\magnification 2000
\parskip 10pt
\parindent 0pt

\def\R{{\bf R}}
\def\C{{\bf C}}
\def\x{{\bf x}}
\def\y{{\bf y}}
\def\e{{\bf e}}
\def\i{{\bf i}}
\def\j{{\bf j}}
\def\k{{\bf k}}
\def\ep{\epsilon}
\def\f{\vskip 10pt}
\def\u{\vskip -8pt}

\def\v{\vskip 5pt}


Equations of lines:
\vskip -5pt

Slope-intercept form:   $x_2 = mx_1 + b$ in $R^2$
\hfil \break
(in $R^3$, this is the equation of a plane)

General form:  $ax_1 + bx_2 = c$ in $R^2$ 
\hfil \break
(in $R^3$, this is the equation of a plane)


Point-Normal Vector Form:  ${\bf n} \cdot ({\bf x - p}) = 0$ in $R^2$ \hfil \break
(in $R^3$, this is the equation of a plane)

\vskip 40pt

Point-Normal Form: $a(x_1 - p_1) + b(x_2 - p_2) = 0$ in $R^2$
\break
(in $R^3$, this is the equation of a plane)


Point-Parallel Vector Form:  ${\bf x} = {\bf p} + t{\bf v}$ in
$R^m$

Parametric equations:  $x_1 = p_1 + tv_1$

\vskip -8pt
\hskip 3.33 truein		         $x_2 = p_2 + tv_2$

\vskip -8pt
\hskip 4 truein		.

\vskip -8pt
\hskip 4 truein		.

\vskip -8pt
\hskip 4 truein		.

\vskip -8pt
\hskip 3.33 truein			$x_m = p_m + tv_m$ in $R^m$


\eject

Equations of planes:

General form:  $ax_1 +bx_2 +cx_3 = d$ in $R^3$

Point-Normal Vector Form:  ${\bf n} \cdot ({\bf x - p}) = 0$ in $R^3$

Point-Normal Form: 

\centerline{$a(x_1 - p_1) + b(x_2 - p_2) + c(x_3 - p_3)= 0$ in $R^3$}

\vfill

Plane containing the vectors  {\bf v} and  {\bf w} anchored at {\bf 0}
\vskip - 8pt
 = plane containing the points
{\bf 0}, {\bf v} and  {\bf w}

\vskip - 8pt
 = plane containing the point
{\bf 0} \&  vectors {\bf v} \&  {\bf w}:

\centerline{${\bf x} = s{\bf v} +
t{\bf w]$}

\centerline{${\bf x} = t{\bf v} +
s{\bf w}$}
\vfill

Plane containing the point {\bf p} and the vectors {\bf v} \&  {\bf
w}: 

\centerline{${\bf x} = {\bf p} + t{\bf v} +
s{\bf w}$}
\vfill

Parametric equations:  $x_1 = p_1 + tv_1 + sw_1$

\vskip -8pt
\hskip 3.33 truein		         $x_2 = p_2 + tv_2 + sw_2$

\vskip -8pt
\hskip 4 truein		.

\vskip -8pt
\hskip 4 truein		.

\vskip -8pt
\hskip 4 truein		.

\vskip -8pt
\hskip 3.33 truein			$x_m = p_m + tv_m + sw_m$ in $R^m$

\eject

2.1 Hyperplanes:

The solution set of a system of linear equations in $R^m$ is
called a
hyperplane.  The number of free variables is the dimension of the
hyperplane.

If  ${\bf n} = (a_1, ..., a_m), {\bf p} = (p_1, ..., p_m),$ 
 $ {\bf x} =
(x_1, ..., x_m)$, then 
\vskip -6pt

\centerline{${\bf n} \cdot ({\bf x - p}) = 0$}

\centerline{${\bf n} \cdot {\bf x} - {\bf n} \cdot {\bf p} = 0$}

\centerline{${\bf n} \cdot {\bf x} = {\bf n} \cdot {\bf p}$}
\vskip -8pt

$(a_1, ..., a_m) \cdot (x_1, ..., x_m) =  (a_1, ..., a_m) \cdot (p_1,
..., p_m)$

\line{$a_1x_1 + a_2x_2 + ... + a_mx_m = a_1p_1 + a_2p_2 + ... +
a_mp_m$}

\vskip -8pt
$\left[\matrix{a_1 & a_2 & ...& a_m & &a_1p_1 + a_2p_2 + ... +
a_mp_m}\right]$

Thus, the solution set of   ${\bf n} \cdot ({\bf x - p}) = 0$  in $R^m$
where ${\bf
n} \not = {\bf 0}$ is $m-1$ dimensional.

Thus in $R^2$, ${\bf n} \cdot ({\bf x - p}) = 0$ is the equation of a
line.

Thus in $R^3$, ${\bf n} \cdot ({\bf x - p}) = 0$ is the equation of a
plane.

