Transcript Plucker Coordinate

Plücker Coordinate of a Line in
3-Space
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Motivation
1  t P  tQ, t  R
Q
P
O
• The other way of representing
L lines in 3-space is parametric
equation
• We are interested in learning
the aspects/features of Plucker
coordinates that make life
easier!
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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
Line in parametric form
x(t )  p  tq
Define q0  p  q
Plucker coordinate of the line (q, q0)
Six coordinate 4 DOFs:
q0  q  0
q, q0   k q, q0  [scale q by k ]
q0
O
p
q
•(q, q0): q0, q00: general line
•(q, q0): q0, q0=0: line through origin
•(q, q0): q=0, (q0=0): [not allowed]
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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: PQ, 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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L={P-Q:PQ}
Determinant
Definition
L
Q
 px 
p 
 y
 pz 
 
1
P
O
qx 
q 
 y
qz 
 
1
… row x
… row y
… row z
… row w
Make all possible determinants of pairs of rows
px
qx
py
qy
pz
qz
py
qy
pz
qz
px
qx
1
1
1
1
1
1
pz
qz
px
qx
py
qy
P–Q
PQ
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L
Q
Example
P=(2,3,7), Q=(2,1,0).
L = {U:V} = {0:2:7:-7:14:-4}.
P
Order does not matter
O
identical
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
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L={U:UQ}
Tangent-Normal Definition
U×Q
O
U
P
Q
L
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}
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Example
y
x
z
U=(1,0,-1)
Q=(0,0,1)
UQ = (0,-1,0)
L={1:0:-1:0:-1:0}
y
If we reverse the tangent:
U=(-1,0,1)
Q=(0,0,1)
UQ = (0,1,0)
L={-1:0:1:0:1:0}
… still get the same line
(but different orientation)
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Remark
L
Q
U
P
P’
P’=Q+kU
U’ = P’– Q = kU
V’ = P’×Q = (Q+kU) ×Q = kU×Q
P’Q {kU:kV}
O
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 (U0, 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 different
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Exercise
y
P=(1,0,0)
Q=(0,1,0)
Q
L={P-Q:PQ}
L={U:UQ}
P=(0,1,0)
Q=(1,0,0)
x
P
z
U=(-1,1,0)
Q=(0,1,0)
U=(1,-1,0)
Q=(0,1,0)
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U=(2,-2,0)
Q=(0,1,0)
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Vector triple product
Distance
to
Origin
L
Q
U
T
T: closest to origin
Any Q on L: Q = T + sU
V = U×Q = U×(T+sU) =U×T
||V|| = ||U|| ||T|| sin90 = ||U|| ||T||
O
L={U:V}
Squared distance: T.T = (V.V) / (U.U)
V×U=(U×T)×U = (U.U)T
Closest point:
T=(V×U:U.U)
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Example
y
U=(1,0,-1)
Q=(0,0,1)
UQ = (0,-1,0)
L={1:0:-1:0:-1:0}
x
z
y
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
[E:e]
[F:f]
P
L
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 = 0
fE-eF defines the normal of an origin plane
through L
direction U = EF
L = {EF: fE – eF}
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Example
y
E = [1:0:0:-1]
F = [0:0:1:0]
L = {EF:fE-eF} = {0:-1:0:0:0:1}
P
x
Q
z
Check
P = (1,1,0), Q = (1,0,0)
L = {P-Q:PQ} = {0:1:0:0:0:-1}
z=0
[0:0:1:0]
x=1
[1:0:0:-1]
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Line as Intersection of Two Planes2
• If both planes do not pass through origin, e0 and
f0, we can normalize both planes to [E:1] and
[F:1].
• The intersecting line then becomes {EF:E-F}
Q
L
{P-Q:PQ}
{EF:E-F}
P
O
[E:1] [F:1]
L
Duality!
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Other Duality
L={U:V}
• (U:0) direction of L
• [V:0] origin plane thru
L (V0)
• T=(VU:U.U) point of
L  (U:0)
• [UV:V.V] plane thru
L  [V:0]
L
Q
U
[UV:V.V]
T
O
P.18
O
[V:0]
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Verify!
