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{\bf Question}

A crystal lattice is generated by the vectors ${\bf a}_1 = 3{\bf
i}$, ${\bf a}_2 = {\bf i} + 2{\bf j}$, ${\bf a}_3 = {\bf i} + {\bf
j} + {\bf k}$, based at the origin.  A general point ${\bf x}$ of
the lattice can be expressed as

$${\bf x} = r{\bf a}_1 + s{\bf a}_2 + t{\bf a}_3$$

where r, s and t are scalars.  Write down the matrix which allows
one to convert these $``{\bf a}_i$-coordinates$"$ into standard
$``{\bf i} ,\ {\bf j} ,\ {\bf k}$ coordinates$"$, and use it to
convert the following vectors into standard form:
\begin{description}
\item[(i)]
${\bf u} = 3{\bf a}_1 + 2{\bf a}_2 + 4{\bf a}_3$
\item[(ii)]
${\bf v} = 2{\bf a}_1 - 4{\bf a}_2 + 5{\bf a}_3$
\item[(iii)]
${\bf w} = -3{\bf a}_1 - 2{\bf a}_2 + 4{\bf a}_3$
\end{description}
Do the vectors ${\bf u}$, ${\bf v}$ and ${\bf w}$ lie on a common
plane through the origin?


{\bf Answer}

The matrix is $\left(\begin{array} {rrr} {3} & {1} & {1}\\ {0} &
{2} & {1}\\ {0} & {0} & {1} \end{array} \right)$, so that a point
of the lattice with ${\bf a}_i$-coordinates $(r,s,t)$ has ${\bf
i},\ {\bf j},\ {\bf k}$-coordinates $(x,y,z)$ where:

$$\left(\begin{array} {c} x\\y\\z \end{array}
\right)=\left(\begin{array} {ccc} {3} & {1} & {1}\\ {0} & {2} &
{1}\\ {0} & {0} & {1} \end{array} \right) \left(\begin{array} {c}
r\\s\\t
\end{array} \right)$$
\begin{description}

\item[(i)]
$${\bf u}=\left(\begin{array} {ccc} {3} & {1} & {1}\\ {0} & {2} &
{1}\\ {0} & {0} & {1} \end{array} \right) \left(\begin{array} {c}
3\\2\\ {-4}
\end{array} \right)=\left(\begin{array} {c}
7\\0\\ {-4}
\end{array} \right) \rm {so\ } {\bf u}=7{\bf i}-4{\bf k}$$

\item[(ii)]
$${\bf v}=\left(\begin{array} {ccc} {3} & {1} & {1}\\ {0} & {2} &
{1}\\ {0} & {0} & {1} \end{array} \right) \left(\begin{array} {c}
2\\ {-4} \\ 5
\end{array} \right)=\left(\begin{array} {c}
7\\ {-3} \\ 5
\end{array} \right) \rm {so\ } {\bf v}=7{\bf i}-3{\bf j}+5{\bf k}$$

\item[(iii)]
$${\bf w}=\left(\begin{array} {ccc} {3} & {1} & {1}\\ {0} & {2} &
{1}\\ {0} & {0} & {1} \end{array} \right) \left(\begin{array} {c}
{-3} \\ {-2} \\ 4
\end{array} \right)=\left(\begin{array} {c}
{-7} \\ 0 \\ 4
\end{array} \right) \rm {so\ } {\bf w}=-7{\bf i}+4{\bf k}$$

\end{description}
Note that ${\bf u}=-{\bf w}$ so that ${\bf u}$ and ${\bf w}$ are
parallel vectors.  Hence ${\bf u}$ and ${\bf w}$ lie on a common
line through the origin, and so ${\bf u},\ {\bf v}$ and ${\bf w}$
lie on a common plane through the origin.

\vspace{0.2in}

\setlength{\unitlength}{.5in}

\begin{center}
\begin{picture}(2,2)
\put(1,1){\vector(2,1){2}} \put(1,1){\vector(-2,-1){2}}
\put(1,1){\vector(-2,1){2}} \put(1.2,0.8){$(0,0,0)$}
\put(3.1,2){\bf u} \put(-1.3,2){\bf v} \put(-1.3,0){\bf w}

\end{picture}
\end{center}



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