A temporary isolation centre is built to manage the increasing number of COVID-19 cases. The roof takes the shape of a triangular prism. Points $(x, y, z)$ are defined relative to an origin, $O$, with unit vectors $\mathbf{i}$ along $\overrightarrow{OA}$, $\mathbf{j}$ along $\overrightarrow{OC}$, and $\mathbf{k}$ along $\overrightarrow{OD}$ (see diagram). The coordinates of $D, E, F, G$ and $I$ are $(0,0,3)$, $(4,0,3)$, $(4,12,4)$, $(0,12,4)$ and $(1,12,7)$ respectively. The units are measured in metres.

1. Find two non-parallel direction vectors lying on the plane $EFIH$.

Using points $E(4, 0, 3)$, $F(4, 12, 4)$, and $I(1, 12, 7)$:
$\overrightarrow{EF} = \begin{pmatrix} 4 \\ 12 \\ 4 \end{pmatrix} - \begin{pmatrix} 4 \\ 0 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 12 \\ 1 \end{pmatrix}$
$\overrightarrow{FI} = \begin{pmatrix} 1 \\ 12 \\ 7 \end{pmatrix} - \begin{pmatrix} 4 \\ 12 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 0 \\ 3 \end{pmatrix}$

2. Calculate the cross product of the two direction vectors to find a normal vector to the plane.

$\mathbf{n} = \overrightarrow{EF} \times \overrightarrow{FI} = \begin{pmatrix} 0 \\ 12 \\ 1 \end{pmatrix} \times \begin{pmatrix} -3 \\ 0 \\ 3 \end{pmatrix} = \begin{pmatrix} (12)(3) - (1)(0) \\ (1)(-3) - (0)(3) \\ (0)(0) - (12)(-3) \end{pmatrix} = \begin{pmatrix} 36 \\ -3 \\ 36 \end{pmatrix}$
A simpler normal vector parallel to $\mathbf{n}$ is obtained by dividing by $3$:
$\mathbf{n_1} = \begin{pmatrix} 12 \\ -1 \\ 12 \end{pmatrix}$

3. Form the cartesian equation of the plane using the normal vector and a point on the plane.

Let the equation of the plane be $\mathbf{r} \cdot \begin{pmatrix} 12 \\ -1 \\ 12 \end{pmatrix} = D$.
Using point $E(4, 0, 3)$:
$D = \begin{pmatrix} 4 \\ 0 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 12 \\ -1 \\ 12 \end{pmatrix} = 4(12) + 0(-1) + 3(12) = 48 + 0 + 36 = 84$
Thus, the cartesian equation of the plane is $12x - y + 12z = 84$.

1. Obtain the cartesian equation of plane $EFIH$ (required from part i).

From the given coordinates, we find two direction vectors on plane $EFIH$:
$\overrightarrow{EF} = \begin{pmatrix} 4 \\ 12 \\ 4 \end{pmatrix} - \begin{pmatrix} 4 \\ 0 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 12 \\ 1 \end{pmatrix}$
$\overrightarrow{EI} = \begin{pmatrix} 1 \\ 12 \\ 7 \end{pmatrix} - \begin{pmatrix} 4 \\ 0 \\ 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 12 \\ 4 \end{pmatrix}$
A normal vector to plane $EFIH$ is:
$\mathbf{n}_1 = \begin{pmatrix} 0 \\ 12 \\ 1 \end{pmatrix} \times \begin{pmatrix} -3 \\ 12 \\ 4 \end{pmatrix} = \begin{pmatrix} 36 \\ -3 \\ 36 \end{pmatrix} = 3\begin{pmatrix} 12 \\ -1 \\ 12 \end{pmatrix}$
Using point $E(4, 0, 3)$, the equation of plane $EFIH$ is:
$\mathbf{r} \cdot \begin{pmatrix} 12 \\ -1 \\ 12 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 12 \\ -1 \\ 12 \end{pmatrix} = 48 - 0 + 36 = 84$
$\implies 12x - y + 12z = 84$

2. Solve the equations of the two planes simultaneously to find the line of intersection $HI$.

The line $HI$ lies on both planes. We solve their equations simultaneously:
Plane $EFIH$: $12x - y + 12z = 84$ --- (1)
Plane $DGIH$: $36x + y - 12z = -36$ --- (2)

Adding equations (1) and (2):
$(12x - y + 12z) + (36x + y - 12z) = 84 - 36$
$48x = 48 \implies x = 1$

3. Formulate the cartesian equation of the line.

Substitute $x = 1$ into equation (1):
$12(1) - y + 12z = 84 \implies y - 12z = -72$
Rearranging to form a standard line equation:
$y + 72 = 12z \implies \frac{y+72}{12} = z$
Since we know the point $I(1, 12, 7)$ lies on this line, we can re-centre the equation around it to match standard format:
$\frac{y-12}{12} = z - 7$
The cartesian equation is therefore $x = 1$, $\frac{y-12}{12} = z - 7$.

