Intersection line

Intersection line (red) of two planes (green and blue)

In geometry, a straight line of intersection is a straight line in which two non- parallel planes intersect in three-dimensional Euclidean space . A straight line in space is usually described by a parametric form of a straight line equation . The way to the straight line equation of the line of intersection of two planes depends on the description of the two planes to be intersected. Since there are two standard descriptions for this ( normal form and parametric form ), there are three possibilities to determine the straight line equation of the intersection line.

If one of the planes to be cut is a coordinate plane , the cutting line is called the track line . If several levels have a common line of intersection, one speaks of a level cluster .

Intersection of a plane in normal form with a plane in parametric form

calculation

Given a plane in normal form,

${\displaystyle \varepsilon _{1}\colon {\vec {n}}\cdot {\vec {x}}=d}$,

and a plane in parametric form,

${\displaystyle \varepsilon _{2}\colon {\vec {x}}={\vec {p}}+r{\vec {u}}+s{\vec {v}}}$.

So that the planes are not parallel, must or be, because otherwise there would also be a normal vector of . We are now looking for a parametric representation of the intersection line ${\displaystyle {\vec {n}}\cdot {\vec {u}}\neq 0}$ ${\displaystyle {\vec {n}}\cdot {\vec {v}}\neq 0}$ ${\displaystyle {\vec {n}}}$ ${\displaystyle \varepsilon _{2}}$

${\displaystyle g\colon {\vec {x}}={\vec {q}}+t{\vec {w}}}$.

Substituting the parametric form into the normal form leads to

${\displaystyle {\vec {n}}\cdot {\vec {p}}+r\,{\vec {n}}\cdot {\vec {u}}+s\,{\vec {n}}\cdot {\vec {v}}=d}$.

Is , then solving the equation for the parameter and then inserting it into the parameter form ${\displaystyle {\vec {n}}\cdot {\vec {u}}\neq 0}$ ${\displaystyle s}$

${\displaystyle g\colon {\vec {x}}=\left({\vec {p}}+{\frac {d-{\vec {n}}\cdot {\vec {p}}}{{\vec {n}}\cdot {\vec {u}}}}{\vec {u}}\right)+t\left({\vec {v}}-{\frac {{\vec {n}}\cdot {\vec {v}}}{{\vec {n}}\cdot {\vec {u}}}}{\vec {u}}\right)}$.

Is , the roles of and are reversed. ${\displaystyle {\vec {n}}\cdot {\vec {u}}=0}$ ${\displaystyle {\vec {u}}}$ ${\displaystyle {\vec {v}}}$

example

The two levels are through

${\displaystyle \varepsilon _{1}\colon (1,2,1)^{T}\cdot {\vec {x}}=1}$

and

${\displaystyle \varepsilon _{2}\colon {\vec {x}}=(1,1,1)^{T}+r(2,1,-3)^{T}+s(-1,1,0)^{T}}$

given. The parameter representation then results for the intersection line

${\displaystyle g\colon {\vec {x}}=(-5,-2,10)^{T}+t(-3,0,3)^{T}}$.

Intersection of two planes in parametric form

calculation

If both plane equations are available in parametric form, one first calculates the normal form for one of the two planes and then applies the procedure from the previous section. For a plane with the support vector and the direction vectors and is obtained from the cross product ${\displaystyle {\vec {p}}}$ ${\displaystyle {\vec {u}}}$ ${\displaystyle {\vec {v}}}$

${\displaystyle {\vec {n}}={\vec {u}}\times {\vec {v}}}$

a normal vector and the plane equation is then

${\displaystyle \varepsilon \colon {\vec {n}}\cdot {\vec {x}}={\vec {n}}\cdot {\vec {p}}}$.

In order to investigate the parallelism of two planes in parametric form, one first determines a normal vector for one of the planes with the help of the cross product. If the scalar products of this normal vector with the direction vectors of the other plane are equal to zero, the two planes are parallel.

example

The two levels are through

${\displaystyle \varepsilon _{1}\colon {\vec {x}}=(1,-1,1)^{T}+r_{1}(2,1,-1)^{T}+s_{1}(-1,1,0)^{T}}$

and

${\displaystyle \varepsilon _{2}\colon {\vec {x}}=(1,1,1)^{T}+r_{2}(2,2,-1)^{T}+s_{2}(1,0,1)^{T}}$

given. The normal vector for results in ${\displaystyle \varepsilon _{1}}$

${\displaystyle {\vec {n}}=(2,1,-1)^{T}\times (-1,1,0)^{T}=(1,1,3)^{T}}$

and thus the normal form

${\displaystyle \varepsilon _{1}\colon {\vec {x}}=(1,1,3)^{T}\cdot {\vec {x}}=3}$.

