\documentclass[a4paper,12pt]{article}
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\begin{document}
\begin{center}
{\bf Vector Algebra and Geometry}
\end{center}

${}$

{\bf Geometry of Planes and lines}

${}$

We assume that each plane has a unique normal direction.  If ${\bf
a} \not=0$ is a normal to a plane $\pi$ so is $k{\bf a}$ for $k
\not=0$.

Let ${\bf a}$ be a non-zero vector $P({\bf p})$ a fixed point and
$R{\bf r})$ a variable point.  Let $\pi$ be the plane through $P$
perpendicular to ${\bf a}$

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Then $R$ lies on $\pi$ if and only if $PR$ is perpendicular to
${\bf a}$.

$\vec{PR} = {\bf r - p}$.  So $R$ lies on $\pi$ if and only if
${\bf a (r - p)} = 0$

i.e. ${\bf a \cdot r = a \cdot p}$

Now if we write ${\bf a \cdot p} = k$ (where $k$ is a constant)
then the equation of $\pi$ can be written as ${\bf a \cdot r} =
k$, and if ${\bf a} =a{\bf i} + b{\bf j} + c{\bf k}$ and ${\bf r}
=x{\bf i} + y{\bf j} + z{\bf k}$ the equation becomes
$$ax+by+cz=k.$$ Notice that the equation of the plane in cartesian
form immediately gives us a normal vector $(a,b,c)$.

${}$

{\underline {Example}}

Find the equation of the plane containing the three points
$A(0,1,-1), \, $

$B(1,1,0), \, C(1,2,0)$

Now $\vec{AB} = (1,0,1), \, \,\ \, \vec{AC} = (1,1,1)$

and the normal to the plane is perpendicular to both.

Thus the normal vector is given by $\vec{AB} \times \vec{AC} =
(-1,0,1).$

So the equation of the plane is $$-x+z=k.$$ To find $K$ notice
that A(0,1,-1) must satisfy the equation of the plane, thus $k=-1$
(check $B,C$ lie in this plane.)

${}$

Notice that if $AB$ is parallel to $AC$ then $\vec{AB} \times
\vec{AC} = {\bf 0}$ which is not a normal vector.  This
corresponds to the situation where $ABC$ are collonear, and there
isn't a unique plane containing $ABC$

${}$

{\bf Angle between two planes}

the angle between the two planes is equal to that between their
normal vectors.

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Suppose we have $x + 2y +z = 3$ and $x+y=-4$

then normal vectors are (1,2,1) and (1,1,0) So $\ds \cos \theta =
\frac{|a\cdot b|}{|a||b|} = \frac{3}{\sqrt 6 \sqrt 2} =
\frac{\sqrt3}{2}$ So $\theta = 30^\circ$


\newpage

{\bf Perpendicular distance}

Suppose we have a point $P({\bf p})$ and a plane ${\bf a \cdot
r}=k$

Let $Q$ be the foot of the perpendicular, then $Q$ lies in the
plane and $PQ$ is parallel to ${\bf a}$

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So \begin{eqnarray} {\bf a \cot q} & = & k \\ {\bf p - q} & = &
t{\bf a} \end{eqnarray}

Now we want $\ds | {\bf p - q}| = |t ||{\bf a}| = \frac{a \cdot p
- k|}{|a|}$

{\underline {Example}}

The distance from P(1,2,1) from the plane $x - y +2z =5 \, $

$({\bf a} = (1,-1,2), \, k = 5)$ $$\frac{|(1,-1,2) \cdot (1,2,1) -
5|}{\sqrt{1^1 + 1^2 + 2^2}} = \frac{4}{\sqrt6}$$


${}$

{\bf Distance between parallel planes.}

Suppose we have planes ${\bf a \cdot r}=k$ and ${\bf \cdot r} =
l$. Let $P$ lie in the second plane and $Q$ in the first.  The
distance between the two planes is $$PQ = \frac{|{\bf a \cdot p} -
k|}{|a|} = \frac{|l-k|}{|a|}$$ as ${\bf a \cdot p} = l$

\newpage

{\bf Equations of a line}

We have already met the parametric equation  of a line in the form
$${\bf r = a} t{\bf b}$$ representing the line through the point
$A$ with position vector ${\bf a}$, in the direction of the vector
${\bf b}$.  Just as the orientation of a plane is specified by a
normal vector, so the direction of a line is determined by a
direction vector.

