1 plücker coordinate of a line in 3-space spring 2013
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Plücker Coordinate of a Line in 3-Space
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Motivation
• The other way of representing lines in 3-space is parametric equation
• We are interested in learning the aspects/features of Plucker coordinates that make life easier!
2
Q
P
L
O
RttQPt ,1
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References
• Plucker coordinate tutorial, K. Shoemake [rtnews]
• Plucker coordinates for the rest of us, L. Brits [flipcode]
• Plucker line coordinate, J. Erickson [cgafaq]
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Introduction
• A line in 3-space has four degree-of-freedom (why so?!)
• Plucker coordinates are concise and efficient for numerous chores
• One special case of Grassmann coordinates– Uniformly manage points, lines, planes and flats in
spaces of any dimension.
– Can generate, intersect, … with simple equations.
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Mason’s Version
tqptx )(
qpq 0
Line in parametric form
Define
Plucker coordinate of the line (q, q0)
Six coordinate 4 DOFs:
]by [scale ,,
0
00
0
kqqqkqq
p q
q0
•(q, q0): q0, q00: general line
•(q, q0): q0, q0=0: line through origin
•(q, q0): q=0, (q0=0): [not allowed]
O
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The following are from Shoemake’s note…
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Summary 1/3
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Summary 2/3
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Summary 3/3
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Notations
• Upper case letter: a 3-vector U = (ux,uy,uz) • Vector U; homogeneous version (U:0)• Point P; homo version (P:1), (P:w)• Cross and dot product: PQ, U.V• Plane equation: ax+by+cz+dw=0
– [a:b:c:d] or [D:d] with D=(a,b,c)– [D:0] origin plane: plane containing origin
• Plucker coordinate: {U:V}• Colon “:” proclaims homogeneity
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Determinant DefinitionQ
P
L
O
1z
y
x
p
p
p
1z
y
x
q
q
q … row x
… row y
… row z
… row w
Make all possible determinants of pairs of rows
11xx qp
11yy qp
11zz qp
yy
xx
qp
qp
zz
yy
qp
qp
xx
zz
qp
qp
P–Q PQ
L={P-Q:PQ}L={P-Q:PQ}
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ExampleQ
P
L
O
P=(2,3,7), Q=(2,1,0). L = {U:V} = {0:2:7:-7:14:-4}.
Order does not matter
Q=(2,3,7), P=(2,1,0). L = {U:V} = {0:-2:-7:7:-14:4}
Identical lines: two lines are distinct IFF their Plucker coordinates are linearly independent
identical
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Tangent-Normal Definition
PQ: {U:V} U = P–Q V = P×Q = (U+Q)×Q = U×Q
(U:0) direction of line[V:0] origin plane through L
Question: any pair of points P,Q gives the same {U:V}? Yes
{p.14}
L={U:UQ}L={U:UQ}
PQ
U
L
O
U×Q
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Example
x
y
z y
U=(1,0,-1)Q=(0,0,1)UQ = (0,-1,0)L={1:0:-1:0:-1:0}
If we reverse the tangent:U=(-1,0,1)Q=(0,0,1)UQ = (0,1,0)L={-1:0:1:0:1:0}… still get the same line(but different orientation)Spring 2013
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Remark
P’=Q+kU
U’ = P’– Q = kUV’ = P’×Q = (Q+kU) ×Q = kU×Q
P’Q {kU:kV}
Q
P
P’
U
L
O
Moving P and/or Q scales U & V together!Similar to homogeneous coordinates
Moving P and/or Q scales U & V together!Similar to homogeneous coordinates
