If you have slope, m, and want to have a direction vector instead,
$$
\mathbf{u}=\left(\begin{array}{c}
1\\
m
\end{array}\right).
$$
For example, if $m=\frac{2}{3},$ then
$$
\mathbf{u=}\left(\begin{array}{c}
1\\
2/3
\end{array}\right)=\left(\begin{array}{c}
3\\
2
\end{array}\right)
$$
If it is pointing the wrong way, then
$$
\mathbf{u}=\left(\begin{array}{c}
-1\\
-m
\end{array}\right)
$$
Converting from a direction vector to a slope can only be done if
$u_x$ is non-zero.
$$
\mathbf{u}=\left(\begin{array}{c}u_x\\u_y \end{array}\right) \qquad
m=\frac{u_{y}}{u_{x}}.
$$
2.2 Vector Dot Product I
Just a reminder: If $\mathbf{u}$ is a 2-d vector, then $|\mathbf{u}|=\sqrt{u_{x}^{2}+u_{y}^{2}}.$
The definition and vertical bars indicate a "modulus'', however,
it is loosely referred to as absolute value.
There is a geometric definition of vector dot product
\[
\mathbf{u\cdot v=|u|\cdot|v|\cdot}\cos\theta
\]
In words, The dot product of two vectors is equal to the product of
their lengths and the included angle. I have proof for this relationship
in a figure in the next section.
This formula shows that if the included angle is a $90^{\circ}$ angle
that the $\cos\theta$ will be zero and thus the dot product of two
orthogonal vectors will be zero.
From standard form,
\[
ax+by+c=0,
\]
the slope can be shown to be $-\frac{a}{b}.$ To make this into a
vector, we will get $\left(\begin{array}{c}
b\\
-a
\end{array}\right)$.
\[
\left(\begin{array}{c}
1\\
-\frac{a}{b}
\end{array}\right)\cdot\left(b\right)=\left(\begin{array}{c}
b\\
-a
\end{array}\right)
\]
All of the lines orthogonal to $ax+by+c=0$ will have slope $\frac{b}{a},$
which becomes vector $\left(\begin{array}{c}
a\\
b
\end{array}\right)$. Now the dot product of these two vectors
\[
\left(\begin{array}{c}
b\\
-a
\end{array}\right)\cdot\left(\begin{array}{c}
a\\
b
\end{array}\right)=ab-ab=0
\]
which proves that $-1/m$ is orthogonal to $m,$ since the length
of the two vectors is non-zero, the cosine of the angle in between
must be zero, thus the angle is $90^{\circ}.$
2.3 Vector Dot Product II
Let's begin with the same relationship that we used previously,
\[
\mathbf{u\cdot v=|u|\cdot|v|}\cos\theta
\]
where $\theta$ is the angle between the two vectors.
However, way more important is
\[
\cos\theta=\frac{\mathbf{u\cdot v}}{\mathbf{|u|\cdot|v|}}.
\]
Why would that be? The intersection point, $E,$ coming from $P$
is given by vector projection, i.e. we will project a new point on
the line from a known point on the line by moving some distance along
the line.
\begin{equation}
E=A+(\text{distance})\cdot(\text{line unit direction vector)}\label{eq:Vector Intersection}
\end{equation}
That distance derived from a right triangle is
\[
\text{distance}=\text{hypotenuse}\cdot\cos\theta.
\]
The hypotenuse is the modulus of vector $\mathbf{|v|}$ and the line's
$\textbf{unit}$ direction vector is vector $\hat{\mathbf{u}}=\frac{\mathbf{u}}{|\mathbf{u}|}.$
The point is that we don't need $\theta,$ only $\cos\theta,$ which
can be had from $\cos\theta=\frac{\mathbf{u\cdot v}}{\mathbf{|u|\cdot|v|}}.$
Lemma The cosine of an angle between two vectors is given by $\frac{\mathbf{u\cdot v}}{\mathbf{|u|\cdot|v|}}.$
The proof is shown in the following figure.
Understanding the Intersection
To find the intersection point coordinates, $E$ in the diagram. Start with
the coordinates of $A,$ and add the distance from $A$ to $E$ multiplied
times the unit direction vector of the line.
What you are about to learn is that we do not have to know the coordinates of $E$ in order to
know how far away from $A$ it is.
\[
E=A+\left(distance \overline{AE}\right)\left(\frac{\mathbf{u}}{|\mathbf{u}|}\right)
\]
The figure shows that the distance from A to E can be calculated using only
the two direction vectors. Vector \( \vec{v} \) has to be (P-A) but vector \( \vec{u} \) can be a unit direction vector.
\[
E=A+\left(|\mathbf{v}|\cdot\cos\theta\right)\cdot\frac{\mathbf{u}}{\mathbf{|u|}}.
\]
and by substitution
\[
E=A+\left(|\mathbf{v}|\cdot\frac{\mathbf{u\cdot v}}{|\mathbf{u}|\cdot|\mathbf{v}|}\right)\cdot\frac{\mathbf{u}}{\mathbf{|u|}}.
\]
It will work nicely as is, but we can cancel things that are not vectors.
