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Vector Geometry
CS5240 Theoretical Foundations in Multimedia
Leow Wee Kheng
Department of Computer Science
School of Computing
National University of Singapore
Leow Wee Kheng (NUS) Vector Geometry 1 / 41
Motivation
Motivation
You are an IT assistant to a surgeon operating on a patient.She wants to know the difference between the left and right side.
? ?
Leow Wee Kheng (NUS) Vector Geometry 2 / 41
Motivation
Need to calculate
◮ symmetric plane
◮ distance of point to plane
◮ average difference between left and right sides
She also wants to know
◮ Is symmetric plane perpendicular to horizontal plane?
◮ Are certain features on a straight line?
◮ etc.
Need a convenient tool: vector geometry.
Leow Wee Kheng (NUS) Vector Geometry 3 / 41
Vector Geometry
Vector Geometry
Vector geometry studies plane geometry using vector algebra.Can be 2-D, 3-D or m-D.
Basic geometrical elements
◮ 0-D: point
◮ 1-D: line
◮ 2-D: plane
◮ m-D: hyperplane
A hyperplane is a m-D linear structure in (m+ 1)-D vector space.
◮ In 2-D space, it is a 1-D line.
◮ In 3-D space, it is a 2-D plane.
Leow Wee Kheng (NUS) Vector Geometry 4 / 41
Vector Geometry
Geometry studies many interesting properties:
◮ Normal of a plane.
◮ Distance of a point to a plane.
◮ Projection of a point on a plane.
◮ Intersection of two planes.
◮ Distance of a point to a line.
◮ Projection of a point on a line.
◮ Intersection of two lines.
Leow Wee Kheng (NUS) Vector Geometry 5 / 41
Vector Geometry Vector Space
Vector Space
A vector space is a structure in which vector addition and scalarmultiplication are defined:
◮ commutative: u+ v = v + u
◮ associative: (u+ v) +w = u+ (v +w)
◮ zero vector: v + 0 = v
◮ negatives: v + (−v) = 0
◮ distributive: s(u+ v) = su+ sv
◮ associative: r(sv) = (rs)v
◮ distributive: (r + s)v = rv + sv
◮ scalar-identity: 1v = v
Leow Wee Kheng (NUS) Vector Geometry 6 / 41
Vector Geometry Vector Space
A Hierarchy of Spaces
a set of vectors
vector addition,scalar multiplication
y
vector space
with norm
y
normed space
with limit
y
Banach space
with inner product
y
Hilbert space(Euclidean space)
Leow Wee Kheng (NUS) Vector Geometry 7 / 41
Vector Geometry Vector Space
◮ length or norm: ‖v‖ =
m∑
j=1
v2j
1/2
◮ distance: d(u,v) = ‖u− v‖ =
m∑
j=1
(uj − vj)2
1/2
◮ dot product: u · v =m∑
j=1
ujvj .
◮ In matrix notation, we denote a vector as a column matrixu = [u1 · · · um]⊤, and vector dot product becomes matrix innerproduct u⊤v.
◮ v · v = ‖v‖2
Leow Wee Kheng (NUS) Vector Geometry 8 / 41
Vector Geometry Vector Space
◮ u · v = ‖u‖‖v‖ cos θ, where θ is the angle between u and v.
vθ
u
◮ If v is a unit vector, then u · v = ‖u‖ cos θ, which is thecomponent of u along v.
◮ In 3-D vector space, cross product u×v is orthogonal to u and v.
u×v =
∣
∣
∣
∣
∣
∣
i j k
u1 u2 u3v1 v2 v3
∣
∣
∣
∣
∣
∣
= (u2v3 − u3v2)i+ (u3v1 − u1v3)j+ (u1v2 − u2v1)k.
(1)
Leow Wee Kheng (NUS) Vector Geometry 9 / 41
Vector Geometry Vector Space
◮ positive: v · v ≥ 0
◮ non-degenerate: v · v = 0 iff v = 0
◮ distributive: u · (v +w) = u · v + u ·w◮ multiplicative: su · v = s(u · v)◮ symmetric: u · v = v · u
◮ positive: ‖v‖ ≥ 0
◮ non-degenerate: ‖v‖ = 0 iff v = 0
◮ multiplicative: ‖sv‖ = |s| ‖v‖
◮ positive: d(u,v) ≥ 0
◮ non-degenerate: d(u,v) = 0 iff u = v
◮ symmetric: d(u,v) = d(v,u)
Leow Wee Kheng (NUS) Vector Geometry 10 / 41
Plane Geometry Normal of Plane
Normal of Plane
A hyperplane π in m-D space is given by the linear equation
a1x1 + a2x2 + · · ·+ amxm + am+1 = 0. (2)
◮ Eq. 2 may be scaled by 1/am+1 to remove am+1.
