\documentclass[a4paper,12pt]{article}
\begin{document}


{\bf Question}

Find the vector equation and standard equation of the following
planes:
\begin{description}
\item[(i)]
The plane $\Pi_1$ which passes through the point $P=(5,7,-6)$ and
has the normal vector ${\bf n}=2{\bf i}+2{\bf j}-7{\bf k}$;
\item[(ii)]
The plane $\Pi_2$ which passes through the three points
$P_1=(4,4,-2)$, $P_2=(1,0,-4)$, $P_3=(7,15,2)$;
\item[(iii)]
The plane $\Pi_3$ which passes through the point $P=(5,0,-8)$ and
is parallel to the plane with equation $4x+4y-14z=11$.
\end{description}
Are any of the planes $\Pi_1$, $\Pi_2$, $\Pi_3$ parallel? Are any
of them equal?



{\bf Answer}


\begin{description}
\item[(i)]
Vector equation of the plane is ${\bf n}\cdot({\bf r} - {\bf r_0})
= 0.$

Where \begin{eqnarray*} {\rm \bf n} & = & \left(
\begin{array}{c} 2 \\ 2 \\ -7 \end{array} \right) {\rm \ is\ a\ normal\ vector\ to\ the\
plane.}\\ {\rm \bf r} & = & \left(
\begin{array}{c} x \\ y \\ z \end{array} \right) {\rm \ is\ the\ possition\ vector\ of\ a\ general\ point\ in\ the\
plane.}\\ {\rm \bf r_0} & = & \left(
\begin{array}{c} 5 \\ 7 \\ -6 \end{array} \right) {\rm \ is\ the\ possition\ vector\ of\ a\ known\ point\ in\ the\
plane.} \end{eqnarray*}

Hence $\left( \begin{array}{c} 2 \\ 2 \\ -7 \end{array} \right)
\cdot \left[ \left(\begin{array}{c} x \\ y \\ z \end{array}
\right) - \left( \begin{array}{c} 5 \\ 7 \\ -6 \end{array} \right)
\right] = 0$ or $\left( \begin{array}{c} 2 \\ 2 \\ -7 \end{array}
\right) \cdot \left(\begin{array}{c} {x - 5} \\ {y - 7} \\ {z + 6}
\end{array} \right) = 0$

This gives the equation \begin{eqnarray*} 2(x - 5) + 2(y - 7) -
7(z + 6) & = & 0 \\ 2x + 2y - 7z & = & 10 + 14 + 42 = 66 \\ {\rm
so \ \ \ }2x + 2y - 7z & = & 66. \end{eqnarray*}

\item[(ii)]
Plane passes through the point $P_2 = (1, 0, -4)$ with position
vector $ \bf r_0  =  \left( \begin{array}{c} 1 \\ 0 \\ -4
\end{array} \right)$

Two vectors in the plane are:

$\stackrel{\longrightarrow}{P_2P_1} =  \left(
\begin{array}{c} 4 \\ 4 \\ -2 \end{array} \right) -  \left(
\begin{array}{c} 1 \\ 0 \\ -4 \end{array} \right) = \left(
\begin{array}{c} 3 \\ 4 \\ 2 \end{array} \right)$

$\stackrel{\longrightarrow}{P_2P_3} =  \left(
\begin{array}{c} 7 \\ 15 \\ 2 \end{array} \right) -  \left(
\begin{array}{c} 1 \\ 0 \\ -4 \end{array} \right) = \left(
\begin{array}{c} 6 \\ 15 \\ 6 \end{array} \right)$


\setlength{\unitlength}{.5in}

\begin{center}
\begin{picture}(3,2)
\put(0,1){\vector(1,0){3}} \put(0,1){\vector(2,1){2}}
\put(-0.5,1){$P_2$} \put(3.1,1){$P_3$} \put(2.1,2){$P_1$}
\put(1.5,0.5){$\stackrel{\longrightarrow}{P_2P_3}$}
\put(0.5,1.7){$\stackrel{\longrightarrow}{P_2P_1}$}

\end{picture}
\end{center}
A normal vector to the plane is

${\bf n}= \stackrel{\longrightarrow}{P_2P_1} \times
\stackrel{\longrightarrow}{P_2P_3}  = \left|
\begin{array}{ccc} {\bf{i}} & {\bf{j}} &
{\bf{k}}\\ 3 & 4 & 2 \\ 6 & 15 & 6 \end{array} \right| = \left(
\begin{array}{c} -6 \\ -6 \\ 21 \end{array} \right)$

Vector equation of the plane is: ${\bf n}\cdot({\bf r} - {\bf
r_0}) = 0.$

$\left( \begin{array}{c} -6 \\ -6 \\ 21 \end{array} \right) \cdot
\left(\begin{array}{c} {x - 1} \\ {y} \\ {z + 4}
\end{array} \right) = 0$

Standard equation of plane: \begin{eqnarray*} -6(x - 1) - 6y +
21(z + 4) & = & 0
\\ -6x - 6y + 21z & = & -6 - 84 = -90 \\ {\rm so \ \ \ }2x + 2y -
7z & = & 30. \end{eqnarray*}


\item[(iii)]
The plane $\pi_3$ passes through the point  with position vector $
\bf r_0  =  \left( \begin{array}{c} 5 \\ 0 \\ -8
\end{array} \right)$A normal to the plane is ${\bf n}= \left( \begin{array}{c} 4 \\ 4
\\ -14 \end{array} \right)$ (i.e. the same normal vector as its parallel plane $4x + 4y - 14z = 11$)

Vector equation of $\pi_3$: $\left( \begin{array}{c} 4 \\ 4 \\ -14
\end{array} \right) \cdot \left(\begin{array}{c} {x - 5} \\ {y}
\\ {z + 8} \end{array} \right) = 0$
Standard equation of $\pi_3$: \begin{eqnarray*} 4(x - 5) + 4y -
14(z + 8) & = & 0 \\ 4x + 4y - 14z & = & 20 + 112 = 132 \\ {\rm so
\ \ \ }2x + 2y - 7z & = & 66.
\end{eqnarray*}

The planes $\pi_1, \pi_2, \pi_3 $ are all parallel, because their
normal vectors are all parallel to ${\bf n} =
\left(\begin{array}{c} 2 \\ 2 \\ -7 \end{array} \right) $ The
planes $\pi_1$ and $\pi_3$ are equal because the are parallel, and
their equations are consistent.
\end{description}

\end{document}