The molecular geometry is the three-dimensional arrangement of atoms in a molecule. Fig. 1 shows the molecular geometry of a particular molecule which is of a tetrahedral form with one central atom lying inside the tetrahedron and 4 other atoms at the vertices of the tetrahedron. The points $A$, $B$, $C$ and $D$ represent the 4 vertices of the tetrahedron and are connected to the central atom, which is denoted by the point $E$, forming 4 pairs of bonding atoms $AE$, $BE$, $CE$ and $DE$. The coordinates of $A$, $B$, $C$, $D$ and $E$ are given by $(2,3,4)$, $(0,0,3)$, $(3,0,0)$, $(0,4,0)$ and $(1,1,2)$ respectively. The plane $p$ passes through the points $A$, $B$ and $C$. The line $l$ passes through the points $D$ and $E$.

1. Find two distinct direction vectors parallel to the plane.

Let the origin be $O$. The position vectors are $\vec{OA} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}$, $\vec{OB} = \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix}$, and $\vec{OC} = \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix}$.

Find the vectors $\vec{BC}$ and $\vec{BA}$:
$\vec{BC} = \vec{OC} - \vec{OB} = \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ -3 \end{pmatrix}$
$\vec{BA} = \vec{OA} - \vec{OB} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$

2. Determine a normal vector to the plane using the cross product.

A normal vector $\mathbf{n}$ to the plane is given by the cross product $\vec{BC} \times \vec{BA}$:
$\mathbf{n} = \begin{pmatrix} 3 \\ 0 \\ -3 \end{pmatrix} \times \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} (0)(1) - (-3)(3) \\ (-3)(2) - (3)(1) \\ (3)(3) - (0)(2) \end{pmatrix} = \begin{pmatrix} 9 \\ -9 \\ 9 \end{pmatrix}$
To simplify, we can use a scalar multiple of this vector as our normal vector:
$\mathbf{n}_1 = \frac{1}{9} \begin{pmatrix} 9 \\ -9 \\ 9 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$

3. Form the cartesian equation of the plane.

Using the normal vector $\mathbf{n}_1 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ and the point $C(3, 0, 0)$, the scalar product equation of the plane $p$ is $\mathbf{r} \cdot \mathbf{n}_1 = \vec{OC} \cdot \mathbf{n}_1$:
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = (3)(1) + (0)(-1) + (0)(1) = 3$
Converting to cartesian form:
$x - y + z = 3$

1. Find the direction vector of the line $l$ passing through points $D$ and $E$.

The position vectors are $\overrightarrow{OD} = \begin{pmatrix} 0 \\ 4 \\ 0 \end{pmatrix}$ and $\overrightarrow{OE} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$.

The direction vector of line $l$ is:
$\overrightarrow{DE} = \overrightarrow{OE} - \overrightarrow{OD} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} - \begin{pmatrix} 0 \\ 4 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}$

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

Using point $D$ as the position vector, the vector equation of the line is:
$\mathbf{r} = \begin{pmatrix} 0 \\ 4 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}, \lambda \in \mathbb{R}$

3. Derive the cartesian equation of the plane $p$ (required from part i) to find the intersection.

Plane $p$ passes through $A(2,3,4)$, $B(0,0,3)$, and $C(3,0,0)$.
$\overrightarrow{AB} = \begin{pmatrix} -2 \\ -3 \\ -1 \end{pmatrix}, \quad \overrightarrow{AC} = \begin{pmatrix} 1 \\ -3 \\ -4 \end{pmatrix}$
A normal vector to $p$ is:
$\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix} (-3)(-4) - (-1)(-3) \\ (-1)(1) - (-2)(-4) \\ (-2)(-3) - (-3)(1) \end{pmatrix} = \begin{pmatrix} 9 \\ -9 \\ 9 \end{pmatrix} = 9 \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$
Using the normal vector $\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ and point $B(0,0,3)$, the plane equation is:
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = 3$

4. Substitute the vector equation of line $l$ into the equation of plane $p$ to find the parameter $\lambda$ at the intersection.

Substitute $\mathbf{r}$ of the line into the plane equation:
$\left[ \begin{pmatrix} 0 \\ 4 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} \right] \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = 3$
$(0 - 4 + 0) + \lambda(1 - (-3) + 2) = 3$
$-4 + 6\lambda = 3$
$6\lambda = 7 \implies \lambda = \frac{7}{6}$

5. Calculate the coordinates of the point of intersection.

Substitute $\lambda = \frac{7}{6}$ back into the vector equation of $l$:
$\mathbf{r} = \begin{pmatrix} 0 \\ 4 \\ 0 \end{pmatrix} + \frac{7}{6} \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} = \begin{pmatrix} 7/6 \\ 4 - 21/6 \\ 14/6 \end{pmatrix} = \begin{pmatrix} 7/6 \\ 1/2 \\ 7/3 \end{pmatrix}$
Therefore, the coordinates of the point of intersection are $\left(\frac{7}{6}, \frac{1}{2}, \frac{7}{3}\right)$.