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Remarks• Six numbers in Plucker coordinate {U:V} are not
independent.– Line in R3 has 4 dof. : six variables, two equations: one from
homogeneity; one from U.V = 0
• Geometric interpretation {U:V}– U: line tangent (U0, by definition)– V: the normal of origin plane containing L (V=0 L
through origin)
• Identical lines: two lines are distinct IFF their Plucker coordinates are linearly independent
Ex: {0:-2:-7:7:-14:4} and {0:4:14:-14:28:-8} are the same (but different orientation); {2:1:0:0:0:0} is differentSpring 2013
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Exercise
x
y
zP
Q P=(1,0,0)Q=(0,1,0)
P=(0,1,0)Q=(1,0,0)
L={P-Q:PQ}L={U:UQ}
L={P-Q:PQ}L={U:UQ}
U=(1,-1,0)Q=(0,1,0)
U=(2,-2,0)Q=(0,1,0)
U=(-1,1,0)Q=(0,1,0)
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Distance to OriginT: closest to origin Any Q on L: Q = T + sU
V = U×Q = U×(T+sU) =U×T||V|| = ||U|| ||T|| sin90 = ||U|| ||T||
T.T = (V.V) / (U.U)
V×U=(U×T)×U = (U.U)T
T=(V×U:U.U)
Q
T
U
L
O Squared distance:
Closest point:
Vector triple product
L={U:V}
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Example
x
y
z y
U=(1,0,-1)Q=(0,0,1)UQ = (0,-1,0)L={1:0:-1:0:-1:0}
T=(V×U:U.U) = (1:0:1:2)= (1/2,0,1/2)Squared distance = (V.V)/(U.U) = 1/2
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Line as Intersection of Two Planes1
L
[E:e] [F:f]
P
Plane equation: ax + by + cz + d = 0 P = (x,y,z), point on L E.P + e = 0 F.P + f = 0 f(E.P+e) – e(F.P+f) = 0 (fE – eF).P = 0fE-eF defines the normal of an origin plane through Ldirection U = EF
L = {EF: fE – eF}
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Example
x
z
y
z = 0[0:0:1:0]
x = 1[1:0:0:-1]
E = [1:0:0:-1]F = [0:0:1:0]L = {EF:fE-eF} = {0:-1:0:0:0:1}
CheckP = (1,1,0), Q = (1,0,0)L = {P-Q:PQ} = {0:1:0:0:0:-1}
Q
P
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Line as Intersection of Two Planes2
• If both planes do not pass through origin, e0 and f0, we can normalize both planes to [E:1] and [F:1].
• The intersecting line then becomes {EF:E-F}
L
[E:1] [F:1]Q
P
L
O
{P-Q:PQ}
{EF:E-F}
Duality!Spring 2013
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• [V:0] origin plane thru L (V0)
• [UV:V.V] plane thru L [V:0]
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Other Duality
• (U:0) direction of L• T=(VU:U.U) point
of L (U:0)
Q
T
U
L
O
O[V:0]
[UV:V.V]
Verify!(next page)P.18
L={U:V}
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Verify P.23R
O[V:0]
[UV:V.V]
L
L={E F:fE-eF}= {(U V) V: -(V.V)V}
(U V) V = -U(V.V)+V(U.V) = -U(V.V)
L = {-U(V.V):-V(V.V)} = {U:V}
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Line-Plane Intersection1
• L and plane [N:0] Points on {VN:0} = (VN:w)Intersection: the point on [UV:V.V]!
O
[V:0]
[N:0]
L{U:V} [N:0] = (VN:U.N)
[UV:V.V]
{VN:0}
(U V).(V N)+w(V.V) = 0w(V.V) = (V U).(V N) = N.((V U) V)=N.(-(U.V)V+(V.V)U)=(V.V)(U.N)w = U.N
Triple product
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Line-Plane Intersection2
• L and plane [N:n]
O
[V:0]
[N:0]
L
{U:V} [N:n] = (VN–nU:U.N)
O
[V:0]
[N:0]
L
Derivation pending
[N:n]
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Example
x
y
z y
U=(1,0,-1)Q=(0,0,1)UQ = (0,-1,0)L={1:0:-1:0:-1:0}
VN–nU = (-1,0,0) – (-2)(1,0,-1) = (1,0,-2)U.N = (1,0,-1).(-1,0,0) = -1Intersection at (-1,0,2)!
