128.135.100.107: /* Conductor */

2014-03-18T02:29:24Z

Conductor

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2014-01-07T05:47:01Z

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In geometry, an [[intersection]] is a point, line, or curve common in two or more objects (such as lines, curves, planes, and surfaces). The most simple case in [[Euclidean geometry]] is the intersection points of two distinct [[line (geometry)|lines]], that is either one [[point (geometry)|point]] or does not exist if lines are [[parallel lines|parallel]].
[[File:Schnittpunkt-2g.png|200px|thumb|intersection point of two lines]]

Determination of the intersection of [[flat (geometry)|flats]] is a simple task of [[linear algebra]], namely a [[system of linear equations]]. In general the determination of an intersection leads to [[non-linear equation]]s, which can be [[numerical solution|solved numerically]], for example using a [[Newton iteration]]. Intersection problems between a line and a [[conic section]] (circle, ellipse, parabola, ...) or a [[quadric]] (sphere, cylinder, hyperboloid, ...) lead to [[quadratic equation]]s that can be easily solved. Intersections between quadrics lead to [[quartic equation]]s that can be solved [[algebraic equation|algebraically]].

== On a plane ==
{{further|plane (geometry)|two-dimensional space}}

=== Two lines ===
{{main|Line–line intersection}}
For the determination of the intersection point of two non-parallel lines
*<math>a_1x+b_1y=c_1, \ a_2x+b_2y=c_2 </math>
one gets from [[Cramer's rule]] for the coordinates of the intersection point <math>(x_s,y_s)</math>
:<math> x_s=\frac{c_1b_2-c_2b_1}{a_1b_2-a_2b_1} , \quad y_s=\frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}. \ </math>
(In case of <math> a_1b_2-a_2b_1=0</math> the lines are parallel.)

If the lines are given by two points each, see next section.

=== Two line segments ===
{{main|Line segment intersection}}
[[File:Is-linesegm.png|300px|thumb|intersection of two line segments]]
For two non-parallel [[line segment]]s <math>(x_1,y_1),(x_2,y_2)</math> and <math>(x_3,y_3),(x_4,y_4)</math> there is no need for an intersection point (see picture), because the intersection point <math>(x_0,y_0)</math> of the corresponding lines need not to be contained in the line segments. In order to check the situation one uses parametric representations of the lines:
: <math> (x(s),y(s))=(x_1+s(x_2-x_1),y_1+s(y_2-y_1)),</math>
: <math> (x(t),y(t))=(x_3+t(x_4-x_3),y_3+t(y_4-y_3)). </math>
The line segments intersect only in a common point <math>(x_0,y_0)</math> of the corresponding lines if the corresponding parameters <math>s_0,t_0</math> fulfill the condition <math> 0\le s_0,t_0 \le 1 </math>.
The parametrs <math>s_0,t_0 </math> are the solution of the linear system
: <math>s(x_2-x_1)-t(x_4-x_3)=x_3-x_1,</math>
:<math> s(y_2-y_1)-t(y_4-y_3)=y_3-y_1 \ .</math>
It can be solved using Cramer's rule (see [[#Two lines|above]]). If the condition <math> 0\le s_0,t_0 \le 1 </math> is fulfilled one inserts <math>s_0</math> or <math>t_0</math> into the corresponding parametric representation and gets the intersection point <math>(x_0,y_0)</math>.

''Example:'' For the line segments <math>(1,1),(3,2)</math> and <math>(1,4),(2,-1)</math> one gets the linear system
:<math> 2s-t=0</math>
:<math>s+5t=3</math>
and <math>s_0=\tfrac{3}{11}, t_0=\tfrac{6}{11}</math>. That means: the lines intersect at point <math>(\tfrac{17}{11},\tfrac{14}{11})</math>.

''Remark:'' Considering lines (not segments!) determined by pairs of points, each, condition <math> 0\le s_0,t_0 \le 1 </math> can be skipped and the method yield the intersection point of the lines (see [[#Two lines|above]]).

