Methane ( CH_4 ) is a chemical compound with a tetrahedral structure. The 4 hydrogen (H) atoms form a regular tetrahedron, and the carbon (C) atom is in the centre. Let the centre of the C-atom be the point P, and the centres of the 4 H-atoms be the points Q, R, S and T. The coordinates of P, Q, R, S and T are (0, 0, 2), (-1, 0, 2), (-2, -1, 2), (-2, 1, 3) and (0, -1, a) respectively. The angle θ subtended by any two C-H bonds at the C-atom, such as angle QPR, is known as the H-C-H bond angle (see diagram above).

1. Form the direction vectors of the two bonds from P

$\overrightarrow{PQ} = \begin{pmatrix}-1\\0\\2\end{pmatrix} - \begin{pmatrix}0\\0\\2\end{pmatrix} = \begin{pmatrix}-1\\0\\0\end{pmatrix}$ $\overrightarrow{PR} = \begin{pmatrix}-2\\-1\\2\end{pmatrix} - \begin{pmatrix}0\\0\\2\end{pmatrix} = \begin{pmatrix}-2\\-1\\0\end{pmatrix}$

2. Compute the scalar product and the magnitudes

$\overrightarrow{PQ}\cdot\overrightarrow{PR} = (-1)(-2)+(0)(-1)+(0)(0) = 2$ $|\overrightarrow{PQ}| = 1, \qquad |\overrightarrow{PR}| = \sqrt{(-2)^2+(-1)^2+0^2} = \sqrt{5}$

3. Apply the cosine formula and solve for the angle

$\cos\theta = \frac{\overrightarrow{PQ}\cdot\overrightarrow{PR}}{|\overrightarrow{PQ}||\overrightarrow{PR}|} = \frac{2}{1\times\sqrt{5}} = \frac{2}{\sqrt{5}}$ $\theta = \cos^{-1}\!\left(\frac{2}{\sqrt{5}}\right) = 26.565^\circ \approx 26.57^\circ$

1. Find two non-parallel vectors lying on the plane $\pi$.

We first find two vectors on the plane passing through points $P(0, 0, 2)$, $Q(-1, 0, 2)$ and $R(-2, -1, 2)$:
$\vec{PQ} = \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix}$
$\vec{PR} = \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ -1 \\ 0 \end{pmatrix}$

2. Find a normal vector to the plane by taking the cross product of $\vec{PQ}$ and $\vec{PR}$.

The normal vector $\mathbf{n}$ is perpendicular to both $\vec{PQ}$ and $\vec{PR}$:
$\mathbf{n} = \vec{PQ} \times \vec{PR} = \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix} \times \begin{pmatrix} -2 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ (-1)(-1) - (0)(-2) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$

3. Use the normal vector and a point on the plane to write the Cartesian equation.

Using point $P(0, 0, 2)$ as the position vector $\mathbf{a}$, the scalar product form of the plane is $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$:
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$
$z = 2$

(Alternatively, by observation, since the $z$-coordinates of $P$, $Q$ and $R$ are identically $2$, the plane containing them is uniquely given by $z = 2$.)

1. State the vector equation of the line $SF$ using the coordinates of $S$ and the normal vector to the plane $\pi$.

From part (c), the points $P(0, 0, 2)$, $Q(-1, 0, 2)$, and $R(-2, -1, 2)$ all lie on the plane $z = 2$. Thus, the normal vector to the plane $\pi$ is $\mathbf{n} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

Since $F$ is the point on $\pi$ closest to $S$, the line $SF$ is perpendicular to $\pi$. Therefore, the direction vector of the line $SF$ is parallel to $\mathbf{n}$.

Given the point $S(-2, 1, 3)$, the vector equation of the line $SF$ is:
$\mathbf{r} = \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \lambda \in \mathbb{R}$

1. Set up the midpoint relationship

Since $T$ is the mirror image of $S$ in $\pi$, the foot of perpendicular $F$ is the midpoint of $S$ and $T$:
$\vec{OF} = \frac{\vec{OS} + \vec{OT}}{2}$
with $F = (0,0,2)$ and $S = (-2,1,3)$.

2. Solve for the position vector of T

Rearranging:
$\vec{OT} = 2\,\vec{OF} - \vec{OS} = 2\begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} - \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 0+2 \\ 0-1 \\ 4-3 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$