(next
page)23
Verify P.23R
L={E  F:fE-eF}
= {(U  V)  V: -(V.V)V}
[UV:V.V]
O
L
(U  V)  V
= -U(V.V)+V(U.V) = -U(V.V)
[V:0]
L = {-U(V.V):-V(V.V)} = {U:V}
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Triple product
Line-Plane Intersection1
• L and plane [N:0]
{VN:0}
(U  V).(V  N)+w(V.V) = 0
w(V.V) = (V  U).(V  N)
= N.((V  U)  V)
=N.(-(U.V)V+(V.V)U)=(V.V)(U.N)
w = U.N
[N:0]
[UV:V.V]
L
Points on {VN:0} = (VN:w)
Intersection: the point on [UV:V.V]!
O
[V:0]
{U:V}  [N:0] = (VN:U.N)
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Line-Plane Intersection2
[N:n]
• L and plane [N:n]
[N:0]
[N:0]
L
O
[V:0]
L
Derivation
pending
O
[V:0]
{U:V}  [N:n] = (VN–nU:U.N)
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{U:V}  [N:n] = (VN–nU:U.N)
U=(1,0,-1)
Q=(0,0,1)
UQ = (0,-1,0)
L={1:0:-1:0:-1:0}
Example
y
VN–nU = (-1,0,0) – (-2)(1,0,-1) = (1,0,-2)
U.N = (1,0,-1).(-1,0,0) = -1
Intersection at (-1,0,2)!
x
z
y
Z=2
[0:0:1:-2]
Intersect with y = 0, [0:1:0:0]
(VN:U.N) = (0:0:0:0),
overlap
Intersect with y = 1, [0:1:0:-1]
(VN–nU:U.N) = (1:0:-1:0)
Intersect at infinity
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Common Plane1
Derivation
pending
{U:V} and (P:w)  [UP-wV:V.P]
y
U = (0,0,1)
V = UQ = (0,0,1) (0,1,0) = (-1,0,0)
(P:w) = (1:1:0:1)
x
[UP-wV:V.P] = [0:1:0:-1]
z
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Common Plane2
{U:V} and (N:0)  [UN:V.N]
y
U = (0,0,1)
V = UQ = (-1,0,0)
N = (1,0,0)
x
z
Derivation
pending
…(-1,0,0) get the same
… (1,0,1) also get the same
(N need not ⊥U)
[UN:V.N] = [0:1:0:-1]
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Also related: p. 18, 35L
Generate Points on Line1
Useful for:
• Computing transformed
Plucker coordinate (p.50)
• Line-in-plane test
[N:0]
L
O
Use {U:V}  [N:0] = (VN:U.N)
[V:0]
N
N
U
O
Any N will do, as long as U.N0
{Take non-zero component of U}
Does not work for line with V=0
(line
origin)
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Example
y
As before: L = {U:V} = {0:0:1:-1:0:0}
Take N = (0,1,1)
{U:V}  [N:0] = (VN:U.N) = (0:1:-1:1)
x
z
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Line in Plane Test
y
1. Generate two points on the line
2. Do point-on-plane test
Is L in [1:1:0:0]? No
x
z
Point-on-Plane Test
[N:n] contain (P:w)
IFF N.P+nw = 0
(1,1,0).(0,1,-1) + 0  0
Is L in [1:0:0:0]? Yes
(1,0,0).(0,1,-1) + 0 = 0
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Point-on-Line Test
1. Generate two independent planes
containing the line.
2. Perform point-on-plane tests twice
N2 y
N1
U
N
z
x
N,N1,N2: three base vectors
Choose N according to nonzero component of U
N1 and N2 are the other two axes
Check point-in-plane with
[UN1:V.N1] and [UN2:V.N2]
(common plane, p.29)
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Example
N2 y
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 = 0
Plane2 [0:1:0:-1] (0,1,0).(0,1,-2) - 1 = 0
P
N1
U
N
x
z
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Duality
• Parametric equation of L
• Weighted sum of (U:0)
and T=(VU:U.U)
Pnt(t) = (VU+tU:U.U)
• Parametric form of
planes through L
– Generate two planes as
page 33…
y
L = {0:0:1:-1:0:0}
Pnt(t) = (0:-1:t:1)
x
z
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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)
L1
• Two lines L1 {U1:V1},
L2 {U2:V2} are coplanar
if
U1.V2+V1.U2 = 0
L1&U2: [U1U2:V1.U2]
Same plane!