[[File:Is-circle-line.png|thumb|line–circle intersection]]

=== A line and a circle ===
For the intersection of
*line <math>ax+by=c</math> and [[circle]] <math>x^2+y^2=r^2</math>
one solves the line equation for {{mvar|x}} or {{mvar|y}} and [[substitution (algebra)|substitutes]] it into the equation of the circle and gets for the solution (using the formula of a quadratic equation) <math>(x_1,y_1),(x_2,y_2)</math> with
:<math>x_{1/2}= \frac{ac\pm b\sqrt{r^2(a^2+b^2)-c^2}}{a^2+b^2} \ ,</math>
:<math>y_{1/2}= \frac{bc\mp a\sqrt{r^2(a^2+b^2)-c^2}}{a^2+b^2} \ , </math>
if <math> r^2(a^2+b^2)-c^2\ge0 \ .</math><br />
If <math> r^2(a^2+b^2)-c^2=0 </math> holds, there exists only one intersection point and the line is tangent to the circle.

''Remark:''
#If the circle's midpoint is not the origin, see.<ref>[http://www.mathematik.tu-darmstadt.de/~ehartmann/cdgen0104.pdf,''Geometry and Algorithms for COMPUTER AIDED DESIGN''], p. 17</ref>
#The intersection of a line and a parabola or hyperbola may be treated analogously.

=== Two circles ===
[[File:Is-circle-circle.png|thumb|circle–circle intersection]]
[[File:Is-circle-ellipse.png|thumb|circle–ellipse intersection]]
The determination of the intersection points of two circles
* <math>(x-x_1)^2+(y-y_1)^2=r_1^2 ,\ \quad (x-x_2)^2+(y-y_2)^2=r_2^2</math>
can be reduced to the previous case of intersecting a line and a circle. By subtraction of the two given equations one gets the line equation:
:<math>2(x_2-x_1)x+2(y_2-y_1)y=r_1^2-x_1^2-y_1^2-r_2^2+x_2^2+y_2^2. </math>

=== Two conic sections ===
The problem of intersection of an ellipse/hyperbola/parabola with another [[conic section]] leads to a [[algebraic system|system of quartic equations]], which can be solved in special cases easily by elimination of one coordinate. In general the intersection points can be determined by solving the equation by a Newton iteration. If a) both conics are given implicitly (by an equation) a 2-dimensional Newton iteration b) one implicitly and the other parametrically given a 1-dimensional Newton iteration is necessary. See next section.

=== Two curves ===
[[File:Schnittp2d-transv.png|250px|thumb|A transversal intersection of two curves]]
[[File:Beruehr-schnitt.png|350px|thumb|touching intersection (left), touching (right)]]
Two curves in <math>\R^2</math>, which are continuously differentiable (i.e. there is no sharp bend),
have an intersection point, if they have a point of the plane in common and have at this point
: a: different tangent lines ('''[[transversality (mathematics)|transversal]] intersection'''), or
: b: the tangent line in common and they are crossing each other ('''touching intersection''', s. picture).

If both the curves have a point {{mvar|S}} and the tangent line there in common but do not cross each other, they are just ''touching'' at point {{mvar|S}}.

Because touching intersection appears rarely and is difficult to deal with, the following considerations omit this case. In any case below all necessary differential conditions are presupposed. The determination of intersection points always lead to 1 or 2 non-linear equations which can be solved by a Newton iteration. A list of the appearing cases follows:

