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{\bf Question}

A methane molecule has one carbon atom $C$ and four hydrogen atoms
$H_1, H_2, H_3, H_4$. If we take the carbon atom to be centred at
the origin $(0,0,0)$ then the hydrogen atoms $H_1, H_2, H_3, H_4$
have position vectors $R\bf {\hat{r}_1}$,\ $R \bf {\hat{r}_2}$,\
$R \bf {\hat{r}_3}$,\ $R \bf {\hat{r}_4}$ respectively, where
$\bf{r_1=i+j+k}$,\ $\bf{r_2=-i-j+k}$,\ $\bf{r_3=-i+j-k}$,\
$\bf{r_4=i-j-k}$ and $R$ is a scalar known as the bond distance of
the molecule. $R\bf {\hat{r}_1}$,\ $R \bf {\hat{r}_2}$,\ $R \bf
{\hat{r}_3}$,\ $R \bf {\hat{r}_4}$ are called the bond vectors of
the molecule.

\begin{description}
\item[(a)]
Calculate the unit vectors $\bf {\hat{r}_1}$,\ $\bf {\hat{r}_2}$,\
$\bf {\hat{r}_3}$,\ $\bf {\hat{r}_4}$ and hence write down the
four bond vectors of the methane molecule.
\item[(b)]
Find the angle between the bond vectors $R\bf {\hat{r}_1}$ and
$R\bf {\hat{r}_2}$
\item[(c)]
Calculate ${R\bf {\hat{r}_1}}-{R\bf {\hat{r}_1}}$ and hence
express $|{{R\bf {\hat{r}_1}}-{R\bf {\hat{r}_1}}}|$, the distance
between the hydrogen atoms $H_1$ and $H_2$, in terms of $R$.
\end{description}



{\bf Answer}


\begin{description}
\item[(a)]
$|{\bf r}_i|=\sqrt{3}$ so we have the unit vectors: $${\bf \hat
r}_1 = \left( \begin{array}{r}
     \frac{1}{\sqrt 3}\\
     \frac{1}{\sqrt 3}\\
     \frac{1}{\sqrt 3}
\end{array}  \right),\ \ {\bf \hat r}_2 = \left( \begin{array}{r}
     -\frac{1}{\sqrt 3}\\
     -\frac{1}{\sqrt 3}\\
     \frac{1}{\sqrt 3}
\end{array}  \right),\ \ {\bf \hat r}_3 = \left( \begin{array}{r}
     -\frac{1}{\sqrt 3}\\
     \frac{1}{\sqrt 3}\\
     -\frac{1}{\sqrt 3}
 \end{array}  \right),\ \ {\bf \hat r}_4 = \left( \begin{array}{r}
     \frac{1}{\sqrt 3}\\
     -\frac{1}{\sqrt 3}\\
     -\frac{1}{\sqrt 3}
 \end{array}  \right)$$

and the corresponding bond vectors:

$$R{\bf \hat r}_1 = \left( \begin{array}{r}
     \frac{R}{\sqrt 3}\\
     \frac{R}{\sqrt 3}\\
     \frac{R}{\sqrt 3}
\end{array}  \right),\ \ R{\bf \hat r}_2 = \left( \begin{array}{r}
     -\frac{R}{\sqrt 3}\\
     -\frac{R}{\sqrt 3}\\
     \frac{R}{\sqrt 3}
\end{array}  \right),\ \ R{\bf \hat r}_3 = \left( \begin{array}{r}
     -\frac{R}{\sqrt 3}\\
     \frac{R}{\sqrt 3}\\
     -\frac{R}{\sqrt 3}
 \end{array}  \right),\ \ R{\bf \hat r}_4 = \left( \begin{array}{r}
     \frac{R}{\sqrt 3}\\
     -\frac{R}{\sqrt 3}\\
     -\frac{R}{\sqrt 3}
 \end{array}  \right).$$

\item[(b)]
Let $\theta$ be the angle between the bond vectors $R{\bf \hat
r}_1$ and $R{\bf \hat r}_2$:

$$R{\bf \hat r}_1 \cdot R{\bf \hat r}_2=\left( \begin{array}{r}
     \frac{R}{\sqrt 3}\\
     \frac{R}{\sqrt 3}\\
     \frac{R}{\sqrt 3}
\end{array}  \right) \cdot \left( \begin{array}{r}
     -\frac{R}{\sqrt 3}\\
     -\frac{R}{\sqrt 3}\\
     \frac{R}{\sqrt 3}
\end{array}  \right)=-\frac{R^2}{3}-\frac{R^2}{3}+\frac{R^2}{3}=-\frac{R^2}{3}$$

$|R{\bf \hat r}_1|=R,\ \ |R{\bf \hat r}_2|=R$ so that:
$$\theta=\cos^{-1} \left (\frac{R{\bf \hat r}_1 \cdot R{\bf \hat
r}_2}{|R{\bf \hat r}_1| |R{\bf \hat r}_2|} \right)=\cos^{-1} \left
( \frac{\frac{-R^2}{3}}{R^2} \right) =\cos^{-1} \left(
\frac{-1}{3} \right) \approx 109.47^{\circ}$$

\item[(c)]
$$R{\bf \hat r}_1 - R{\bf \hat r}_2=\left( \begin{array}{r}
     \frac{R}{\sqrt 3}\\
     \frac{R}{\sqrt 3}\\
     \frac{R}{\sqrt 3}
\end{array}  \right) - \left( \begin{array}{r}
     -\frac{R}{\sqrt 3}\\
     -\frac{R}{\sqrt 3}\\
     \frac{R}{\sqrt 3}
\end{array}  \right) = \left( \begin{array}{c}
     2\frac{R}{\sqrt 3}\\
     2\frac{R}{\sqrt 3}\\
     0
\end{array}  \right)$$
so the distance between the atoms $H_1$ and $H_2$ is given by

\begin{eqnarray*}
|R{\bf \hat r}_1 - R{\bf \hat r}_2| & = &
\sqrt{\left(\frac{2R}{\sqrt 3} \right )^2 + \left(\frac{2R}{\sqrt
3} \right )^2 + 0}\\ & = & \sqrt{\frac{8R^2}{3}} =
2R\sqrt{\frac{2}{3}}
\end{eqnarray*}
\end{description}

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