Z = 2[0:0:1:-2]
Intersect with y = 0, [0:1:0:0](VN:U.N) = (0:0:0:0), overlap
Intersect with y = 1, [0:1:0:-1](VN–nU:U.N) = (1:0:-1:0)Intersect at infinity
{U:V} [N:n] = (VN–nU:U.N)
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Common Plane1
{U:V} and (P:w) [UP-wV:V.P]
U = (0,0,1)V = UQ = (0,0,1) (0,1,0) = (-1,0,0)
(P:w) = (1:1:0:1)
[UP-wV:V.P] = [0:1:0:-1]
Derivation pending
x
y
z
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Common Plane2
{U:V} and (N:0) [UN:V.N]
U = (0,0,1)V = UQ = (-1,0,0)
N = (1,0,0) …(-1,0,0) get the same … (1,0,1) also get the same (N need not U)⊥
[UN:V.N] = [0:1:0:-1]
Derivation pending
x
y
z
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Generate Points on Line1
Useful for:
• Computing transformed Plucker coordinate (p.50)
• Line-in-plane testO
[V:0]
[N:0]
LUse {U:V} [N:0] = (VN:U.N)
U
NAny N will do, as long as U.N0{Take non-zero component of U}N
O
Also related: p. 18, 35L
Does not work for line with V=0(line through origin)Spring 2013
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Example
As before: L = {U:V} = {0:0:1:-1:0:0}
Take N = (0,1,1)
{U:V} [N:0] = (VN:U.N) = (0:1:-1:1)x
y
z
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Line in Plane Test
Point-on-Plane Test [N:n] contain (P:w) IFF N.P+nw = 0
Is L in [1:1:0:0]? No
(1,1,0).(0,1,-1) + 0 0
Is L in [1:0:0:0]? Yes
(1,0,0).(0,1,-1) + 0 = 0
x
y
z
1. Generate two points on the line 2. Do point-on-plane test
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Point-on-Line Test
x
y
z
N
N1
N2
N,N1,N2: three base vectorsChoose N according to nonzero component of UN1 and N2 are the other two axes
Check point-in-plane with [UN1:V.N1] and [UN2:V.N2](common plane, p.29)
U
1. Generate two independent planes
containing the line.
2. Perform point-on-plane tests twice
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Example
x
y
z
N
N1
N2
U
L = {0:0:1:-1:0:0}, P = (0:1:-2:1)N = (0,0,1), N1 = (0,1,0), N2 = (1,0,0)Plane1 [-1:0:0:0] (-1,0,0).(0,1,-2)+0 = 0Plane2 [0:1:0:-1] (0,1,0).(0,1,-2) - 1 = 0
P
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• Parametric equation of L• Weighted sum of (U:0)
and T=(VU:U.U)
Pnt(t) = (VU+tU:U.U)
35
Duality
• Parametric form of planes through L– Generate two planes as
page 33…
L = {0:0:1:-1:0:0}Pnt(t) = (0:-1:t:1)
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Two Lines Can Be …
• Identical– Linearly dependent Plucker coordinate
• Coplanar: find common plane– Intersecting: find intersection– Parallel: find distance
• Skewed: find distance, closest points
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Coplanarity Test(intersect)
• Two lines L1 {U1:V1}, L2 {U2:V2} are coplanar if
U1.V2+V1.U2 = 0
37
L1&U2: [U1U2:V1.U2]
L2 &U1: [U2U1:V2.U1]
Same plane!