◮ We keep am+1 in Eq. 2 to retain the scaling factor.
Denote vectors a = (a1, . . . , am), x = (x1, . . . , xm).Then, equation of π can be written as
a · x+ am+1 = 0. (3)
◮ a and am+1 represent the plane π.
◮ x is a point on π.
Leow Wee Kheng (NUS) Vector Geometry 11 / 41
Plane Geometry Normal of Plane
Plane π has a unit normal vector n.
p2
p1
p2p1
π
n
−
Let p1 and p2 be two points on π. Then,
a · p1 + am+1 = 0, (4)
a · p2 + am+1 = 0. (5)
Subtracting Eq. 4 and 5 yields
a · (p1 − p2) = 0. (6)
Leow Wee Kheng (NUS) Vector Geometry 12 / 41
Plane Geometry Normal of Plane
Since p1 − p2 is on π, a must be normal to π.Then, the unit normal n of π is
n =a
‖a‖ . (7)
Leow Wee Kheng (NUS) Vector Geometry 13 / 41
Plane Geometry Distance to Plane
Distance to Plane
Now, consider a point q not on π.
π
O
d
x
nq
Let x denote the (perpendicular) projection of q on π along n.Then, the signed (perpendicular, shortest) distance d of q to π is
d = q · n− x · n. (8)
Leow Wee Kheng (NUS) Vector Geometry 14 / 41
Plane Geometry Distance to Plane
d = q · n− x · n
=q · a− x · a
‖a‖(9)
From Eq. 3, for any point x on π,
a · x = −am+1. (10)
So,
d =a · q+ am+1
‖a‖ . (11)
◮ For a point p on π, d = 0.
◮ For a point q on the +n side of π, d > 0.
◮ For a point q on the −n side of π, d < 0.
Leow Wee Kheng (NUS) Vector Geometry 15 / 41
Plane Geometry Projection on Plane
Projection on Plane
π
O
d
x
nq
The projection x of q on π along n is given by
x = q− dn. (12)
Leow Wee Kheng (NUS) Vector Geometry 16 / 41
Plane Geometry Projection on Plane
So,
x = q− a · q+ am+1
‖a‖2 a. (13)
For q at the origin, q = 0, and its projection is
x = −am+1
‖a‖2 a. (14)
Leow Wee Kheng (NUS) Vector Geometry 17 / 41
Plane Geometry Representations
Representations
A hyperplane can be represented in several ways.
The implicit equation represents a hyperplane as
a1x1 + a2x2 + · · ·+ amxm + am+1 = 0, (15)
where aj are the plane’s coefficients and x is a point on the plane.
In contrast, an explicit equation has the form xj = · · · .
The point-normal form represents a hyperplane by a known point pand unit normal n on the plane, with
n · (x− p) = 0, (16)
where x is a point on the plane.
Leow Wee Kheng (NUS) Vector Geometry 18 / 41
Plane Geometry Representations
A line in 2-D can be represented in the previous forms.
A line in m-D, m ≥ 3, can only be represented as a parametric equation
x = p+ su, (17)
◮ x is a point on the line.
◮ p is a known point on the line.
◮ u is the unit vector along the line.
◮ s is a scalar parameter.
Leow Wee Kheng (NUS) Vector Geometry 19 / 41
Plane Geometry Representations
Consider a point p = (x1, . . . , xm) in m-D vector space.
x2
x1
b1
b2
p
O
Let bj denote m mutually non-parallel unit vectors in the same space.
Then, p can be represented as a linear combination of bj :
p = c1b1 + · · ·+ cmbm. (18)
The vectors bj form a basis.
Leow Wee Kheng (NUS) Vector Geometry 20 / 41
Plane Geometry Representations
In particular, if bj are orthogonal to each other, they form anorthonormal basis.
x2
x1
b1
b2
p
O
In this case,p = p · b1b1 + · · ·+ p · bmbm. (19)
The coordinates of p in this basis is (p · b1, . . . ,p · bm).
Leow Wee Kheng (NUS) Vector Geometry 21 / 41
Plane Geometry Conversion of Representations
Conversion of Representations
Consider a plane π represented by
a1x1 + · · ·+ amxm + am+1 = 0. (20)
As shown previously, π can be represented in point-normal form by
◮ its unit normal n =a
‖a‖ ,
◮ the projected point p of the origin on π, where p = −am+1
‖a‖2 a.