(The modulus of a vector is a scalar!)
\[
\begin{aligned}E= & A+\left(\cancel{|\mathbf{v}|}\cdot\frac{\mathbf{u\cdot v}}{|\mathbf{u}|\cdot\cancel{|\mathbf{v}|}}\right)\cdot\frac{\mathbf{u}}{\mathbf{|u|}}\\
= & A+\left(\frac{\mathbf{u\cdot v}}{|\mathbf{u}|^{2}}\right)\cdot\mathbf{u}\\
= & A+\left(\frac{\mathbf{u\cdot v}}{\mathbf{u^{2}}}\right)\cdot\mathbf{u}
\end{aligned}
\]
2.4 Slopes in Polar Coordinates
A line in polar coordinates is still a line and it will have a slope.
Since the line is written as a function of $r(\theta),$ we can use
the Cartesian transform to get
\[
\begin{aligned}x= & r(\theta)\cdot\cos\theta\\
y= & r(\theta)\cdot\sin\theta
\end{aligned}
\]
The first equation defines $\theta$ implicitly. So we can differentiate
the second one and get the slope.
\[
\frac{dy}{dx}=\frac{dy}{d\theta}\cdot\frac{d\theta}{dx}.
\]
Furthermore, since $\frac{d\theta}{dx}=1/(dx/d\theta),$we can write
$(\text{Use the multiplication rule }d(uv)=du\cdot v+v\cdot du$)
\[
\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{r^{\prime}(\theta)\sin\theta+r(\theta)\cos\theta}{r'(\theta)\cos\theta-r(\theta)\sin\theta}
\]
Example: Let's find the slope of the line connecting $Q=(3;\pi/2)$ and $P=(4,\pi/12).$ Answer Here we do not have $r$ described as $r(\theta),$ so all we need
do is convert $Q$ and $P.$
\[
Q=\begin{cases}
x & 3\cdot\cos(\pi/2)\\
y & 3\cdot\sin(\pi/2)
\end{cases}=(0,3)
\]
\[
P=\begin{cases}
x & 4\cdot\cos(\pi/12)\\
y & 4\cdot\sin(\pi/12)
\end{cases}=(3.8637,1.0353)
\]
So the slope is $\Delta y/\Delta x=\frac{3-1.0353}{0-3.8637}=-0.5085$.
Example:
Suppose we have a line defined as
\[
r(t)=\frac{4}{\cos(t-\frac{\pi}{3})}.
\]
Here is a sketch of the line, determine its slope and direction vector.
Answer:
\[
\frac{dy}{dx}=\frac{r^{\prime}(t)\sin t+r(t)\cos t}{r'(t)\cos t-r(t)\sin t}
\]
\[
r'(t)=4\cdot\frac{\sin\left(t-\frac{1}{3}\;\pi\right)}{\cos^{2}\left(t-\frac{1}{3}\;\pi\right)}
\]
\[
\frac{dy}{dx}=\frac{4\cdot\frac{\sin\left(t-\frac{1}{3}\;\pi\right)}{\cos^{2}\left(t-\frac{1}{3}\;\pi\right)}\sin t+\frac{4}{\cos(t-\frac{\pi}{3})}\cos t}{4\cdot\frac{\sin\left(t-\frac{1}{3}\;\pi\right)}{\cos^{2}\left(t-\frac{1}{3}\;\pi\right)}\cos t-\frac{4}{\cos(t-\frac{\pi}{3})}\sin t}
\]
which will simplify to
\[
\frac{-\cos\left(t\right)\cos\left(\frac{1}{3}\left(3t-\pi\right)\right)-\sin\left(t\right)\sin\left(\frac{1}{3}\left(3t-\pi\right)\right)}{-\cos\left(t\right)\sin\left(\frac{1}{3}\left(3t-\pi\right)\right)+\cos\left(\frac{1}{3}\left(3t-\pi\right)\right)\sin\left(t\right)}.
\]
Let $\alpha=\frac{1}{3}(3t-\pi)$ and rearrange the terms:
\[
\frac{-1(\cos t\cdot\cos\alpha+\sin t\cdot\sin\alpha)}{\sin t\cdot\cos\alpha-\cos t\cdot\sin\alpha}
\]
which is by the sum identity rule,
\[
\frac{-\cos(t-\alpha)}{\sin(t-\alpha)}
\]
but $\alpha$ is $(t-\pi/3),$so substitute:
\[
\frac{-\cos(t-t+\pi/3)}{\sin(t-t+\pi/3)}=\frac{-\cos(\pi/3)}{\sin(\pi/3)}=-\frac{\sqrt{3}}{3}=SLOPE
\]
The direction vector will be
\[
u=\left(\begin{array}{c}
1\\
-\frac{\sqrt{3}}{3}
\end{array}\right)=\left(\begin{array}{c}
3\\
-\sqrt{3}
\end{array}\right)
\]
Now that was a lot of fun, but it would have been quicker to note
that the form of the equation gave us point $P(4,\pi/3)$ and that
means that the line is 4 units from the origin and its perpendicular
angle is $\pi/3.$ Thus the perpendicular line has slope $\tan(\pi/3)=$
and thus the "line'' has slope
\[
\frac{-1}{m}=\frac{-1}{\tan\left(\frac{\pi}{3}\right)}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}=SLOPE
\]
