{j}$ are parallel to $OA$ and $OC$ respectively, and $\mathbf{k}$ is perpendicular to the plane $OABC$.

1. Determine the position vectors of the relevant points B, D, and E from the given geometric description.

Let the origin be $O(0, 0, 0)$. With $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ parallel to $OA$, $OC$ and perpendicular to the base respectively:
$\vec{OA} = \begin{pmatrix} 12 \\ 0 \\ 0 \end{pmatrix}$, $\vec{OC} = \begin{pmatrix} 0 \\ 8 \\ 0 \end{pmatrix}$
Since $OABC$ is a rectangle, $\vec{OB} = \vec{OA} + \vec{OC} = \begin{pmatrix} 12 \\ 8 \\ 0 \end{pmatrix}$.
$M$ is the midpoint of $OC$, so $\vec{OM} = \begin{pmatrix} 0 \\ 4 \\ 0 \end{pmatrix}$. $D$ is $7\text{ m}$ vertically above $M$, so $\vec{OD} = \begin{pmatrix} 0 \\ 4 \\ 7 \end{pmatrix}$.
$N$ is the midpoint of $AB$, so $\vec{ON} = \begin{pmatrix} 12 \\ 4 \\ 0 \end{pmatrix}$. $E$ is $5\text{ m}$ vertically above $N$, so $\vec{OE} = \begin{pmatrix} 12 \\ 4 \\ 5 \end{pmatrix}$.

2. Find the vector equation of the line $DB$.

The direction vector of line $DB$ is:
$\vec{DB} = \vec{OB} - \vec{OD} = \begin{pmatrix} 12 \\ 8 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ 4 \\ 7 \end{pmatrix} = \begin{pmatrix} 12 \\ 4 \\ -7 \end{pmatrix}$
Thus, a vector equation of the line $DB$ is:
$\mathbf{r} = \begin{pmatrix} 0 \\ 4 \\ 7 \end{pmatrix} + \lambda \begin{pmatrix} 12 \\ 4 \\ -7 \end{pmatrix}, \lambda \in \mathbb{R}$

3. Express the position vector of the foot of the perpendicular, $F$, and the vector $\vec{EF}$ in terms of $\lambda$.

Let $F$ be the foot of the perpendicular from $E$ to $DB$. Then $F$ lies on the line $DB$, so its position vector is:
$\vec{OF} = \begin{pmatrix} 12\lambda \\ 4+4\lambda \\ 7-7\lambda \end{pmatrix}$ for some $\lambda \in \mathbb{R}$.
The displacement vector from $E$ to $F$ is:
$\vec{EF} = \vec{OF} - \vec{OE} = \begin{pmatrix} 12\lambda \\ 4+4\lambda \\ 7-7\lambda \end{pmatrix} - \begin{pmatrix} 12 \\ 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 12\lambda - 12 \\ 4\lambda \\ 2-7\lambda \end{pmatrix}$

4. Use the dot product to solve for $\lambda$.

Since $\vec{EF}$ is perpendicular to line $DB$, the dot product of $\vec{EF}$ with the direction vector of $DB$ is zero:
$\vec{EF} \cdot \vec{DB} = 0$
$\begin{pmatrix} 12\lambda - 12 \\ 4\lambda \\ 2-7\lambda \end{pmatrix} \cdot \begin{pmatrix} 12 \\ 4 \\ -7 \end{pmatrix} = 0$
$12(12\lambda - 12) + 4(4\lambda) - 7(2-7\lambda) = 0$
$144\lambda - 144 + 16\lambda - 14 + 49\lambda = 0$
$209\lambda - 158 = 0 \implies \lambda = \frac{158}{209}$

5. Substitute $\lambda$ back to find the final position vector of the foot of the perpendicular.

Substitute $\lambda = \frac{158}{209}$ into $\vec{OF}$:
$\vec{OF} = \begin{pmatrix} 0 \\ 4 \\ 7 \end{pmatrix} + \frac{158}{209} \begin{pmatrix} 12 \\ 4 \\ -7 \end{pmatrix} = \begin{pmatrix} 1896/209 \\ 4 + 632/209 \\ 7 - 1106/209 \end{pmatrix} = \frac{1}{209} \begin{pmatrix} 1896 \\ 1468 \\ 357 \end{pmatrix}$

1. Find the coordinates of the relevant points and the direction vector of the line $DB$.

Let the origin be $O(0, 0, 0)$.
Since $OABC$ is a rectangle with $OA = 12$ and $OC = 8$, the coordinates of $B$ are $(12, 8, 0)$.
$M$ is the mid-point of $OC$, so $M$ is $(0, 4, 0)$. $D$ is $7\text{ m}$ vertically above $M$, giving $D(0, 4, 7)$.
The direction vector of line $DB$ is:
$\vec{DB} = \vec{OB} - \vec{OD} = \begin{pmatrix} 12 \\ 8 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ 4 \\ 7 \end{pmatrix} = \begin{pmatrix} 12 \\ 4 \\ -7 \end{pmatrix}$

