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{\bf Question}

A plane passes through the points $A$, $B$ and $C$ given by the
vectors $\vec{OA} = (1,0,-1)$, $\vec{OB} = (1,2,3)$ and $\vec{OC}
= (0,1,2)$.
\begin{description}
\item[(a)] Write down the vectors $\vec{AB}$ and $\vec{AC}$;
\item[(b)] Obtain a vector ${\bf n}$ perpendicular to the plane P;
\item[(c)] By writing the equation of the plane in the form $({\bf r} -  {\bf r_0})  \cdot {\bf n}
= 0$, show that the equation of the plane is $2x - 4y + 2z = 0$.
Verify that the points $A$, $B$ and $C$ all satisfy this equation
\end{description}


\vspace{.25in}

{\bf Answer}


\begin{description}
\item[(a)]
$\vec{AB} = (0,2,4) = {\bf m}$ $\vec{AC} = \vec{OC} - \vec{OA} =
(0,1,2) - (1,0,-1) = (-1,1,3) = {\bf l}$
\item[(b)]
The plane P passes through $A$, $B$ and $C$.  Hence the vector
${\bf n}$ is perpendicular to all vectors in the plane.  Hence in
particular $ {\bf n}$ is perpendicular to $\vec{AB}$ and
$\vec{AC}$ (or ${\bf l}$ and ${\bf m}$).  Thus ${\bf n} = {\bf
l}\times {\bf m}$ is perpendicular to ${\bf l}$ and ${\bf m}$.

\setlength{\unitlength}{.5in}

\begin{picture}(9,3)

\put(0,1.4){Then ${\bf n}=$}

\put(2,2.1){$-1$} \put(3,2.1){$1$} \put(4,2.1){$3$}
\put(5,2.1){$-1$} \put(6,2.1){$1$} \put(7,2.1){$3$}

\put(2,.7){$0$} \put(3,.7){$2$} \put(4,.7){$4$} \put(5,.7){$0$}
\put(6,.7){$2$} \put(7,.7){$4$}

\put(3.1,2){\line(1,-1){1}} \put(4.1,2){\line(1,-1){1}}
\put(5.1,2){\line(1,-1){1}}

\put(4.1,2){\line(-1,-1){1}} \put(5.1,2){\line(-1,-1){1}}
\put(6.1,2){\line(-1,-1){1}}

\put(8,1.4){$=(-2,4,-2)$}

\end{picture}

\item[(c)]
Then the equation of the plane is $({\bf r} -  {\bf r_0})  \cdot
{\bf n} = 0$ with $\bf r_0$ any point in the plane.

Then \begin{eqnarray*} [{\bf r} - (1,0,-1)] \cdot [-2,4,-2] &amp; = &amp;
0 \\ {\rm or \ \ \ } [(x,y,z) - (1,0,-1)] \cdot [-2,4,-2] &amp; = &amp; 0
\\ \Rightarrow -2x + 4y + 2z + 2 - 2 &amp; = &amp; 0 \\ \Rightarrow 2x -
4y + 2z &amp; = &amp; 0 \end{eqnarray*}

Check that this is right by putting the $A$, $B$, $C$ into the
equation.

$\begin{array}{lcr} (1,0,-1) &amp; \hspace{,2in} &amp; 2(1) + 2(-2) = 0 \\
(1,2,3) &amp; &amp; 2(1) - 4(2) + 2(3) = 2 - 8 + 6 = 0 \\ (0,1,2) &amp; &amp;
-4(1) + 2(2) = 0 \end{array}$

Hence the plane does indeed pass through all the points.
\end{description}



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