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Ingeometry,Plücker coordinates, introduced byJulius Plücker in the 19th century, are a way to assign sixhomogeneous coordinates to eachline inprojective 3-space,⁠${\displaystyle {\mathbb{P}}^3}$⁠. Because they satisfy a quadratic constraint, they establish aone-to-one correspondence between the 4-dimensional space of lines in⁠${\displaystyle {\mathbb{P}}^3}$⁠ and points on aquadric in⁠${\displaystyle {\mathbb{P}}^5}$⁠ (projective 5-space). A predecessor and special case ofGrassmann coordinates (which describek-dimensional linear subspaces, orflats, in ann-dimensionalEuclidean space), Plücker coordinates arise naturally ingeometric algebra. They have proved useful forcomputer graphics, and also can be extended to coordinates for thescrews and wrenches in the theory ofkinematics used forrobot control.

A lineL in 3-dimensionalEuclidean space is determined by two distinct points that it contains, or by two distinct planes that contain it (aplane-plane intersection). Consider the first case, with points${\displaystyle x=(x_1,x_2,x_3)}$ and${\displaystyle y=(y_1,y_2,y_3).}$ Thevector displacement fromx toy is nonzero because the points are distinct, and represents thedirection of the line. That is, every displacement between points on the lineL is ascalar multiple ofd =y –x. If a physical particle of unit mass were to move fromx toy, it would have amoment about the origin of the coordinate system. The geometric equivalent to this moment is a vector whose direction is perpendicular to the plane containing the lineL and the origin, and whose length equals twice the area of the triangle formed by the displacement and the origin. Treating the points as displacements from the origin, the moment ism =x ×y, where "×" denotes the vectorcross product. For a fixed line,L, the area of the triangle is proportional to the length of the segment betweenx andy, considered as the base of the triangle; it is not changed by sliding the base along the line, parallel to itself. By definition the moment vector is perpendicular to every displacement along the line, sod ⋅m = 0, where "⋅" denotes the vectordot product.

Although neither directiond nor momentm alone is sufficient to determine the lineL, together the pair does so uniquely, up to a common (nonzero) scalar multiple which depends on the distance betweenx andy. That is, the coordinates

${\displaystyle (\mathbf{d}:\mathbf{m})=(d_1:d_2:d_3:m_1:m_2:m_3)}$

may be consideredhomogeneous coordinates forL, in the sense that all pairs(λd : λm), forλ ≠ 0, can be produced by points onL and onlyL, and any such pair determines a unique line so long asd is not zero andd ⋅m = 0. Furthermore, this approach extends to includepoints,lines, and aplane "at infinity", in the sense ofprojective geometry. In addition a point${\displaystyle x}$ lies on the lineL if and only if${\displaystyle x\times d=m}$.

Example. Letx = (2, 3, 7) andy = (2, 1, 0). Then(d :m) = (0 : −2 : −7 : −7 : 14 : −4).

Alternatively, let the equations for pointsx of two distinct planes containingL be

Then their respective planes are perpendicular to vectorsa andb, and the direction ofL must be perpendicular to both. Hence we may setd =a ×b, which is nonzero becausea,b are neither zero nor parallel (the planes being distinct and intersecting). If pointx satisfies both plane equations, then it also satisfies the linear combination

That is,

${\displaystyle \mathbf{m}=a\mathbf{b}-b\mathbf{a}}$

is a vector perpendicular to displacements to points onL from the origin; it is, in fact, a moment consistent with thed previously defined froma andb.