1. Determine the displacement vector $\vec{EF}$ from the central atom $E$ to the intersection point $F$ on the plane $p$.

From part (ii), the intersection point $F$ corresponds to the parameter $\lambda = \frac{1}{6}$ when using the line equation $\mathbf{r} = \vec{OE} + \lambda \vec{DE}$.
Thus, the vector from $E$ to $F$ is:
$\vec{EF} = \frac{1}{6}\vec{DE} = \frac{1}{6} \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}$

(Note: Alternatively, using the exact coordinates of $F\left(\frac{7}{6}, \frac{1}{2}, \frac{7}{3}\right)$ and $E(1, 1, 2)$ yields the same displacement $\vec{EF} = \vec{OF} - \vec{OE} = \begin{pmatrix} 1/6 \\ -1/2 \\ 1/3 \end{pmatrix}$)

2. Calculate the shortest distance from $E$ to $p$ by projecting $\vec{EF}$ onto the normal vector of $p$.

From part (i), the normal vector to plane $p$ is $\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$.
The shortest distance from $E$ to $p$ is the length of the projection of $\vec{EF}$ onto $\mathbf{n}$:
$\text{Distance} = |\vec{EF} \cdot \hat{\mathbf{n}}|$
$= \left| \frac{1}{6} \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} \cdot \frac{1}{\sqrt{1^2 + (-1)^2 + 1^2}} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \right|$
$= \frac{1}{6\sqrt{3}} \left| (1)(1) + (-3)(-1) + (2)(1) \right|$
$= \frac{1}{6\sqrt{3}} (1 + 3 + 2)$
$= \frac{6}{6\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \text{ units}$

1. Find the displacement vectors for the bonds from the central atom $E$ to atoms $A$ and $C$.

$\overrightarrow{EA} = \overrightarrow{OA} - \overrightarrow{OE} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$

$\overrightarrow{EC} = \overrightarrow{OC} - \overrightarrow{OE} = \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix}$

2. Use the scalar product to find the angle between the two vectors.

Let $\theta$ be the bond angle. Using the scalar product formula $\cos \theta = \frac{\overrightarrow{EA} \cdot \overrightarrow{EC}}{|\overrightarrow{EA}| |\overrightarrow{EC}|}$:

$\cos \theta = \frac{ \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix} }{ \sqrt{1^2 + 2^2 + 2^2} \sqrt{2^2 + (-1)^2 + (-2)^2} }$

$\cos \theta = \frac{(1)(2) + (2)(-1) + (2)(-2)}{3 \times 3}$

$\cos \theta = \frac{-4}{9}$

$\theta = \cos^{-1}\left(-\frac{4}{9}\right) \approx 116.387^{\circ} \approx 116.4^{\circ} \text{ (to 1 d.p.)}$

1. Separate the variables

$x\frac{dy}{dx}=y+6xy=y(1+6x)$
Divide through by $xy$:
$\frac{1}{y}\,dy=\frac{1+6x}{x}\,dx=\left(\frac{1}{x}+6\right)dx$

2. Integrate both sides and solve for y

$\int\frac{1}{y}\,dy=\int\left(\frac{1}{x}+6\right)dx$
$\ln|y|=\ln|x|+6x+c$
Exponentiating:
$y=e^{\ln|x|+6x+c}=A x\,e^{6x}\quad(A=\pm e^{c})$

1. Recall the solution of the differential equation

From part (ii), solving $x\dfrac{\mathrm{d}y}{\mathrm{d}x}=y+6xy=y(1+6x)$ gives
$\int\frac{1}{y}\,\mathrm{d}y=\int\left(\frac{1}{x}+6\right)\mathrm{d}x$
$\ln|y|=\ln|x|+6x+C\;\Rightarrow\; y=Ax e^{6x}.$

2. Interpret the long-term behaviour

As the situation continues for many months, $x \to \infty$. Since $e^{6x} \to \infty$, we have $y = Axe^{6x} \to \infty$.

Hence the number of people infected keeps increasing without bound — it grows indefinitely. In context this is unrealistic, as the number infected cannot exceed the total population.