If ${\bf b} \not-0$ is such a direction vector, so is $k{\bf b} \,
\, (k \not=0)$.  If $\hat {b}$ is a unit direction vector of a
line its components are called direction \underline {cosines} of
the line.

Now ${\bf r} = (x,y,z) , \, {\bf a} = (a,b,c) , \, {\bf b} =
(l,m,n)$ the equation can be written in terms of components
\begin{eqnarray*} x & = & a +tl \\ y & = & b +tm \\ z & = & c+tn
\end{eqnarray*} So $$\frac{x-a}{l} = \frac{y-b}{m} = \frac{z-c}{n}
(=t)$$ These are the cartiesian equations of the line.

\underline {Example}

Find equation for the line through $P(1,0,2)$ and $Q(2,1,0)$.

Now a direction vector is $\vec{PQ} = (1,1,-2) \, \, (={\bf b})$

taking ${\bf a = P}$ we obtain $$\frac{x-1}{1} = \frac{y}{1} =
\frac{z-2}{-2}$$

taking ${\bf a = q}$ we obtain $$\frac{x-2}{1} = \frac{y-1}{1} =
\frac{z}{2}$$

So we have non-uniqueness in the set of equations.

${}$

Two lines in space will, in general, not meet.  If they do not
meet and are not parallel they are said to be skew.

Suppose we have two lines $${\bf r = a}_1 + t{\bf b}_1
\hspace{.5in} {\bf r = a}_2 + t{\bf b}_2$$

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The lines will meet if and only if $\vec{A_1A_2}, \, _1, \, b_2$
are coplanar.  A condition is therefore $$(a_1-a_2) \cdot (b_1
\times b_2 ) = 0$$

${}$

${}$

{\bf Distance of a point from a line}

${}$

Suppose we hve a line ${\bf r  = a}+t{\bf b}$ and a point ${\bf
p}$


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$d = AP \sin \theta = |{\bf p-a}| \sin \theta = \frac{|{\bf (p-a)
\times b}|}{|{\bf b}|}$

${}$

\underline {Example}

Find the distance of the point $P(-1,2,1)$ from the line
$$\frac{x-1}{3} = \frac{y}{2} = \frac{z+3}{4}$$

The equation of the line in vector form is $${\bf r} = (1,0,-3) +
t(3,2,4) = {\bf a} + t{\bf b}$$ So $\ds {\bf p - a} = (-2,2,4)$

$\ds {\bf (p-a) \times b} = \left| \begin{array}{ccc} {\bf i} &
{\bf j} & {\bf k} \\ -2 & 2 & a \\ 3 & 2 & 4 \end{array} \right| =
0{\bf i} + 20{\bf j} - 10 {\bf k}$

So $|(p-a) \times {\bf b}| = \sqrt{20^2 + 10^2} = 10 \sqrt 5
\hspace{.2in} |{\bf b}| = \sqrt{29}$

So the required distance is $\ds \frac{10 \sqrt5}{\sqrt{29}} =
4.152$

${}$

\newpage

{\bf Intersections of two planes}

Given two planes which are not parallel then we expect them to
intersect in a line.  We can find the equation of the line in
standard from as in the following example.

