\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
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\begin{document}


{\bf Question}

Using vector methods
\begin{description}
\item[(a)] find the equation of a circle on $AB$ as diameter,
\item[(b)] prove that the altitudes of a triangle are concurrent,
\item[(c)] show that the diagonals of a rhombus are orthogonal.
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]

\begin{center}
\epsfig{file=v-9-1.eps, width=45mm}
\end{center}

$A, B, R$ have position vectors ${\bf a, b, r}$  R lies on the
circle.

Diameter $AB$ iff $RA$ is perpendicular to $RB$.

${\bf (r - a) \cdot (r-b)} = 0$

${\bf r \cdot r - r \cdot (a+b) + (a \cdot b)} = 0$

${}$
\item[(b)]
${}$

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\put(0,0){\line(5,1){2.5}}

\put(2.5,0.5){\line(-1,1){1.5}}

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\put(0,0){\line(1,1){1.5}}

\put(2.5,0.5){\line(-2,1){1.8}}

\put(1,2){\line(1,-5){.35}}

\put(1.3,0.45){\line(5,1){0.2}}

\put(1.55,0.3){\line(-1,5){.04}}

\put(1.5,1.1){\line(-1,1){0.2}}

\put(1.7,1.3){\line(-1,-1){.2}}

\put(1,2.15){\makebox(0,0){$A$}}

\put(2.7,0.5){\makebox(0,0){$B$}}

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

\put(1.3,0.1){\makebox(0,0){$K$}}

\put(.6,1.5){\makebox(0,0){$L$}}

\put(1.75,1.5){\makebox(0,0){$M$}}

\put(0.9,1.1){\makebox(0,0){$H$}}
\end{picture}

$CM$ is perpendicular to $AB$.

$AK$ is perpendicular to $BC$

$H$ is $AK \cap CM$

$L$ if $BH \cap AC$

prove $BL$ is perpendicular to $AC$

Let $H$ be the origin and let the position vectors of $ABC$ be
${\bf a \, b \, c}$.

$\begin{array}{rcc} {\rm Then\ } M = m {\bf c} & {\rm for\ some\ }
& m \not= 0 \\  K = k {\bf a} & & k \not= 0 \\ L = l {\bf b} & & l
\not= 0 \end{array}$

Since $HM$ is perpendicular to $AB$ so $m{\bf c} \cdot ({\bf b-a})
= 0$

therefore ${\bf c \cdot b = c \cdot a}$

Since $HK$ is perpendicular to $BC$ so $k{\bf a} \cdot ({\bf c-b})
= 0$

therefore ${\bf a \cdot c = a \cdot b}$

Thus ${\bf c \cdot b = a \cdot b} = 0$ so $l{\bf b \cdot (a - c)}
= 0$

i.e. $HL$ is perpendicular to $AC$




${}$
\item[(c)]
${}$

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\put(2,1.5){\line(1,0){3}}

\put(0,0.5){\line(1,0){3}}

\put(3,0.5){\line(2,1){2}}

\put(-0.2,0.5){\makebox(0,0){$O$}}

\put(1.7,1.75){\makebox(0,0){$A$}}

\put(5.2,1.75){\makebox(0,0){$C$}}

\put(3.25,0.3){\makebox(0,0){$B$}}

\put(1,1.25){\makebox(0,0){${\bf a}$}}

\put(1.75,0.2){\makebox(0,0){${\bf b}$}}
\end{picture}

$\begin{array}{rcl} \vec{OC} & = & {\bf a+b} \\ \vec{AB} & = &
{\bf  b -a} \\ \vec{OC} \cdot \vec{AB} & = & (a+b) \cdot(b-a) \\ &
= & |b|^2 - |a|^2 = 0 \\  & & {\rm Since\ } |OA|  =  |OB|
\end{array}$





\end{description}

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