parallel lines
(U1U2=0) are (by
definition) coplanar
L2
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L2 &U1: [U2U1:V2.U1]
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L1: {U1:V1}
L2: {U2:V2}
L1 & L2 Coplanar
• Intersecting point (non-parallel)
– Find the common plane: [U1U2:V1.U2]
– ((V1.N)U2-(V2.N)U1-(V1.U2)N:(U1U2).N)
– Where N is unit basis vector, independent of U1 and U2,
(U1U2).N ≠0)
• Parallel (distinct) lines (U1U2 = 0)
Common plane:
– [(U1.N)V2-(U2.N)V1:(V1V2).N] with N.U10
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Example
L1={1:1:0:0:0:1}
L2={2:2:0:0:0:-4}
L1={1:1:0:0:0:1}
Pick N = (1,0,0)
L2={0:1:0:0:0:-1}
[(U1.N)V2-(U2.N)V1:(V1V2).N]
Pick N = (0,0,1)
=[0:0:-6:0]
((V1.N)U2-(V2.N)U1-(V1.U2)N:(U1U2).N)
=(1:2:0:1)
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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
Point
(x2:y2:z2:1)
Point
(x1:y1:z1:1)
Line {U:V}
Line {U:V}
x1  x2 2  y1  y2 2  z1  z2 2
(1)
(2a): parallel
(2b): skewed
3
R
Plane [D:d]
ax1  by1  cz1  d
a2  b2  c2
If no
intersection,
generate a point
on line & pointplane distance
d1  d 2
Plane [D:d]
a2  b2  c2
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(1) Line-Point Distance
L
D
p
P2
1. Generate P1 containing L & p as [D:d]
2. Generate P2 containing L & D
3. Compute distance from p to P2
P1=[D:d]
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(2a) Parallel Line Distance
D U
[D:d]
D
L1
L2
P1
P2
Find
Find
Find
Find
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the common plane [D:d]
P1 containing L1 and D
P2 containing L2 and D
distance between P1 & P2
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(2b) Skewed Line Distance
L2
P2
U2
U1
P1
L1
Generate P1 containing L1 and U2
Generate P2 containing L2 and U1
Find distance between P1 & P2
How to find the pair
of points that are
closest?
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Application
• Ray-polygon and ray-convex volume
intersection
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Here, the lines are “oriented”!!
{orientation defined by U}
Relative Position Between 2 Lines
Looking from tail of L1 …
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Note here the line is “oriented”; L and –L are
not the same
Example
y
R
L3
L2
x
L1
z
L1={1:0:0:0:0:0}
L2={-1:1:0:0:0:-1}
L3={0:-1:0:0:0:0}
P:(1/3,1/3,0)
Q:(1/3,1/3,1)
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
R vs. L3: (0:0:-3).(0:0:0) + (1:-1:0).(0:-1:0) = 1 > 0
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Example
R
y
L3
L2
x
L1
z
L1={1:0:0:0:0:0}
L2={-1:1:0:0:0:-1}
L3={0:-1:0:0:0:0}
P:(1,1,0)
Q:(1,1,1)
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
R vs. L3: (0:0:-1).(0:0:0) + (1:-1:0).(0:-1:0) = 1 > 0
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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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Index
Constructors,
two points
tangent-normal
two planes
Distance (closest pt) to origin
Line-plane intersect
Line-line intersect
Common plane
line, point
line, dir
Generate points on line
Parametric equation of line
Parametric plane of line
11
13
20
18
25,26
38
28
29
30, 35L
35L
Line in plane
Point in line
Point on plane
30,32
33
32
Line-line configuration
Parallel (distance, common plane)
Intersect (point, common plane)
Skew
Distance (point-line-plane)
Winding
37-40
38
38
44
41-44
45
35R
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Triple product
Vector triple product
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[Example]
y
U=(1,0,-1)
V=(0,-1,0)
L={1:0:-1:0:-1:0}
x
z
Different normal gives different line
L’ = {1:0:-1: 0:-2:0}
y
x
z
Reverse normal gives different line
U=(1,0,-1)
V=(0,1,0)
L’={1:0:-1:0:1:0}
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