1. Find the coordinates of $M$, the midpoint of $EF$.

$\overrightarrow{OM} = \frac{\overrightarrow{OE} + \overrightarrow{OF}}{2}$
$\overrightarrow{OM} = \frac{1}{2} \left[ \begin{pmatrix} 4 \\ 0 \\ 3 \end{pmatrix} + \begin{pmatrix} 4 \\ 12 \\ 4 \end{pmatrix} \right] = \begin{pmatrix} 4 \\ 6 \\ 3.5 \end{pmatrix}$

2. Formulate the vector equations of the lines representing the two cables.

For cable 1 (line $FD$):
$\overrightarrow{DF} = \begin{pmatrix} 4 \\ 12 \\ 4 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 12 \\ 1 \end{pmatrix}$
$l_1: \mathbf{r} = \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 12 \\ 1 \end{pmatrix}, \quad \lambda \in \mathbb{R}$

For cable 2 (line $GM$):
$\overrightarrow{GM} = \begin{pmatrix} 4 \\ 6 \\ 3.5 \end{pmatrix} - \begin{pmatrix} 0 \\ 12 \\ 4 \end{pmatrix} = \begin{pmatrix} 4 \\ -6 \\ -0.5 \end{pmatrix}$
Scaling the direction vector by 2 yields $\begin{pmatrix} 8 \\ -12 \\ -1 \end{pmatrix}$:
$l_2: \mathbf{r} = \begin{pmatrix} 0 \\ 12 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} 8 \\ -12 \\ -1 \end{pmatrix}, \quad \mu \in \mathbb{R}$

3. Set up a system of equations to find the parameter values at the intersection point.

Equating the $x$ and $y$ components of $l_1$ and $l_2$:
$4\lambda = 8\mu \implies \lambda = 2\mu \quad \text{--- (1)}$
$12\lambda = 12 - 12\mu \quad \text{--- (2)}$
Substitute (1) into (2):
$12(2\mu) = 12 - 12\mu$
$24\mu + 12\mu = 12 \implies 36\mu = 12 \implies \mu = \frac{1}{3}$
Thus, $\lambda = 2\left(\frac{1}{3}\right) = \frac{2}{3}$.

(Checking with the $z$-component: $3 + \frac{2}{3} = \frac{11}{3}$ and $4 - \frac{1}{3} = \frac{11}{3}$, which are consistent)

4. Evaluate the coordinates of intersection point $J$.

Substitute $\lambda = \frac{2}{3}$ into $l_1$:
$\overrightarrow{OJ} = \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} + \frac{2}{3} \begin{pmatrix} 4 \\ 12 \\ 1 \end{pmatrix} = \begin{pmatrix} 8/3 \\ 8 \\ 11/3 \end{pmatrix}$

1. State the vector equation of the roof ridge $HI$ (derivable from earlier parts).

From earlier parts, the roof ridge $HI$ is parallel to the base edge $DG$, so its direction vector is $\overrightarrow{DG} = \begin{pmatrix} 0 \\ 12 \\ 4 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 12 \\ 1 \end{pmatrix}$.
Passing through $I(1, 12, 7)$, the vector equation of the line containing $HI$ is:
$\mathbf{r} = \begin{pmatrix} 1 \\ 12 \\ 7 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 12 \\ 1 \end{pmatrix}$

2. State the coordinates of the cable intersection point $J$ (calculated from part iii).

Cable 1 passes through $F(4,12,4)$ and $D(0,0,3)$. Cable 2 passes through $G(0,12,4)$ and the midpoint of $EF$, $M(4,6,3.5)$.
Equating their respective line equations gives:
$\begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} + s\begin{pmatrix} 4 \\ 12 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 12 \\ 4 \end{pmatrix} + t\begin{pmatrix} 4 \\ -6 \\ -0.5 \end{pmatrix}$
Solving yields $s = t = \frac{2}{3}$. Substituting $s = \frac{2}{3}$ into the first line gives the coordinates of intersection $J$:
$J\left(\frac{8}{3}, 8, \frac{11}{3}\right)$

3. Express the displacement vector $\overrightarrow{JP}$ from $J$ to a general point $P$ on $HI$.

Let $P$ be a point on the line $HI$.
$\overrightarrow{JP} = \overrightarrow{OP} - \overrightarrow{OJ} = \begin{pmatrix} 1 \\ 12 + 12\lambda \\ 7 + \lambda \end{pmatrix} - \begin{pmatrix} 8/3 \\ 8 \\ 11/3 \end{pmatrix} = \begin{pmatrix} -5/3 \\ 4 + 12\lambda \\ 10/3 + \lambda \end{pmatrix}$

4. Use the dot product to locate the foot of the perpendicular from $J$ to $HI$.

For the shortest distance, $\overrightarrow{JP}$ must be perpendicular to the direction vector of $HI$:
$\begin{pmatrix} -5/3 \\ 4 + 12\lambda \\ 10/3 + \lambda \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 12 \\ 1 \end{pmatrix} = 0$
$0 + 12(4 + 12\lambda) + 1\left(\frac{10}{3} + \lambda\right) = 0$
$48 + 144\lambda + \frac{10}{3} + \lambda = 0 \implies 145\lambda = -\frac{154}{3} \implies \lambda = -\frac{154}{435} \approx -0.354$
(Note: The segment $HI$ spans from $y=0$ at $\lambda=-1$ to $y=12$ at $\lambda=0$. Since $-1 < -0.354 < 0$, the closest point lies strictly on the roof ridge segment itself, rather than beyond its ends.)

5. Calculate the required length of cable and conclude.

Substitute $\lambda = -\frac{154}{435}$ back into $\overrightarrow{JP}$:
$\overrightarrow{JP} = \begin{pmatrix} -5/3 \\ 4 - \frac{1848}{435} \\ \frac{10}{3} - \frac{154}{435} \end{pmatrix} = \begin{pmatrix} -5/3 \\ -36/145 \\ 432/145 \end{pmatrix}$
The shortest distance is the magnitude $|\overrightarrow{JP}|$:
$|\overrightarrow{JP}| = \sqrt{ \left(-\frac{5}{3}\right)^2 + \left(-\frac{36}{145}\right)^2 + \left(\frac{432}{145}\right)^2 } = \sqrt{\frac{15289}{1305}} \approx 3.42 \text{ m}$
Since $3.42 \text{ m} > 3.2 \text{ m}$, the remaining steel cable is not long enough.