[[File:Schnittp2d-pi.png|thumb|intersection of a parametric curve and an implicit curve]]
[[File:Schnittp2d-ii.png|thumb|intersection of two implicit curves]]
*If ''both curves are explicitly'' given: <math> y=f_1(x), \ y=f_2(x)</math>, equalizing yields the equation
:: <math>f_1(x)=f_2(x) \ .</math>
*If ''both curves are parametrically'' given: <math>C_1: (x_1(t),y_1(t)), \ C_2: (x_2(s),y_2(s)).</math>
: Equalizing yields two equations for two variables:
:: <math>x_1(t)=x_2(s), \ y_1(t)=y_2(s) \ .</math>
*If ''one curve is parametrically and the other implicitly'' given: <math>C_1: (x_1(t),y_1(t)), \ C_2: f(x,y)=0.</math>
:This is beside the explicit case the simplest case. One has to insert the parametric representation of <math>C_1</math> into the equation <math>f(x,y)=0</math> of curve <math>C_2</math> and one gets the equation:
::<math>f(x(t),y(t))=0 \ .</math>
*If ''both curves are implicitly'' given: <math>C_1: f_1(x,y)=0, \ C_2: f_2(x,y)=0.</math>
: Here, an intersection point is a solution of the system
::<math>f_1(x,y)=0, \ f_2(x,y)=0 \ .</math>
Any Newton iteration needs convenient starting values, which can be derived by a visualization of both the curves. A parametrically or explicitly given curve can easily be visualized, because to any parameter {{mvar|t}} or {{mvar|x}} respectively it is easy to calculate the corresponding point. For implicitly given curves this task is not as easy. In this case one has to determine a curve point with help of starting values and an iteration. See
.<ref>[http://www.mathematik.tu-darmstadt.de/~ehartmann/cdgen0104.pdf, ''Geometry and Algorithms for COMPUTER AIDED DESIGN''], p. 33</ref>

''Examples:''
:1: <math>C_1: (t,t^3)</math> and circle <math>C_2: (x-1)^2+(y-1)^2-10=0</math> (s. picture).
:: The Newton iteration <math>t_{n+1}:=t_n-\frac{f(t_n)}{f'(t_n)}</math> for function
:::<math>f(t)=(t-1)^2+(t^3-1)^2-10</math> has to be done. As startvalues one can choose −1 and 1.5.
::The intersection points are: (−1.1073, −1.3578), (1.6011, 4.1046)
:2:<math>C_1: f_1(x,y)=x^4+y^4-1=0,</math>
:: <math>C_2: f_2(x,y)=(x-0.5)^2+(y-0.5)^2-1=0 </math> (s. picture).
:: The Newton iteration
:::<math>{x_{n+1}\choose y_{n+1}}={x_{n}+\delta_x\choose y_n+\delta_y}</math> has to be performed, where <math>{\delta_x \choose \delta_y}</math> is the solution of the linear system
:::<math>\begin{pmatrix}
\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\
\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y}
\end{pmatrix}{\delta_x \choose \delta_y}={-f_1\choose -f_2}
</math> at point <math>(x_n,y_n)</math>. As starting values one can choose(−0.5, 1) and (1, −0.5).
:: The linear system can be solved by Cramer's rule.
::The intersection points are (−0.3686, 0.9953) and (0.9953, −0.3686).

=== Two polygons ===
[[File:Is-polygpolyg.png|300px|thumb|intersection of two polygons: window test]]
If one wants to determine the intersection points of two polygons, one can check the intersection of any pair of line segments of the polygons (see [[#Two line segments|above]]). For polygons with a lot of segments this method is rather time consuming. In praxis one accelerates the intersection algorithm by using ''window tests''. In this case one divides the polygons into small sub-polygons and determines the smallest window (rectangle with sides parallel to the coordinate axes) for any sub-polygon. Before starting the time consuming determination of the intersection point of two line segments any pair of windows is tested for common points. See.<ref>[http://www.mathematik.tu-darmstadt.de/~ehartmann/cdg-skript-1998.pdf CDKG: Computerunterstützte Darstellende und Konstruktive Geometrie (TU Darmstadt)] (PDF; 3,4 MB), p. 79</ref>

== In space (three dimensions) ==
{{more information|three-dimensional space}}
In 3-dimensional space there are intersection points (common points) between curves and surfaces. In the following sections we consider ''[[transversality (mathematics)|transversal]] intersection'' only.

