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! Represenation of a Line and other linear geometrical elements
!! Line on a plane
On a plane line is a set of __x__=(x,y) points, where the following expression holds:
{EQUATION(size="80") }$
ax+by=c
{EQUATION} where {EQUATION(size="80") }$ \sqrt{a^{2}+b^{2}}\neq 0 {EQUATION}.
Equivalently:
{EQUATION(size="80") }$
\frac{a}{\sqrt{a^{2}+b^{2}}}x + \frac{b}{\sqrt{a^{2}+b^{2}}}y = \frac{c}{\sqrt{a^{2}+b^{2}}}
{EQUATION}
Now the coefficients of ''x'' and of ''y'' squared add up to 1, we can define the
{EQUATION(size="80") }$\boldsymbol{\omega} {EQUATION}unit vector:
{EQUATION(size="80") }$
\boldsymbol{\omega}= \left (\frac{a}{\sqrt{a^{2}+b^{2}}} , \frac{b}{\sqrt{a^{2}+b^{2}}} \right )
{EQUATION}
and a ''t'' scalar:
{EQUATION(size="80") }$
t= \frac{c}{\sqrt{a^{2}+b^{2}}} {EQUATION}
With these notations the expression of a line is:
{EQUATION(size="80") }$
\boldsymbol{\omega}\boldsymbol{x}=t
{EQUATION} (1)
Thus, we are looking for the set of o points, where the projection of the location vector to a given {EQUATION(size="80") }$ \boldsymbol{\omega}{EQUATION} vector is constant.These points are located on a line perpendicular to vector {EQUATION(size="80") }$ \boldsymbol{\omega}{EQUATION}, and the distance of this line from the origin is ''t''.
{img fileId="280" thumb="mousesticky" button="y"}
In order to parametrize the points of this line we look for the {EQUATION(size="80") }$ \boldsymbol{\omega}_{\perp}{EQUATION} unit vector perpendicular to vector {EQUATION(size="80") }$ \boldsymbol{\omega}{EQUATION}. For unique solution let us choose the sign of the determinant of these vectors, now we chose positive:
{EQUATION(size="80") }$
\begin{vmatrix}
\omega_{1} & \omega_{\perp1}\\
\omega_{2} & \omega_{\perp2}
\end{vmatrix}=1
{EQUATION} (2)
Let us have the variable of integration the point ''s'', with that we obtain the __l__ points of an ''L'' line as follows:
{EQUATION(size="80") }$
\mathbf{l}_{t,\boldsymbol{\omega}}\left ( s \right )=t\boldsymbol{\omega}+s\boldsymbol{\omega}_{\perp}
{EQUATION}
{img fileId="282" thumb="mousesticky" button="y"}
This description still does not constitute a unique description, as when
{EQUATION(size="80") }$ t\in \left \{ -\infty,\infty \right \} {EQUATION}, then {EQUATION(size="80") }$\mathbf{l}_{t,\boldsymbol{\omega}}=\mathbf{l}_{-t,\boldsymbol{-\omega}} {EQUATION}. We should limit either ''t'' to positive numbers, or limit {EQUATION(size="80") }$\boldsymbol{\omega} {EQUATION} to one of the half-spaces. E.g., when {EQUATION(size="80") }$\boldsymbol{\omega}= \left ( \cos \left ( \vartheta \right ),\sin \left ( \vartheta \right ) \right ){EQUATION}, then either
{EQUATION(size="80") }$ t\in \left \{ -\infty,\infty \right \} {EQUATION} and {EQUATION(size="80") }$ \vartheta\in \left \{ 0,\pi \right \} {EQUATION}
or
{EQUATION(size="80") }$ t\in \left \{ 0,\infty \right \} {EQUATION} and {EQUATION(size="80") }$ \vartheta\in \left \{ 0,2\pi \right \} {EQUATION}
In the literature both conventions are present.