Let $\pi_1, \, \pi_2$ be the planes

\begin{eqnarray*} x - 2y +3z & = & 1 \\ 2x + y + z & = & 3
\end{eqnarray*}

To find the equation  in the form ${\bf r = a} + t{\bf b}$ we need
a point on the line ${\bf a}$ and $a$ direction vector ${\bf b}$.
The line will meet at least one of the planes $x=0, \, y=0, \,
z=0$ so we try these values in the two equations.

Putting $x=0$ gives $\begin{array}{c} -2 + y = 1 \\ y + z = 3
\end{array}$ these give $y = \frac{8}{5}$ and $z = \frac{7}{5}$

So $\ds {\bf a} = \left(0, \frac{8}{5}, \frac{7}{5}\right)$ lies
in both planes.  Now since the line lies in both planes its
direction is perpendicular to the normals to both planes and so is
parallel to the vector product of these normals.

So we have ${\bf b} = (1,-2,3) \times (2,1,1) = (-5,5,5)$

So the equation for the line is $${\bf r} = \left(0, \,
\frac{8}{5} , \, \frac{7}{5} \right) + t(-5,5,5)$$ or in cartisean
form $$-\frac{x}{5} = \frac{y-\frac{8}{5}}{5} =
\frac{z-\frac{7}{5}}{5}$$ or $$-x = y - \frac{8}{5}=
z-\frac{7}{5}$$ or $$-5x = 5y-8 = 5z-7$$


${}$

{\bf Angles between lines}

The angle between two lines is defined to be the angle between
their direction vectors.  The angles between a line and a plane is
defined as follows:

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If $\theta$ is the angle between the line and the plane and ${\bf
a}$ is the normal to the plane then $\theta = 90 - \phi$

$\ds \cos \phi = \frac{a \cdot b}{|a||b|}$

So $\ds \sin \theta \frac{a \cdot b}{|a||b|}$

Notice that if the other normal is taken then $a \cdot b<0.$  We
want the acute angle so $\ds \sin \theta  = \frac{a \cdot
b}{|a||b|}$

${}$

\underline{Problems}

The following are examples of problems on lines and planes.  One
should aim to reduce them to some standard results, and so first
one needs to think about the sequence of steps to be undertaken.
\begin{description}
\item[1.] A plane contains the line $l_1$ $$2x-2 = 4-2y = z-1$$
and is parallel to the line $l_2$ $$6x = 3y+21 = 2z$$ Find its
equation.

Firstly to read information about the lines we reduce them to
standard form: $$l_1: \, \, \frac{x-1}{1} = \frac{y-2}{-1} =
\frac{z-2}{2}$$ $$ l_2:  \, \, \frac{x}{1} = \frac{y+7}{2} =
\frac{z}{3}$$ If ${\bf n}$ is normal to the plane $\pi$ then ${\bf
n}$ is perpendicular to ${\bf a} $ and ${\bf b}$ so we need one
point on $\pi$, and so any point on $l_1$ will do.

So $\ds{\bf n} = (1,-1,2)\times(1,2,3) = (-7,-1,3)$

So $\pi$ has equations $\ds -7x - y +3z = k.$

Since (1,2,1) lies on $l$ and thus gives $k=-6$ so the required
equation is $$-7x-y+3z=-6$$

${}$

\item[2.] Let $O$ be the origin and $L$ be the line $$x =
\frac{y-1}{2} = z-3$$ Find a point $P$ on $l$ such that $OP$ makes
an angle of $45^\circ$ with $L$.

