Plücker Coordinate of a Line in
3-Space
Spring 2013
1
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!
Spring 2013
2
References
• Plucker coordinate tutorial, K. Shoemake
[rtnews]
• Plucker coordinates for the rest of us, L.
Brits [flipcode]
• Plucker line coordinate, J. Erickson [cgafaq]
Spring 2013
3
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.
Spring 2013
4
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]
Spring 2013
5
The following are
from Shoemake’s note…
Spring 2013
6
Summary 1/3
Spring 2013
7
Summary 2/3
Spring 2013
8
Summary 3/3
Spring 2013
9
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
Spring 2013
10
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
1
1
1
1
1
1
pz
qz
P–Q
pz
px
qz
qx
px
py
qx
qy
PQ
Spring 2013
11
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
Spring 2013
12
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}
Spring 2013
13
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)
Spring 2013
14
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
Spring 2013
15
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
Spring 2013
16
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)
Spring 2013
U=(2,-2,0)
Q=(0,1,0)
17
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)
Spring 2013
18
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
Spring 2013
19
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}
Spring 2013
20
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]
Spring 2013
21
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!
Spring 2013
22
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]
Spring 2013
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}
Spring 2013
24
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)
Spring 2013
25
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)
Spring 2013
26
{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
Spring 2013
27
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
Spring 2013
28
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]
Spring 2013
29
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)
Springthrough
2013
30
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
Spring 2013
31
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
Spring 2013
32
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)
Spring 2013
33
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
Spring 2013
34
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
Spring 2013
35
Two Lines Can Be …
• Identical
– Linearly dependent Plucker coordinate
• Coplanar: find common plane
– Intersecting: find intersection
– Parallel: find distance
• Skewed: find distance, closest points
Spring 2013
36
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
Spring 2013
L2 &U1: [U2U1:V2.U1]
37
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
Spring 2013
38
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)
Spring 2013
39
L1 & L2 Skewed
• Not coplanar IFF skewed
• Find distance
• Find pair of closet points
Spring 2013
40
Distance Computation in
Point
(x2:y2:z2:1)
Point
(x1:y1:z1:1)
Line {U:V}
Line {U:V}
x1  x2    y1  y2   z1  z2 
2
2
(1)
2
(2a): parallel
(2b): skewed
3
R
Plane [D:d]
ax1  by1  cz 1  d
a2  b2  c2
If no
intersection,
generate a point
on line & pointplane distance
d1  d 2
Plane [D:d]
a2  b2  c2
Spring 2013
41
(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]
Spring 2013
42
(2a) Parallel Line Distance
D U
[D:d]
D
L1
L2
P1
P2
Find
Find
Find
Find
Spring 2013
the common plane [D:d]
P1 containing L1 and D
P2 containing L2 and D
distance between P1 & P2
43
(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?
Spring 2013
44
Application
• Ray-polygon and ray-convex volume
intersection
Spring 2013
45
Here, the lines are “oriented”!!
{orientation defined by U}
Relative Position Between 2 Lines
Looking from tail of L1 …
Spring 2013
46
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
Spring 2013
47
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
Spring 2013
48
Discussion
• Plucker coordinate of transformed line
– More efficient by computing the Plucker
coordinates of the transformed points (p.30)
Spring 2013
49
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
Spring 2013
50
Triple product
Vector triple product
Spring 2013
51
[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}
Spring 2013
52