L1
L2parallel lines
(U1U2=0) are (by definition) coplanar
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L1 & L2 Coplanar
• Intersecting point (non-parallel)– Find the common plane: [U1U2:V1.U2]
– ((V1.N)U2-(V2.N)U1-(V1.U2)N:(U1U2).N)
– Where N is unit basis vector, independent of U1 and U2, (U1U2).N ≠0)
• Parallel (distinct) lines (U1U2 = 0)Common plane:
– [(U1.N)V2-(U2.N)V1:(V1V2).N] with N.U10
L1: {U1:V1}L2: {U2:V2}
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Example
39
L1={1:1:0:0:0:1}L2={0:1:0:0:0:-1}Pick N = (0,0,1)((V1.N)U2-(V2.N)U1-(V1.U2)N:(U1U2).N)=(1:2:0:1)
L1={1:1:0:0:0:1}L2={2:2:0:0:0:-4}
Pick N = (1,0,0)[(U1.N)V2-(U2.N)V1:(V1V2).N]
=[0:0:-6:0]
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L1 & L2 Skewed
• Not coplanar IFF skewed
• Find distance
• Find pair of closet points
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Distance Computation in R3
221
221
221 zzyyxx 222
111
cba
dczbyax
222
21
cba
dd
Point (x2:y2:z2:1)
Line {U:V} Plane [D:d]
Point (x1:y1:z1:1)
(1)
Line {U:V} (2a): parallel
(2b): skewed
If no intersection,
generate a point on line & point-plane distance
Plane [D:d]
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(1) Line-Point Distance
L
p
D
1. Generate 1 containing L & p as [D:d]2. Generate 2 containing L & D3. Compute distance from p to 2
1=[D:d]
2
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(2a) Parallel Line Distance
[D:d]
D U
1 2
L1 L2Find the common plane [D:d] Find 1 containing L1 and DFind 2 containing L2 and DFind distance between 1 & 2
D
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(2b) Skewed Line Distance
1
2
L1
L2
U1
U2
Generate 1 containing L1 and U2
Generate 2 containing L2 and U1
Find distance between 1 & 2
How to find the pair of points that are
closest?Spring 2013
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Application
• Ray-polygon and ray-convex volume intersection
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Relative Position Between 2 Lines
Looking from tail of L1 …
Here, the lines are “oriented”!!{orientation defined by U}
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Example
x
y
zL1
L2L3
R
L1={1:0:0:0:0:0}L2={-1:1:0:0:0:-1}
L3={0:-1:0:0:0:0}
R={0:0:-1:1/3:-1/3:0} = {0:0:-3:1:-1:0}
R vs. L1: (0:0:-3).(0:0:0) + (1:-1:0).(1:0:0) = 1 > 0
R vs. L2: (0:0:-3).(0:0:-1) + (1:-1:0).(-1:1:0) = 1 > 0
P:(1/3,1/3,0)Q:(1/3,1/3,1)
R vs. L3: (0:0:-3).(0:0:0) + (1:-1:0).(0:-1:0) = 1 > 0
Note here the line is “oriented”; L and –L are not the same
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Example
x
y
zL1
L2L3
R
L1={1:0:0:0:0:0}L2={-1:1:0:0:0:-1}
L3={0:-1:0:0:0:0}
R={0:0:-1:1:-1:0}
R vs. L1: (0:0:-1).(0:0:0) + (1:-1:0).(1:0:0) = 1 > 0
R vs. L2: (0:0:-1).(0:0:-1) + (1:-1:0).(-1:1:0) = -1 < 0
P:(1,1,0)Q:(1,1,1)
R vs. L3: (0:0:-1).(0:0:0) + (1:-1:0).(0:-1:0) = 1 > 0Spring 2013
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Discussion
• Plucker coordinate of transformed line– More efficient by computing the Plucker
coordinates of the transformed points (p.30)
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IndexConstructors,
two points 11
tangent-normal 13
two planes 20
Distance (closest pt) to origin 18
Line-plane intersect25,26
Line-line intersect 38
Common plane
line, point 28
line, dir 29
Generate points on line 30, 35L
Parametric equation of line 35L
Parametric plane of line 35R
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Line in plane30,32
Point in line 33
Point on plane 32
Line-line configuration37-40
Parallel (distance, common plane) 38
Intersect (point, common plane) 38
Skew 44
Distance (point-line-plane)41-44
Winding 45
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Vector triple product
Triple product
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[Example]
x
y
z
U=(1,0,-1)V=(0,-1,0)L={1:0:-1:0:-1:0}
Different normal gives different lineL’ = {1:0:-1: 0:-2:0}
Reverse normal gives different lineU=(1,0,-1)V=(0,1,0)L’={1:0:-1:0:1:0}
x
y
z Spring 2013