\eject

The smallest hyperplane containing the vectors \break
 ${\bf  v_1},  {\bf
v_2},  ...,
{\bf  v_k}$ anchored at {\bf 0}
\vskip -6pt

= the hyperplane containing the points \hfil \break
     ${\bf 0}, {\bf  v_1},  {\bf  v_2},
..., {\bf  v_k}$
\vskip -6pt

= the hyperplane containing the point ${\bf 0}$ \hfil \break
      and the vectors ${\bf  v_1},  {\bf  v_2},
..., {\bf  v_k}$:

\vskip 6pt
\centerline{${\bf x} =  t_1{\bf  v_1} + t_2{\bf  v_2} +  ... + t_k{\bf  v_k}$}


\vfill

The smallest hyperplane  containing the point {\bf p} and the vectors ${\bf  v_1},  {\bf  v_2},
..., {\bf  v_k}$: 

\vskip 6pt
\centerline{${\bf x} = {\bf p} + t_1{\bf  v_1} + t_2{\bf  v_2} +  ... + t_k{\bf  v_k}$}
\vfill

Parametric equations:  

\centerline{$x_1 = p_1 + t_1v_{11} + t_2v_{21} + ... + t_kv_{k1}$~~~}

\centerline{$x_2 = p_2 + t_1v_{12} + t_2v_{22} + ... + t_kv_{k2}$~~~}

\centerline{.~~~}

\centerline{.~~~}

\centerline{.~~~}

\centerline{~~~~~$x_m = p_m + t_1v_{1m} + t_2v_{2m} + ... +
t_kv_{km}$in
$R^m$}


Defn: If {\bf u}, {\bf v}, and {\bf w} are vectors in 3-space, then:

the {\bf scalar triple product} of {\bf u}, {\bf v}, and {\bf w} = 

${\bf u} \cdot ({\bf v} \times {\bf w}) = det\left[\matrix{u_1 & u_2 & u_3 \cr v_1
& v_2 & v_3 \cr w_1 & w_2 & w_3 \cr}\right]$

Thus, 
${\bf u} \cdot ({\bf v} \times {\bf w}) = {\bf w} \cdot ({\bf u} \times {\bf v}) = 
{\bf v} \cdot ({\bf w} \times {\bf u})$.


Given ${\bf x} = t(2, 4, 1) + s(-1, 0, 3)$.  
\f
Note this is a plane in $R^3$ in point-parallel vector form.

1.)  Find the line containing the point (7, 8, 9) and perpendicular to
this plane given above.

First find a vector perpendicular to the plane, \hfil \break  ${\bf x}
= t(2, 4,
1) + s(-1, 0, 3)$.

\vskip -5pt
This is equivalent to finding a vector perpendicular to the vectors 
(2, 4, 1) and (-1, 0, 3).  Thus we can use the cross product.




$
det\left[\matrix{{\bf i} & {\bf j} & {\bf k} \cr 2 & 4 & 1 \cr -1 & 0
& 3
\cr}\right]$

$= {\bf i}~det\left[\matrix{4 & 1 \cr 0 & 3}\right] -
{\bf j}~det\left[\matrix{2 &
1
\cr -1 & 3}\right] + {\bf k}~det\left[\matrix{2 & 4 \cr -1 & 0
\cr}\right] 
$


$= 12{\bf i} -7{\bf j} + 4{\bf k} = (12, -7, 4)$

Since the line is perpendicular to the plane, \hfil \break
{\bf x} = t(2, 4, 1) +
s(-1, 0, 3), the line is perpendicular to the vectors
(2, 4, 1) and (-1, 0, 3) and thus the vector (12, -7, 4) describes the
direction of the line.
Since \break (7, 8, 9) is a point in the line,
an equation of the line in point parallel vector form is
\vskip 5pt
\centerline{{\bf x} = (7, 8, 9) + t(12, -7, 4)}


Given ${\bf x} = t(2, 4, 1) + s(-1, 0, 3)$.
\f
Note this is a plane in $R^3$ in point-parallel vector form.  

2.)  Write the above plane in point-normal vector form.

First find a vector perpendicular to the plane, \hfil \break  ${\bf x}
= t(2, 4,
1) + s(-1, 0, 3)$.  We did this for problem 1.  The vector (12, -7, 4)
is perpendicular to this plane.

We also need a point in the plane.  Setting $t = 0, ~s = 0$, we
note that $(0, 0, 0) = 0(2, 4, 1) + 0(-1, 0, 3)$ is a point in
the plane.

Thus, an equation of this
plane in point normal vector form is

\centerline{$(12, -7, 4) \cdot ({\bf x} - (0, 0, 0)) = 0$}
\vskip -5pt

or 
\vskip -5pt
\centerline{$(12, -7, 4) \cdot {\bf x} = 0$}

Note there are many equivalent equations that also describe this
plane.  For example, since (12, -7, 4) is perpendicular to the
plane, so is (-120, 70, -40).

Setting $t = 1, ~s = 0$, we
note that \hfil \break
$(2, 4, 1) = 1(2, 4, 1) + 0(-1, 0, 3)$ is a point in
the plane.

Thus 
\centerline{$(-120, 70, -40) \cdot ({\bf x} - (2, 4, 1)) =
0$~~~~~~~~~~~~~~~}

is also an equation of this plane in point-normal form.


\end