The parameter representation is then obtained for the intersection line

${\displaystyle g\colon {\vec {x}}=(-3,-3,3)^{T}+t(-7,-8,5)^{T}}$.

Intersection of two planes in normal form

calculation

There are now two levels

${\displaystyle \varepsilon _{1}\colon {\vec {n}}_{1}\cdot {\vec {x}}=d_{1}}$

and

${\displaystyle \varepsilon _{2}\colon {\vec {n}}_{2}\cdot {\vec {x}}=d_{2}}$

So that the planes are not parallel, the two normal vectors must be linearly independent , that is, must not be a multiple of . We are looking for a parametric representation of the intersection line ${\displaystyle {\vec {n}}_{1},{\vec {n}}_{2}}$ ${\displaystyle {\vec {n}}_{1}}$ ${\displaystyle {\vec {n}}_{2}}$

${\displaystyle g\colon {\vec {x}}={\vec {q}}+t{\vec {w}}}$.

The direction vector of the line of intersection results from the cross product of the normal vectors:

${\displaystyle {\vec {w}}={\vec {n}}_{1}\times {\vec {n}}_{2}}$.

A support vector of the line of intersection is obtained by connecting the planes and with the plane perpendicular to them ${\displaystyle {\vec {q}}}$ ${\displaystyle \varepsilon _{1}}$ ${\displaystyle \varepsilon _{2}}$

${\displaystyle \varepsilon _{3}\colon {\vec {x}}=s_{1}{\vec {n}}_{1}+s_{2}{\vec {n}}_{2}}$

cuts. The parameters and are found by inserting them into the equations of the levels and and are thus obtained ${\displaystyle s_{1}}$ ${\displaystyle s_{2}}$ ${\displaystyle \varepsilon _{1}}$ ${\displaystyle \varepsilon _{2}}$

${\displaystyle {\vec {q}}={\frac {d_{1}{\vec {n}}_{2}^{2}-d_{2}({\vec {n}}_{1}\cdot {\vec {n}}_{2})}{{\vec {n}}_{1}^{2}{\vec {n}}_{2}^{2}-({\vec {n}}_{1}\cdot {\vec {n}}_{2})^{2}}}{\vec {n}}_{1}+{\frac {d_{2}{\vec {n}}_{1}^{2}-d_{1}({\vec {n}}_{1}\cdot {\vec {n}}_{2})}{{\vec {n}}_{1}^{2}{\vec {n}}_{2}^{2}-({\vec {n}}_{1}\cdot {\vec {n}}_{2})^{2}}}{\vec {n}}_{2}}$.

example

The two levels are through

${\displaystyle \varepsilon _{1}\colon (1,2,1)^{T}\cdot {\vec {x}}=1}$

and

${\displaystyle \varepsilon _{2}\colon (2,-3,2)^{T}\cdot {\vec {x}}=2}$

given. This gives the direction vector of the line of intersection as

${\displaystyle {\vec {w}}={\vec {n}}_{1}\times {\vec {n}}_{2}=(7,0,-7)^{T}}$.

For the support vector it follows from and from the above formula ${\displaystyle {\vec {p}}}$ ${\displaystyle {\vec {n}}_{1}^{2}=6,{\vec {n}}_{2}^{2}=17,{\vec {n}}_{1}\cdot {\vec {n}}_{2}=-2}$

${\displaystyle {\vec {q}}={\tfrac {1}{2}}(1,0,1)^{T}}$.

So is

${\displaystyle g\colon {\vec {x}}={\tfrac {1}{2}}(1,0,1)^{T}+t(7,0,-7)^{T}}$

a parametric representation of the intersection of both planes.

annotation

The above formula provides a parameter representation of the line of intersection without any case distinctions, but it is computationally expensive. In the case of specifically specified plane equations , it may be better to use the Gaussian algorithm to determine a parameter representation of the straight line of intersection. For the above example is the linear system of equations

${\displaystyle x+2y+z=1}$

${\displaystyle 2x-3y+2z=2}$

to solve. 2 times the first equation minus 1 times the second equation results in the system of equations in line step form:

${\displaystyle {\begin{array}{rrrcl}x&+2y&+z&=&1\\&7y&&=&0\end{array}}}$

The unknown can be chosen: . After is an insertion into the first equation yields . This gives the (slightly different) parametric representation of the intersection line: ${\displaystyle z}$ ${\displaystyle z=t}$ ${\displaystyle y=0}$ ${\displaystyle x=1-t}$

${\displaystyle {\vec {x}}=(1,0,0)^{T}+t(-1,0,1)^{T}}$.

See also

• Intersection curve
• Intersection
• Cutting angle (geometry)
• Positional relationship