The equation of $l$ in vector form is $${\bf r} = (0,1,3) +
t(1,2,1) = (t, 2t+1, 3+t)$$For the angle between $\vec{OP} = {\bf
r}$ and the direction ${\bf b} = (1,2,1)$ to be $45^\circ$.  We
require $${\bf b \cdot r} = \frac{|b||r|}{\sqrt2}$$ This gives
$\ds 6t +5 = \frac{\sqrt 6\sqrt{6t^2+10t + 10}}{\sqrt 2}$

This simplifies to $$ 18t^2 + 30t - 5 =0$$ this gives two values
for $t$, which are then used to give two possible points $P$

\item[3.] A line $N$ contains the point (1,0,2) and meets each of
the lines: $$x = y = z+2 \hspace{.4in}-L$$ $$x+3 = -\frac{y}{2} =
\frac{z}{3} \hspace{.4in} -M$$ Find its equations

${}$

Suppose (a,b,c) is a direction vector for $N$ then all popints on
$N$ are the form $${\bf r} = (1+at, \, bt , \, 2+ct)$$

For $t=k$ say this point satisfies the equations $L$

For $t=l$ say the point satisfies the equations $M$.

This gives four equations in five unknowns $a,\,  b,\,  c,\,  k,\,
l$.

Eliminating $k,l$ will give \begin{eqnarray*} 4a - 3b -c & = & 0
\\ 2a - 5b - 4c & = & 0 \end{eqnarray*} Now it is only the ratio
$a:b:c$ we need since if $(a,b,c)$ is a direction vector, so is
$(ma,mb,mc)$ for any $m\not=0$.  We find that $a:b:c = 1:2:-2$ so
that the line $N$ has equations $$\frac{x-1}{1} = \frac{y}{2} =
\frac{z-2}{-2}$$

${}$

${}$

\item[4.]

Find the shortest distance between the two skew lines $$L: \, \,
\, x-1 = y = z +5 $$ $$ M:\, \, \, \frac{x}{3} = \frac{y+2}{2} =
z+2$$ Find the equations on the common perpendicular and the
points where this meets $l$ and $m$.

${}$

\underline{Plan 1}

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Let $P$ be an arbitrary point on $L$ - parameter $k$

Let $P$ be an arbitrary point on $M$ - parameter $l$

we want to choose $P$ and $Q$ so that $PQ cdot b_1 = 0$ and $PQ
\cdot b_2 = 0$ this will give two equations for $k$ and $l$, which
will therefore determine $P$ and $Q$ so we can find $\vec{PQ}$ and
from this both $|\vec{PQ}|$ adn the equation of the line (throught
$P$ with direct vector $\vec{PQ}$)

\underline{solution} \begin{eqnarray*} {\bf P} & = & (1,0,-5) +
k(1,1,1) = (1+k, k, -5+l) \\ {\bf Q} & = & (0,-2,-2) + l(3,2,1) =
(3l, -2-l, 2+l)\\ {} \\ \vec{PQ} & = & (-1-+3l, -2-k+2lm 3-k+l) \\
\vec{PQ} \cdot {\bf b}_1 & = & -3k + 6l =7 \\ \vec{PQ} \cdot {\bf
b}_2 & = & -4-6+14l =0 \end{eqnarray*} Solving these equations
gives $=4$ and $l=2$

So $P=(5,4,-1)$ and $Q=(6,2,0)$

So $\vec{PQ} = (1,-2,1) \hspace{.2in} |\vec{PQ}\ = \sqrt6$

The equation $PQ$ is $$\frac{x-5}{1} = \frac{y-4}{-2} =
\frac{z+1}{1}$$


\underline{Plan 2}

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\put(4.5,3.5){$ M$}

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\put(4.5,4.75){$ M'$}
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Now $PL$ is parallel to $QL'$ and $PM'$ is parallel to $QM$

Let $\pi_1$ be the plane $PQLL'$

Let $\pi_2$ be the plane $PQMM'$

\begin{enumerate}
\item Work out a direction vector for $PQ$  ${\bf c = b_1 \times
b_2}$ will do.
\item Work out the equation of $\pi_1$ (contains $L$ and $PQ$)
\item Work out the equation of $\pi_2$ (contains $M$ and $PQ$)
\item Obtain $Q$ as intersection of $\pi_1$ and $M$
\item Obtain $P$ as intersection of $\pi_2$ and $L$
\end{enumerate}
\end{description}
\end{document}
