Editing Plücker coordinates (section)

== Geometric intuition ==
[[File:Plücker line coordinate geometry.png|thumb|right|Displacement ''d'' (yellow arrow) and moment ''m'' (green arrow) of two points ''x'',''y'' on a line (in red)]]

A line {{mvar|L}} in 3-dimensional [[Euclidean space]] is determined by two distinct points that it contains, or by two distinct planes that contain it (a [[plane-plane intersection]]). Consider the first case, with points <math>x=(x_1,x_2,x_3)</math> and  <math>y=(y_1,y_2,y_3).</math> The [[vector displacement]] from {{mvar|x}} to {{mvar|y}} is nonzero because the points are distinct, and represents the ''[[direction (geometry)|direction]]'' of the line. That is, every displacement between points on the line {{mvar|L}} is a [[scalar multiplication|scalar multiple]] of {{math|1=''d'' = ''y'' − ''x''}}. If a physical particle of unit mass were to move from {{mvar|x}} to {{mvar|y}}, it would have a [[moment (physics)|moment]] 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 line {{mvar|L}} 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 is {{math|1='''m''' = '''x''' × '''y'''}}, where "×" denotes the vector [[cross product]]. For a fixed line, {{mvar|L}}, the area of the triangle is proportional to the length of the segment between {{mvar|x}} and {{mvar|y}}, 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, so {{math|1='''d''' ⋅ '''m''' = 0}}, where "⋅" denotes the vector [[dot product]].

Although neither direction {{mvar|d}} nor moment {{mvar|m}} alone is sufficient to determine the line {{mvar|L}}, together the pair does so uniquely, up to a common (nonzero) scalar multiple which depends on the distance between {{mvar|x}} and {{mvar|y}}. That is, the coordinates

: <math>(\mathbf d : \mathbf m ) = (d_1:d_2:d_3\ :\ m_1:m_2:m_3)</math>

may be considered [[homogeneous coordinates]] for {{mvar|L}}, in the sense that all pairs {{math|(λ'''d''' : λ'''m''')}}, for {{math|λ ≠ 0}}, can be produced by points on {{mvar|L}} and only {{mvar|L}}, and any such pair determines a unique line so long as {{math|'''d'''}} is not zero and {{math|1='''d''' ⋅ '''m''' = 0}}. Furthermore, this approach extends to include [[point at infinity|points]], [[line at infinity|lines]], and a [[plane at infinity|plane "at infinity"]], in the sense of [[projective geometry]]. In addition a point <math>x</math> lies on the line {{mvar|L}} if and only if <math>x \times d = m</math>.

: '''Example.''' Let {{math|1='''x''' = (2, 3, 7)}} and {{math|1='''y''' = (2, 1, 0)}}. Then {{math|1=('''d''' : '''m''') = (0 : −2 : −7 : −7 : 14 : −4)}}.

Alternatively, let the equations for points {{math|'''x'''}} of two distinct planes containing {{mvar|L}} be

: <math>\begin{align}
0 &= a + \mathbf a \cdot \mathbf x, \\ 
0 &= b + \mathbf b \cdot \mathbf x.
\end{align}</math>

Then their respective planes are perpendicular to vectors {{math|'''a'''}} and {{math|'''b'''}}, and the direction of {{mvar|L}} must be perpendicular to both. Hence we may set {{math|1='''d''' = '''a''' × '''b'''}}, which is nonzero because {{math|'''a''', '''b'''}} are neither zero nor parallel (the planes being distinct and intersecting). If point {{math|'''x'''}} satisfies both plane equations, then it also satisfies the linear combination

:<math>\begin{align}
0 &= a (b + \mathbf b \cdot \mathbf x) - b(a+ \mathbf a \cdot \mathbf x) \\
  &= (a \mathbf b - b \mathbf a) \cdot \mathbf x
\end{align}</math>

That is,

:<math>\mathbf m = a \mathbf b - b \mathbf a</math>

is a vector perpendicular to displacements to points on {{mvar|L}} from the origin; it is, in fact, a moment consistent with the {{math|'''d'''}} previously defined from {{math|'''a'''}} and {{math|'''b'''}}.

{{Collapse top|title=Proof of geometric formulation}}

''Proof 1'': Need to show that 
:<math>\mathbf m = a \mathbf b - b \mathbf a = \mathbf r \times \mathbf d = \mathbf r \times (\mathbf a \times \mathbf b).</math><sup>''what is "'''r'''"?''</sup>

[[Without loss of generality]], let

:<math>\mathbf a \cdot \mathbf a = \mathbf b \cdot \mathbf b = 1.</math>

[[File:BaiduShurufa 2018-9-29 20-38-43.png|thumb|Plane orthogonal to line {{mvar|L}} and including the origin.]]
Point {{mvar|B}} is the origin. Line {{mvar|L}} passes through point {{mvar|D}} and is orthogonal to the plane of the picture. The two planes pass through {{mvar|CD}} and {{mvar|DE}} and are both orthogonal to the plane of the picture. Points {{mvar|C}} and {{mvar|E}} are the closest points on those planes to the origin {{mvar|B}}, therefore angles {{math|∠ ''BCD''}} and {{math| ∠ ''BED''}} are right angles and so the points {{mvar|B, C, D, E}} lie on a circle (due to a corollary of [[Thales's theorem]]). {{mvar|BD}} is the diameter of that circle.

