# Representations ## [[Vector]] Form > [!info] Why we don't use function form > We don't use function form because we see lines as *lines*, not necessarily as an input and output ## $\R^2$ ![[../../00 Excalidraw/Vector Form of Lines .excalidraw.dark.svg]] $p$ is a point on $l$ $\vec v$ is parallel to $l$ $ \huge \begin{align*} l:\pa{x, y} = (P_{x,}, P_{y}) + t \mat{\vec v_{x} \\ \vec v_y} \end{align*} $ $t$ can be described as a [[Free Parameter]] This form of the line is *a* form where when $-\infty < t < \infty$, all points on $l$ are created. However, as long as the point $P$ is some point on the line, and $\vec v$ ($\ne \vec 0$) has same direction (or is [[Parallel#Examples#Parallel|Anti Parallel]], you can create any You can generate any point on $l$ by setting the free parameter to another value. Vector form can be thought of as a point *generator* $\in \R^n$. ## $\R^3$ $ \huge \begin{align*} l:\pa{x, y, z} = (P_{x,}, P_{y},P_z) + t \mat{\vec v_{x} \\ \vec v_y\\v_z} \end{align*} $ ## Parametric Form ## $\R^2$ $ \huge l : \begin{cases} x = P_{x}+ t\vec v_{x}\\ y = P_{y}+ t\vec v _y \end{cases} $ Parametric form can be thought of as a point *generator* $\in \R^n$. > [!example] > *Vector Form*: $l: (x,y) = (3, 1) + t\mat{1\\-2}$ > > *Parametric Form*: ${l: \begin{cases} > x = 3 + t \\ > y = 1 - 2t > \end{cases}}$ ## Normal Form ## $\R^2$ $\huge \begin{align*} l&: n_{x}x + n_{y}y = c \\ \vec n &= \mat { n_{x}\\n_{y}}\perp l \end{align*} $ This is a form of representing a line through a different vector, a [[Normal Vector]] of the line. This form offers a *more limited* amount of chaos between different representations, while the [[Lines#Vector Form|Vector Form]] has two degrees of ambiguity (the *point* and the *vector*), while there is less ambiguity between different normal form representations of the same line. Normal form struggles more with point *generation*, but works well for point *testing*. > [!tip] > If the coefficient in 0 in any $\R$, then the line passes through the original ## $\R^3$ There is **no** normal form for lines in $\R^3$, as lines in $\R^3$ have a *continuum* of normal vectors, meaning just having one does not carry enough information. > [!example] Express from normal form to vector form > > $\huge\begin{align*} > \let k \in \R^{2} \text{ be }& k: 5x+y =-3 \\ > > \vec n &= \mat{5 \\ 1} \\ > \therefore \vec v &= \mat{1 \\ -5} \\ > > P &= \pa{0, -3 - 5(0)} =\pa{0,-3} \\ > > l : (x, y) &= \pa{0, -3} + t\mat{1 \\ - 5} > \end{align*}$ > [!example] Vector form to normal form > $\huge\begin{align*} > \let m \in \R^{2}\text{ be }& m : (x,y) =\pa{-2,1}+t\mat{3\\-2} \\ > > > \perp m &= \mat{2 \\ 3} \\ > > \\ > m: 2x + 3y &= c \\ > (-2)x + (1)y &= c \therefore c=-1 \\ > \\ > m: 2x+3y&= -1 > > \end{align*}$ # Getting the angle between lines ![[../../00 Excalidraw/Lines .excalidraw.dark.svg]] > [!tip] The assumption when finding the angle between lines is to pick the smaller of the two angles > To find the angle between the lines, you can find the angle between two [[Vector|Vectors]] parallel to each line respectively, or find the angle between the line's normal vector. Where $l$ and $k$ are lines in $\R^n$ Finding the angle between the two lines can be done with the [[../../02 Areas/Math/Dot Product]]. $\huge \begin{align*} \vec u &\parallel l\\ \vec v &\parallel k\\ \end{align*}$ $\huge\arccos\paren{\frac{\vec u \cdot \vec v } {\norm{\vec u}\norm{\vec v}}} = \theta_{\angle lk} \iff \R^n$ > [!note] Dimensions > Comparing parallel vectors of the lines of the lines extends to $\R^3$ ($\R^n$ when $n\ge2$), however comparing the normals of two lines only works in $\R^2$ because there is no normal form for lines in $\R^3$ and higher dimensions. *Using the normal form* $\huge \arccos\paren{\frac{\paren{\vec l_{\perp} \cdot \vec k_{\perp}}} {\norm{\vec l_\perp}\norm{\vec k_{\perp}}}}=\theta_{\angle{lk}} \iff \R^2 $ > [!abstract] Why comparing the normals works > > The reason why comparing the normals of the lines is ### Angle between lines using a parallel vector and a normal vector If you are given two lines in normal form and vector form, it is valid to convert them to the same form - but you can also get information by just comparing a vector parallel to one to a normal of the other. $\huge\begin{align*} l: 5x-y &= 0\\ k: (x,y) &= (0,0)+t\mat{1\\-1} \end{align*}$ ![[../../00 Excalidraw/Lines _0.excalidraw.dark.svg]] $\huge \theta_{\angle lk} = 90\degree - \arccos\paren{ \frac{\vec l_{\perp} \cdot \vec v}{\norm{\vec l_\perp}\norm{\vec v}} } $ $\huge \theta_{\angle lk}= \arcsin\paren{ \frac{\vec l_{\perp} \cdot \vec v}{\norm{\vec l_\perp}\norm{\vec v}} } $ > [!warning] > The above formula will only work if the angle between $\vec l_\perp$ and $\vec v$ is $\le 90\degree$, if they are above instead use the $180\degree$ supplement to that angle ($180\degree - \theta$)