2. Determine a normal vector to the plane $OBE$.

$N$ is the mid-point of $AB$, so $N$ is $(12, 4, 0)$. $E$ is $5\text{ m}$ vertically above $N$, giving $E(12, 4, 5)$.
The plane $OBE$ contains the vectors $\vec{OB} = \begin{pmatrix} 12 \\ 8 \\ 0 \end{pmatrix}$ and $\vec{OE} = \begin{pmatrix} 12 \\ 4 \\ 5 \end{pmatrix}$.
A normal vector $\mathbf{n}$ to the plane $OBE$ is given by their cross product:
$\mathbf{n} = \vec{OB} \times \vec{OE} = \begin{pmatrix} 12 \\ 8 \\ 0 \end{pmatrix} \times \begin{pmatrix} 12 \\ 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 8(5) - 0(4) \\ 0(12) - 12(5) \\ 12(4) - 8(12) \end{pmatrix} = \begin{pmatrix} 40 \\ -60 \\ -48 \end{pmatrix}$
We can take a simpler parallel vector by factoring out $4$, giving $\mathbf{n_1} = \begin{pmatrix} 10 \\ -15 \\ -12 \end{pmatrix}$.

3. Calculate the acute angle between the line $DB$ and the plane $OBE$.

Let $\theta$ be the acute angle between the line $DB$ and the plane $OBE$.
$\sin \theta = \frac{|\vec{DB} \cdot \mathbf{n_1}|}{|\vec{DB}||\mathbf{n_1}|}$
$\sin \theta = \frac{\left| \begin{pmatrix} 12 \\ 4 \\ -7 \end{pmatrix} \cdot \begin{pmatrix} 10 \\ -15 \\ -12 \end{pmatrix} \right|}{\sqrt{12^2 + 4^2 + (-7)^2} \sqrt{10^2 + (-15)^2 + (-12)^2}}$
$\sin \theta = \frac{|120 - 60 + 84|}{\sqrt{209} \sqrt{469}}$
$\sin \theta = \frac{144}{\sqrt{98021}}$
$\theta = \sin^{-1}\left(\frac{144}{\sqrt{98021}}\right) \approx 27.38^\circ$
The acute angle is $27.4^\circ$ (to 1 d.p.).

1. Identify the coordinates of $D$, $E$, and $B$ to find vectors $\vec{DE}$ and $\vec{DB}$.

From the given dimensions and taking $O$ as the origin:
$D = (0, 4, 7)$
$E = (12, 4, 5)$
$B = (12, 8, 0)$

$\vec{DE} = \begin{pmatrix} 12 \\ 4 \\ 5 \end{pmatrix} - \begin{pmatrix} 0 \\ 4 \\ 7 \end{pmatrix} = \begin{pmatrix} 12 \\ 0 \\ -2 \end{pmatrix}$
$\vec{DB} = \begin{pmatrix} 12 \\ 8 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ 4 \\ 7 \end{pmatrix} = \begin{pmatrix} 12 \\ 4 \\ -7 \end{pmatrix}$

2. Calculate the dot product $\vec{DE} \cdot \vec{DB}$ and the magnitude $|\vec{DB}|$.

$\vec{DE} \cdot \vec{DB} = \begin{pmatrix} 12 \\ 0 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 12 \\ 4 \\ -7 \end{pmatrix}$
$= (12)(12) + (0)(4) + (-2)(-7)$
$= 144 + 0 + 14 = 158$

$|\vec{DB}| = \sqrt{12^2 + 4^2 + (-7)^2} = \sqrt{144 + 16 + 49} = \sqrt{209}$

3. Use the scalar projection formula to find the length of projection.

Length of projection of $DE$ onto $DB$ is given by $\frac{|\vec{DE} \cdot \vec{DB}|}{|\vec{DB}|}$.

Length $= \frac{158}{\sqrt{209}} \approx 10.9\text{ m}$ (to 3 s.f.)

1. State whether the lines are coplanar by analysing their intersection and parallel properties.

Line $AC$ lies entirely on the base plane $OABC$ ($z=0$).
Line $DB$ connects point $D$ (which is above the plane) to point $B$ (which is on the plane), so it intersects the plane $OABC$ exactly at point $B$.
Since $B$ is a vertex of the rectangle and does not lie on the diagonal $AC$, lines $DB$ and $AC$ do not intersect.
Furthermore, the lines are not parallel since $AC$ is horizontal but $DB$ has a vertical component.
Two lines that are neither parallel nor intersecting are skew lines. Therefore, lines $DB$ and $AC$ are not coplanar.