=== A line and a plane ===
{{main|Line–plane intersection}}
[[File:Schnittp-ger-eb.png|thumb|Line–plane intersection]]
The intersection of a line and a plane ''in [[general position]]'' in three dimensions is a point.

Commonly a line in space is represented parametrically <math> (x(t),y(t),z(t)) </math> and a plane by an equation <math>ax+by+cz=d</math>. Inserting the parameter representation into the equation yields the linear equation
:<math>ax(t)+by(t)+cz(t)=d\ ,</math>
for parameter <math>t_0</math> of the intersection point <math>(x(t_0),y(t_0),z(t_0))</math>.

If the linear equation has no solution, the line either lies on the plane or is parallel to it.

=== Three planes ===
If a line is defined by two intersecting planes <math>\varepsilon_i: \ \vec n_i\cdot\vec x=d_i, \ i=1,2</math> and should be intersected by a third plane <math>\varepsilon_3: \ \vec n_3\cdot\vec x=d_3 </math>, the common intersection point of the three planes has to be evaluated.

Three planes <math>\varepsilon_i: \ \vec n_i\cdot\vec x=d_i, \ i=1,2,3 </math> with linear independent normal vectors <math> \vec n_1,\vec n_2, \vec n_3</math> have the intersection point
:<math> \vec p_0=\frac{d_1(\vec n_2\times \vec n_3) +d_2(\vec n_3\times \vec n_1) + d_3(\vec n_1\times \vec n_2)}{\vec n_1\cdot(\vec n_2\times \vec n_3)} \ .</math>
For the proof one should establish <math>\vec n_i\cdot\vec p_0=d_i, \ i=1,2,3 , </math> using the rules of a [[scalar triple product]]. If the scalar triple product equals to 0, then planes either do not have the triple intersection or it is a line (or a plane, if all three planes are the same).

=== A curve and a surface ===
[[File:Is-pcurve-isurface.png|250px|thumb|intersection of curve <math>(t,t^2,t^3)</math> with surface <math>x^4+y^4+z^4=1 </math>]]
Analogously to the plane case the following cases lead to non-linear systems, which can be solved using a 1- or 3-dimensional Newton iteration.<ref>[http://www.mathematik.tu-darmstadt.de/~ehartmann/cdgen0104.pdf, ''Geometry and Algorithms for COMPUTER AIDED DESIGN''], p. 93</ref>
*parametric curve <math>C: (x(t),y(t),z(t) </math> and
:parametric surface <math>S: (x(u,v),y(u,v),z(u,v))\ ,</math>
*parametric curve <math>C: (x(t),y(t),z(t) </math> and
:implicit surface <math> S: f(x,y,z)=0\ .</math>

'''Example:'''
:parametric curve <math>C: (t,t^2,t^3)</math> und
:implicit surface <math>S: x^4+y^4+z^4-1=0</math> (s. picture).
:The intersection points are: (−0.8587, 0.7374, −0.6332), (0.8587, 0.7374, 0.6332).

A [[line–sphere intersection]] is a simple special case.

Like the case of a line and a plane, the intersection of a curve and a surface ''in [[general position]]'' consists of discrete points, but a curve may be partly or totally contained in a surface.

=== A line and a polyhedron ===
{{main|Intersection of a polyhedron with a line}}

=== Two surfaces ===
{{main|Intersection curve}}
Two transversally intersecting surfaces give an [[intersection curve]]. The most simple case the intersection line of two non-parallel planes.

== See also ==
* [[Computational geometry]]
* [[Equation of a line]]

== References ==
<references/>

[[Category:Euclidean geometry]]
[[Category:Linear algebra]]
[[Category:Computational geometry]]

Cyclic decomposition theorem - Revision history

128.135.100.107: /* Conductor */

en>BG19bot: WP:CHECKWIKI error fix. Section heading problem. Violates WP:MOSHEAD.