!! Linear elements in higher dimensions
The expression for the 2D line on {EQUATION(size="80") }$\mathbf{x}\in \mathbb{R}^{2}{EQUATION} determines sets of points such, that for a {EQUATION(size="80") }$ t\in \mathbb{R}{EQUATION} scalar and for a unit vector {EQUATION(size="80") }$\boldsymbol{\omega}\in \mathbb{S}^{1}{EQUATION} of a sphere of one degree of freedom ( {EQUATION(size="80") }$\mathbb{S}{EQUATION}), the equation holds: {EQUATION(size="80") }$
\boldsymbol{\omega}\boldsymbol{x}=t
{EQUATION} and with that {EQUATION(size="80") }$ \left(t,\boldsymbol{\omega} \right ) \in \mathbb{R}\times \mathbb{S}^{1}{EQUATION} equations determine a line with a direction. When we look at the parametrization of it in Eq. (2), ''s'' and ''t'' are interchangeable, since {EQUATION(size="80") }$\boldsymbol{\omega} {EQUATION} és {EQUATION(size="80") }$\boldsymbol{\omega}_{\perp}{EQUATION} determine each other apart from a sign. We could also say, that the parameter of our line is ''s'' and {EQUATION(size="80") }$\boldsymbol{\omega_{\perp}} {EQUATION}, the variable of integration is ''t'', in the direction of {EQUATION(size="80") }$\boldsymbol{\omega} {EQUATION} .
In an ''n'' dimension space, Eq. (1) given that {EQUATION(size="80") }$ \left(t,\boldsymbol{\omega} \right ) \in \mathbb{R}\times \mathbb{S}^{n-1}{EQUATION} is an expression of a hyperplane perpendicular to the direction vector {EQUATION(size="80") }$\boldsymbol{\omega}{EQUATION}. Now to specify a single point on this plane, we need a set of direction vectors of a complete base of unit vectors {EQUATION(size="80") }$\boldsymbol{\omega_{\perp,i}} {EQUATION}, that we now with an off-hand notation order into matrix {EQUATION(size="80") }$\boldsymbol{\Omega_{\perp} }{EQUATION}, so now multiplied by a vector of {EQUATION(size="80") }$\mathbf{s}\in \mathbb{R}^{n-1}{EQUATION} we arrive into a point of the plane as follows:
{EQUATION(size="80") }$
t\boldsymbol{\omega}+\mathbf{s}\boldsymbol{\Omega_{\perp} }
{EQUATION}
If we choose, like we did before, for the parameters of the linear set {EQUATION(size="80") }$ \left(t,\boldsymbol{\omega} \right ){EQUATION}, then our expression describes points of the __H__ hyperplane:
{EQUATION(size="80") }$
\mathbf{H}_{\boldsymbol{\omega},t}\left ( \mathbf{s},\boldsymbol{\Omega_{\perp} } \right )=t\boldsymbol{\omega}+\mathbf{s}\boldsymbol{\Omega_{\perp} }=\mathbf{x}_{0}+\mathbf{s}\boldsymbol{\Omega_{\perp}}
{EQUATION}
On the contrary, if we chose as the parameters the elements of the product {EQUATION(size="80") }$ \mathbf{s}\boldsymbol{\Omega_{\perp} } {EQUATION} , we obtain a line, with points along unit vector {EQUATION(size="80") }$\boldsymbol{\omega} {EQUATION} with variable of integration ''t'':
{EQUATION(size="80") }$\mathbf{L}_{\mathbf{s},\boldsymbol{\Omega_{\perp}}} \left ( t,\boldsymbol{\omega} \right )=t\boldsymbol{\omega}+\mathbf{s}\boldsymbol{\Omega_{\perp} }=t\boldsymbol{\omega}+\mathbf{x}_{0}
{EQUATION}
Note, that the points of the __H__ hyperplane is determined by ''n'' independent information contained in {EQUATION(size="80") }$ \left(t,\boldsymbol{\omega} \right ){EQUATION} together, while the __L__ line is determined by the product {EQUATION(size="80") }$ \mathbf{s}\boldsymbol{\Omega_{\perp} } {EQUATION}
with 2(n-1) independent elements, since additionally to the unit vector{EQUATION(size="80") }$ \boldsymbol{\omega} {EQUATION} we need the values of vector ''s'' as well. To reach a point in space we still need to define the vector base of{EQUATION(size="80") }$ \boldsymbol{\Omega_{\perp}} {EQUATION}, bearing no information on the object, it only constitutes the coordinate system choice.