Proof of geometric formulation

Proof 1: Need to show that ${\displaystyle \mathbf{m}=a\mathbf{b}-b\mathbf{a}=\mathbf{r}\times \mathbf{d}=\mathbf{r}\times (\mathbf{a}\times \mathbf{b}).}$what is "r"? Without loss of generality, let ${\displaystyle \mathbf{a}\cdot \mathbf{a}=\mathbf{b}\cdot \mathbf{b}=1.}$ PointB is the origin. LineL passes through pointD and is orthogonal to the plane of the picture. The two planes pass throughCD andDE and are both orthogonal to the plane of the picture. PointsC andE are the closest points on those planes to the originB, therefore angles∠BCD and ∠BED are right angles and so the pointsB, C, D, E lie on a circle (due to a corollary ofThales's theorem).BD is the diameter of that circle. Angle∠BHF is a right angle due to the following argument. Letε := ∠BEC. Since△BEC ≅ △BFG (by side-angle-side congruence), then∠BFG = ε. Since∠BEC + ∠CED = 90°, letε' := 90° – ε = ∠CED. By theinscribed angle theorem,∠DEC = ∠DBC, so∠DBC = ε'.∠HBF + ∠BFH + ∠FHB = 180°;ε' + ε + ∠FHB = 180°,ε + ε' = 90°; therefore,∠FHB = 90°. Then∠DHF must be a right angle as well. Angles∠DCF, ∠DHF are right angles, so the four pointsC, D, H, F lie on a circle, and (by theintersecting secants theorem) ${\displaystyle ||BF||||BC||=||BH||||BD||}$ that is, Proof 2: Let ${\displaystyle \mathbf{a}\cdot \mathbf{a}=\mathbf{b}\cdot \mathbf{b}=1.}$ This implies that ${\displaystyle a=-||BE||,b=-||BC||.}$ According to thevector triple product formula, ${\displaystyle \mathbf{r}\times (\mathbf{a}\times \mathbf{b})=(\mathbf{r}\cdot \mathbf{b})\mathbf{a}-(\mathbf{r}\cdot \mathbf{a})\mathbf{b}.}$ Then When${\displaystyle ||\mathbf{r}||=0,}$ the lineL passes the origin with directiond. If${\displaystyle ||\mathbf{r}||>0,}$ the line has directiond; the plane that includes the origin and the lineL has normal vectorm; the line is tangent to a circle on that plane (normal tom and perpendicular to the plane of the picture) centered at the origin and with radius${\displaystyle ||\mathbf{r}||.}$ Example. Leta0 = 2,a = (−1, 0, 0) andb0 = −7,b = (0, 7, −2). Then(d :m) = (0 : −2 : −7 : −7 : 14 : −4). Although the usual algebraic definition tends to obscure the relationship,(d :m) are the Plücker coordinates ofL.

In a 3-dimensional projective space⁠${\displaystyle {\mathbb{P}}^3}$⁠, letL be a line through distinct pointsx andy withhomogeneous coordinates(x₀ :x₁ :x₂ :x₃) and(y₀ :y₁ :y₂ :y₃).

The Plücker coordinatesp_ij are defined as follows:

(the skew symmetric matrix whose elements arep_ij is also called thePlücker matrix )
This impliesp_ii = 0 andp_ij = −p_ji, reducing the possibilities to only six (4choose 2) independent quantities. The sextuple

${\displaystyle (p_{01}:p_{02}:p_{03}:p_{23}:p_{31}:p_{12})}$

is uniquely determined byL up to a common nonzero scale factor. Furthermore, not all six components can be zero.Thus the Plücker coordinates ofL may be considered as homogeneous coordinates of a point in a 5-dimensional projective space, as suggested by the colon notation.

To see these facts, letM be the 4×2 matrix with the point coordinates as columns.

The Plücker coordinatep_ij is the determinant of rowsi andj ofM.Becausex andy are distinct points, the columns ofM arelinearly independent;M hasrank 2. LetM′ be a second matrix, with columnsx′,y′ a different pair of distinct points onL. Then the columns ofM′ arelinear combinations of the columns ofM; so for some 2×2nonsingular matrixΛ,

${\displaystyle M^{\prime }=M\mathrm{\Lambda }.}$

In particular, rowsi andj ofM′ andM are related by

Therefore, the determinant of the left side 2×2 matrix equals the product of the determinants of the right side 2×2 matrices, the latter of which is a fixed scalar,det Λ. Furthermore, all six 2×2 subdeterminants inM cannot be zero because the rank ofM is 2.

Denote the set of all lines (linear images of⁠${\displaystyle {\mathbb{P}}^1}$⁠) in⁠${\displaystyle {\mathbb{P}}^3}$⁠ byG_1,3. We thus have a map:

where

${\displaystyle L^{\alpha }=(p_{01}:p_{02}:p_{03}:p_{23}:p_{31}:p_{12}).}$

Alternatively, a line can be described as the intersection of two planes. LetL be a line contained in distinct planesa andb with homogeneous coefficients(a⁰ :a¹ :a² :a³) and(b⁰ :b¹ :b² :b³), respectively. (The first plane equation is$\sum _k{a}^k{x}_k=0,$ for example.) The dual Plücker coordinatep^ij is

Dual coordinates are convenient in some computations, and they are equivalent to primary coordinates:

${\displaystyle (p_{01}:p_{02}:p_{03}:p_{23}:p_{31}:p_{12})=(p^{23}:p^{31}:p^{12}:p^{01}:p^{02}:p^{03})}$

