\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

The four points $A, B, C, D$ have position vectors ${\bf a, b, c,
d}$.
\begin{description}
\item[(a)] Show that the four points $A, B, C, D$ are coplanar if
and only if

${\bf \alpha a + \beta b + \gamma c + \delta d = 0}$ and $\alpha
+\beta +\gamma +\delta = 0$ with not all
$\alpha,\beta,\gamma,\delta$ zero.
\item[(b)] What conditions must be placed on ${\bf a, b, c,
d}$ such that $A,B,C,D$ form

\begin{description}
\item[(i)] a parallelogram
\item[(ii)] a tetrahedron.
\end{description}
\item[(c)] When ${\bf d = 0}$ find the position vector of
intersection of

\begin{description}
\item[(i)] the diagonals of the parallelogram $ABCD$
\item[(ii)] the lines joining the midpoints of opposite edges of
the tetrahedron $ABCD$.
\end{description}

\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] $ABCD$ are coplanar if and only if $\vec{AD} =
k\vec{AB} + l\vec{AC}$  for some $k,l$

if and only if ${\bf d-a} = k({\bf b-a}) + l({\bf c-a})$

if and only if ${\bf a} (k+l-1) + {\bf b} (-k) + {\bf c}(-l) +
{\bf d}=0$

$\Rightarrow \alpha{\bf a}+\beta{\bf b}+\gamma{\bf c}+\delta{\bf
d}=0$ and $\alpha+\beta+\gamma+\delta=0$

${ }$

Conversely if $\alpha{\bf a}+\beta{\bf b}+\gamma{\bf c}+\delta{\bf
d}=0$ and $\alpha+\beta+\gamma+\delta=0$

If $\delta=0$

$\alpha{\bf a}+\beta{\bf b}+\gamma{\bf c}=0$ and
$\alpha+\beta+\gamma=0$ So $A, B, C$ are colinear

Thus $ABCD$ are coplanar.

If $\delta\not=0$

$\frac{\alpha}{\delta}{\bf a}+\frac{\beta}{\delta}{\bf
b}+\frac{\gamma}{\delta}{\bf c}+{\bf d}=0$

Let $\frac{\beta}{\delta} = -k \hspace{.2in} \frac{\gamma}{\delta}
= -l$ then $\frac{\alpha}{\delta} = k+l-1$

Hence $ABCD$ are coplanar.

${ }$

\item[(b)] $ABCD$ form a parallelogram in that order if ${\bf b-a =
c-d}$ with none of ${\bf a, \, b, \, c, \, d,}$ equal.

$ABCD$ form a tetrahedron if the are not coplanar.

${}$

\item[(c)]
\begin{description}
\item[(i)]
${}$

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\begin{picture}(4,2)
\put(0,0){\line(1,2){1}}

\put(1,2){\line(1,0){3}}

\put(4,2){\line(-1,-2){1}}

\put(0,0){\line(1,0){3}}

\put(0,0){\line(2,1){4}}

\put(1,2){\line(1,-1){2}}

\put(1,2.2){\makebox(0,0){$A$}}

\put(4,2.2){\makebox(0,0){$B$}}

\put(3.2,0){\makebox(0,0){$C$}}

\put(-0.2,0){\makebox(0,0){$D$}}

\put(2,1.3){\makebox(0,0){$P$}}

\put(5,1){\makebox(0,0){$P = \frac{1}{2}{\bf b}$}}

\end{picture}

${ }$


\item[(ii)]

${}$

\setlength{\unitlength}{.25in}
\begin{picture}(9,7)
\put(0,0){\line(1,0){4}}

\put(-.3,0){\makebox(0,0){$D$}}

\put(2,0){\circle*{.15}}

\put(2,-.4){\makebox(0,0){$R$}}

\put(0,0){\line(4,1){8}}

\put(8.3,2){\makebox(0,0){$B$}}

\put(4,1){\circle*{.15}}

\put(4.2,1.4){\makebox(0,0){$S$}}

\put(0,0){\line(1,3){2}}

\put(2,6.4){\makebox(0,0){$A$}}

\put(1,3){\circle*{.15}}

\put(0.8,3.2){\makebox(0,0){$U$}}

\put(4,0){\line(-1,3){2}}

\put(3,3){\circle*{.15}}

\put(3.4,3){\makebox(0,0){$T$}}

\put(4,0){\line(2,1){4}}

\put(6,1){\circle*{.15}}

\put(6,0.5){\makebox(0,0){$Q$}}

\put(2,6){\line(3,-2){6}}

\put(5,4){\circle*{.15}}

\put(5,4.5){\makebox(0,0){$P$}}

\put(4,-0.4){\makebox(0,0){$C$}}

\end{picture}


${}$

${}$

${\bf S} = \frac{1}{2}{\bf b} \hspace{.5in} t = \frac{1}{2}({\bf
a+c})$

The mid point of $ST$ is $$\frac{1}{2}\left(\frac{1}{2}{\bf b} +
\frac{1}{2}{\bf a} + \frac{1}{2}{\bf c}\right) = \frac{1}{4}({\bf
a + b + c})$$ which lies on the other lines by symmetry.

\end{description}
\end{description}
\end{document}