: <math>\begin{align}
& \mathbf a := \frac{BE}{||BE||}, \quad  \mathbf b := \frac{BC}{||BC||}, \quad \mathbf r := BD; \\[4pt]
& - \! a = ||BE|| = ||BF||, \quad -b =  ||BC|| = ||BG||; \\[4pt]
& \mathbf m = a \mathbf b - b \mathbf a = FG \\[4pt]
& || \mathbf d || = || \mathbf a \times \mathbf b || = \sin\angle FBG
\end{align}</math>

Angle {{math|∠ ''BHF''}} is a right angle due to the following argument. Let {{math|1=ε := ∠ ''BEC''}}. Since {{math|△ ''BEC'' ≅ △ ''BFG''}} (by side-angle-side congruence), then {{math|1= ∠ ''BFG'' = ε}}. Since {{math|1=∠ ''BEC'' + ∠ ''CED'' = 90°}}, let {{math|1=ε' := 90° − ε = ∠ ''CED''}}. By the [[inscribed angle theorem]], {{math|1=∠ ''DEC'' = ∠ ''DBC''}}, so {{math|1= ∠ ''DBC'' = ε'}}. {{math|1=∠ ''HBF'' + ∠ ''BFH'' + ∠ ''FHB'' = 180°}}; {{math|1=ε' + ε + ∠ ''FHB'' = 180°}}, {{math|1=ε + ε' = 90°}}; therefore, {{math|1= ∠ ''FHB'' = 90°}}. Then {{math|∠ ''DHF''}} must be a right angle as well.

Angles {{math|∠ ''DCF'', ∠ ''DHF''}} are right angles, so the four points {{mvar|C, D, H, F}} lie on a circle, and (by the [[intersecting secants theorem]])

:<math>||BF|| \, ||BC|| = ||BH|| \, ||BD||</math>
that is, 
:<math>\begin{align}
&ab \sin\angle FBG = ||BH|| \, || \mathbf r || \sin\angle FBG , \\[4pt]
& 2 \, \text{Area}_{\triangle BFG} = ab \sin\angle FBG = ||BH|| \, ||FG|| = ||BH|| \, || \mathbf r || \sin\angle FBG, \\[4pt]
& || \mathbf m || = ||FG|| = || \mathbf r || \sin\angle FBG = || \mathbf r || \, || \mathbf d ||, \\[4pt]
& \mathbf m = \mathbf r \times \mathbf d. \blacksquare
\end{align}</math>

''Proof 2'':

Let

:<math>\mathbf a \cdot \mathbf a = \mathbf b \cdot \mathbf b = 1.</math>

This implies that

: <math>a = -||BE||, \quad b = -||BC||.</math>

According to the [[vector triple product]] formula,

: <math>\mathbf r \times (\mathbf a \times \mathbf b) = (\mathbf r \cdot \mathbf b) \mathbf a - (\mathbf r \cdot \mathbf a) \mathbf b.</math>

Then

:<math>\begin{align}
\mathbf r \times (\mathbf a \times \mathbf b) 
 &= \mathbf a \, || \mathbf r || \, || \mathbf b || \cos\angle DBC - \mathbf b \, ||\mathbf r || \, || \mathbf a || \cos\angle DBE \\[4pt]
 &= \mathbf a \, || \mathbf r || \cos\angle DBC - \mathbf b \, || \mathbf r || \cos\angle DBE \\[4pt]
 &= \mathbf a \, || BC || - \mathbf b \, || BE ||  \\[4pt]
 &= -b \mathbf a - (-a) \mathbf b \\[4pt]
 &= a \mathbf b - b \mathbf a\ \  \blacksquare
\end{align}</math>

When <math>|| \mathbf r || = 0,</math> the line {{mvar|L}} passes the origin with direction {{math|'''d'''}}. If <math>|| \mathbf r || > 0,</math>  the line has direction {{math|'''d'''}}; the plane that includes the origin and the line {{mvar|L}} has normal vector {{math|'''m'''}}; the line is tangent to a circle on that plane (normal to {{math|'''m'''}} and perpendicular to the plane of the picture) centered at the origin and with radius <math>|| \mathbf r ||.</math>

: '''Example.''' Let {{math|1=''a''<sub>0</sub> = 2}}, {{math|1='''a''' = (−1, 0, 0)}} and {{math|1=''b''<sub>0</sub> = −7}}, {{math|1='''b''' = (0, 7, −2)}}. Then {{math|1=('''d''' : '''m''') = (0 : −2 : −7 : −7 : 14 : −4)}}.

Although the usual algebraic definition tends to obscure the relationship, {{math|('''d''' : '''m''')}} are the Plücker coordinates of {{mvar|L}}.

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