Here, equality between the two vectors in homogeneous coordinates means that the numbers on the right side are equal to the numbers on the left side up to some common scaling factorλ. Specifically, let(i,j,k,ℓ) be aneven permutation of(0, 1, 2, 3); then

${\displaystyle p_{ij}=\lambda p^{k\ell }.}$

To relate back to the geometric intuition, takex₀ = 0 as the plane at infinity; thus the coordinates of pointsnot at infinity can be normalized so thatx₀ = 1. ThenM becomes

and setting${\displaystyle x=(x_1,x_2,x_3)}$ and${\displaystyle y=(y_1,y_2,y_3)}$, we have${\displaystyle d=(p_{01},p_{02},p_{03})}$and${\displaystyle m=(p_{23},p_{31},p_{12})}$.

Dually, we have${\displaystyle d=(p^{23},p^{31},p^{12})}$ and${\displaystyle m=(p^{01},p^{02},p^{03}).}$

If the point${\displaystyle \mathbf{z}=(z_0:z_1:z_2:z_3)}$ lies onL, then the columns of

arelinearly dependent, so that the rank of this larger matrix is still 2. This implies that all 3×3 submatrices have determinant zero, generating four (4 choose 3) plane equations, such as

The four possible planes obtained are as follows.

Using dual coordinates, and letting(a⁰ :a¹ :a² :a³) be the line coefficients, each of these is simplyaⁱ =p^ij, or

${\displaystyle 0=\sum _{i=0}^3{p}^{ij}{z}_i,j=0,\dots ,3.}$

Each Plücker coordinate appears in two of the four equations, each time multiplying a different variable; and as at least one of the coordinates is nonzero, we are guaranteed non-vacuous equations for two distinct planes intersecting inL. Thus the Plücker coordinates of a line determine that line uniquely, and the map α is aninjection.

The image ofα is not the complete set of points in⁠${\displaystyle {\mathbb{P}}^5}$⁠; the Plücker coordinates of a lineL satisfy the quadratic Plücker relation

For proof, write this homogeneous polynomial as determinants and useLaplace expansion (in reverse).

Since both 3×3 determinants have duplicate columns, the right hand side is identically zero.

Another proof may be done like this:Since vector

${\displaystyle d=\left(p_{01},p_{02},p_{03}\right)}$

is perpendicular to vector

${\displaystyle m=\left(p_{23},p_{31},p_{12}\right)}$

(see above), the scalar product ofd andm must be zero. q.e.d.

Letting(x₀ :x₁ :x₂ :x₃) be the point coordinates, four possible points on a line each have coordinatesx_i =p_ij, forj = 0, 1, 2, 3. Some of these possible points may be inadmissible because all coordinates are zero, but since at least one Plücker coordinate is nonzero, at least two distinct points are guaranteed.

If${\displaystyle (q_{01}:q_{02}:q_{03}:q_{23}:q_{31}:q_{12})}$ are the homogeneous coordinates of a point in⁠${\displaystyle {\mathbb{P}}^5}$⁠, without loss of generality assume thatq₀₁ is nonzero. Then the matrix

has rank 2, and so its columns are distinct points defining a lineL. When the⁠${\displaystyle {\mathbb{P}}^5}$⁠ coordinates,q_ij, satisfy the quadratic Plücker relation, they are the Plücker coordinates ofL. To see this, first normalizeq₀₁ to 1. Then we immediately have that for the Plücker coordinates computed fromM,p_ij =q_ij, except for

${\displaystyle p_{23}=-q_{03}{q}_{12}-q_{02}{q}_{31}.}$

But if theq_ij satisfy the Plücker relation

${\displaystyle q_{23}+q_{02}{q}_{31}+q_{03}{q}_{12}=0,}$

thenp₂₃ =q₂₃, completing the set of identities.

Consequently,α is asurjection onto thealgebraic variety consisting of the set of zeros of the quadratic polynomial

${\displaystyle p_{01}{p}_{23}+p_{02}{p}_{31}+p_{03}{p}_{12}.}$

And sinceα is also an injection, the lines in⁠${\displaystyle {\mathbb{P}}^3}$⁠ are thus inbijective correspondence with the points of thisquadric in⁠${\displaystyle {\mathbb{P}}^5}$⁠, called the Plücker quadric orKlein quadric.

Plücker coordinates allow concise solutions to problems of line geometry in 3-dimensional space, especially those involvingincidence.

