pl ü cker coordinate of a line in 3-space
DESCRIPTION
Pl ü cker Coordinate of a Line in 3-Space. L. Q. P. O. 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 !. References. - PowerPoint PPT PresentationTRANSCRIPT
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Plücker Coordinate of a Line in 3-Space
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Lines in 3-Space
• You know how to represent a 3-space line in parametric equations
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Q
P
L
O
RttQPt ,1
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Motivation
• We are interested in learning the aspects/features of Plucker coordinates that make life easier!
3Fall 2015
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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• 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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For a similar reason, a line in 2-space has two DOF’s
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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, 0): q0, q0=0: line through origin
•(0, 0): 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 (不同 ) IFF their Plucker coordinates are linearly independent(獨
立 )
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.16}
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)Fall 2015
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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 differentFall 2015
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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 Origin
T: closest to origin (to be determined) 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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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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Q (0,0,1)
z
x
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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:e] E.P + e = 0 … (1) [F:f] F.P + f = 0 …. (2)
(1)f – (2)e = 0: f(E.P+e) – e(F.P+f) = 0 (fE – eF).P = 0fE-eF defines the normal of an origin plane through L
direction U = EF L = {EF: fE – eF}Fall 2015
O
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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!Fall 2015
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• [V:0] origin plane thru L
(V0)
• [UV:V.V] plane thru L [V:0]
24
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.24R
O[V:0]
[UV:V.V]
L
Find intersection of [UV:V.V] & [V:0]
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{U:V} and plane [N:0]Points on {VN:0} = (VN:w)Intersection: the point on [UV:V.V]!
{U:V} [N:0] = (VN:U.N)
(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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O[V:0]
[N:0]
L
[UV:V.V]
{VN:0} (p.24)
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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}
Intersect with Z=2, [0:0:1:-2]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), overlapIntersect 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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{U:V} and (P:w) [UP – wV:V.P]
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Common Plane1
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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L={U
:V}
P
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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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L={U
:V}
N
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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)Fall 2015
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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 the 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)
36
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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y
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Two Lines Can Be …
• Identical– Linearly dependent Plucker coordinates
• Coplanar: find the 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
38
L1&U2: [U1U2:V1.U2]
L2 &U1: [U2U1:V2.U1]
Same plane!
L1
L2parallel lines
(U1U2=0) are automatically 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
40
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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x
y
y
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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?Fall 2015
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Application
• Ray-polygon and ray-convex volume intersection
Fall 2015
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Relative Position Between 2 Lines
Looking from tail of L1 …
Here, the lines are “oriented”!!{orientation defined by U}
Fall 2015
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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
Fall 2015
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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 > 0Fall 2015
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Discussion
• Plucker coordinate of transformed line– More efficient by computing the Plucker
coordinates of the transformed points (p.31)
Fall 2015
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Extra
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Vector Review
Fall 2015 52
kww
vvj
ww
vvi
ww
vv
www
vvv
kji
wv
21
21
31
31
32
32
321
321
sinwvwv
20cos
cos
vvvvv
wvwv
Vector triple product
Triple product
kwv :dependencylinear
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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 Fall 2015
(because Q is different!)
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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
54
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
Fall 2015