Two lines in⁠${\displaystyle {\mathbb{P}}^3}$⁠ are eitherskew orcoplanar, and in the latter case they are either coincident or intersect in a unique point. Ifp_ij andp′_ij are the Plücker coordinates of two lines, then they are coplanar precisely when

${\displaystyle \mathbf{d}\cdot {\mathbf{m}}^{\prime }+\mathbf{m}\cdot {\mathbf{d}}^{\prime }=0,}$

as shown by

When the lines are skew, the sign of the result indicates the sense of crossing: positive if a right-handed screw takesL intoL′, else negative.

The quadratic Plücker relation essentially states that a line is coplanar with itself.

In the event that two lines are coplanar but not parallel, their common plane has equation

${\displaystyle 0=(\mathbf{m}\cdot {\mathbf{d}}^{\prime })x_0+(\mathbf{d}\times {\mathbf{d}}^{\prime })\cdot \mathbf{x},}$

where${\displaystyle x=(x_1,x_2,x_3).}$

The slightest perturbation will destroy the existence of a common plane, and near-parallelism of the lines will cause numeric difficulties in finding such a plane even if it does exist.

Dually, two coplanar lines, neither of which contains the origin, have common point

${\displaystyle (x_0:\mathbf{x})=(\mathbf{d}\cdot {\mathbf{m}}^{\prime }:\mathbf{m}\times {\mathbf{m}}^{\prime }).}$

To handle lines not meeting this restriction, see the references.

Given a plane with equation

${\displaystyle 0=a^0{x}_0+a^1{x}_1+a^2{x}_2+a^3{x}_3,}$

or more concisely,

${\displaystyle 0=a^0{x}_0+\mathbf{a}\cdot \mathbf{x};}$

and given a line not in it with Plücker coordinates(d :m), then their point of intersection is

${\displaystyle (x_0:\mathbf{x})=(\mathbf{a}\cdot \mathbf{d}:\mathbf{a}\times \mathbf{m}-a_0\mathbf{d}).}$

The point coordinates,(x₀ :x₁ :x₂ :x₃), can also be expressed in terms of Plücker coordinates as

${\displaystyle x_i=\sum _{j\ne i}{a}^j{p}_{ij},i=0\dots 3.}$

Dually, given a point(y₀ :y) and a line not containing it, their common plane has equation

${\displaystyle 0=(\mathbf{y}\cdot \mathbf{m})x_0+(\mathbf{y}\times \mathbf{d}-y_0\mathbf{m})\cdot \mathbf{x}.}$

The plane coordinates,(a⁰ :a¹ :a² :a³), can also be expressed in terms of dual Plücker coordinates as

${\displaystyle a^i=\sum _{j\ne i}{y}_j{p}^{ij},i=0\dots 3.}$

Because theKlein quadric is in⁠${\displaystyle {\mathbb{P}}^5}$⁠, it contains linear subspaces of dimensions one and two (but no higher). These correspond to one- and two-parameter families of lines in⁠${\displaystyle {\mathbb{P}}^3}$⁠.

For example, supposeL, L′ are distinct lines in⁠${\displaystyle {\mathbb{P}}^3}$⁠ determined by pointsx,y andx′,y′, respectively. Linear combinations of their determining points give linear combinations of their Plücker coordinates, generating a one-parameter family of lines containingL andL′. This corresponds to a one-dimensional linear subspace belonging to the Klein quadric.

If three distinct and non-parallel lines are coplanar; their linear combinations generate a two-parameter family of lines, all the lines in the plane. This corresponds to a two-dimensional linear subspace belonging to the Klein quadric.

If three distinct and non-coplanar lines intersect in a point, their linear combinations generate a two-parameter family of lines, all the lines through the point. This also corresponds to a two-dimensional linear subspace belonging to the Klein quadric.

Aruled surface is a family of lines that is not necessarily linear. It corresponds to a curve on the Klein quadric. For example, ahyperboloid of one sheet is a quadric surface in⁠${\displaystyle {\mathbb{P}}^3}$⁠ ruled by two different families of lines, one line of each passing through each point of the surface; each family corresponds under the Plücker map to aconic section within the Klein quadric in⁠${\displaystyle {\mathbb{P}}^5}$⁠.

During the nineteenth century,line geometry was studied intensively. In terms of the bijection given above, this is a description of the intrinsic geometry of the Klein quadric.

Line geometry is extensively used inray tracing application where the geometry and intersections of rays need to be calculated in 3D. An implementation is described inIntroduction to Plücker Coordinates written for the Ray Tracing forum by Thouis Jones.

• Flat projective plane
• Plücker matrix

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• Projective geometry
• Multilinear algebra
• Geometric algebra
• Coordinate systems

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