Leow Wee Kheng (NUS) Vector Geometry 22 / 41
Plane Geometry Conversion of Representations
Now, consider a plane π represented by a point p and unit normal n.
From Eq. 6,n · (x− p) = 0. (21)
So, we obtain the implicit equation
n1x1 + · · ·+ nmxm − n · p = 0. (22)
Leow Wee Kheng (NUS) Vector Geometry 23 / 41
Plane Geometry Conversion of Representations
Example
Consider a 1-D hyperplane, i.e., a line, in a 2-D space represented by
x+ y − 1 = 0. (23)
x
y
O 1
1 n
p
In this case, a1 = 1, a2 = 1, a3 = −1.
Leow Wee Kheng (NUS) Vector Geometry 24 / 41
Plane Geometry Conversion of Representations
Line normal n is
n =(a1, a2)
‖(a1, a2)‖=
(1, 1)√2
=
(
1√2,1√2
)
(24)
The projected point p of the origin on the line is
p = − a3‖(a1, a2)‖2
(a1, a2) =1
2(1, 1) =
(
1
2,1
2
)
. (25)
The signed distance d of the origin to the line is
d =a · 0+ a3‖(a1, a2)‖
= − 1√2. (26)
That is, the origin is on the negative side of the line.
Leow Wee Kheng (NUS) Vector Geometry 25 / 41
Plane Geometry Plane Intersection
Plane Intersection
Consider two non-parallel (infinite) hyperplanes πa and πb
π1 : a · x+ am+1 = 0,
π2 : b · x+ bm+1 = 0.(27)
The intersection of π1 and π2 is a hyperplane of one fewer dimension.Any point x on the intersection satisfies both plane equations.
Caution!
Do not subtract the two equations to get
(a− b) · x+ am+1 − bm+1 = 0. (28)
Why?
Leow Wee Kheng (NUS) Vector Geometry 26 / 41
Plane Geometry Plane Intersection
Now, let us consider 2-D planes in 3-D space.
π2
π1
n1 n2
n1 n2x
Denote the planes’ point-normal forms as
π1 : n1 · x− n1 · p1 = 0,
π2 : n2 · x− n2 · p2 = 0.(29)
In 3-D space, n1×n2 is well-defined.n1, n2, and n1×n2 form a basis.
Leow Wee Kheng (NUS) Vector Geometry 27 / 41
Plane Geometry Plane Intersection
So, any point p can be represented as
p = c1n1 + c2n2 + c3n1×n2 (30)
for some appropriate c1, c2, c3.
In 3-D space, intersection of 2 planes is a line.
n1×n2 is parallel to the line of intersection.So, the line can be represented by the parametric equation
x = c1n1 + c2n2 + sn1×n2 (31)
where s is a scalar parameter.
Leow Wee Kheng (NUS) Vector Geometry 28 / 41
Plane Geometry Plane Intersection
Let’s denote h1 = n1 · p1 and h2 = n2 · p2.
Substituting Eq. 31 into Eq. 29 gives
c1 + c2n1 · n2 = h1,
c1n1 · n2 + c2 = h2.(32)
Solving Eq. 32 yields (Homework)
c1 =h1 − h2n1 · n2
1− (n1 · n2)2,
c2 =h2 − h1n1 · n2
1− (n1 · n2)2.
(33)
Leow Wee Kheng (NUS) Vector Geometry 29 / 41
Plane Geometry Plane Intersection
Example
Consider these two 2-D planes 3-D space:
π1 : x+ y − 1 = 0,
π2 : x− y = 0.
Want to find the intersectionof the planes.
π2
1π
1n
n2
O
x
y
1
1
z
Leow Wee Kheng (NUS) Vector Geometry 30 / 41
Plane Geometry Plane Intersection
Convert to point-normal form with the projected point of the origin.
plane π1 : n1 =
(
1√2,1√2, 0
)
, p1 =
(
1
2,1
2, 0
)
,
plane π2 : n2 =
(
1√2,− 1√
2, 0
)
, p2 = (0, 0, 0).
(34)
So,
h1 = n1 · p1 =1√2, h2 = n2 · p2 = 0. (35)
Note: n1 · n2 = 0, n1×n2 = (0, 0,−1). (Homework)
Then, the intersection line is given by
x = c1 · n1 + c2 · n2 + sn1×n2 =
(
1
2,1
2, 0
)
+ s(0, 0,−1). (36)
Leow Wee Kheng (NUS) Vector Geometry 31 / 41
Line Geometry Projection on Line
Projection on Line
A line l in m-D space with m ≥ 3 is given by the parametric equation
x = p+ su. (37)
O
p
q
ux l
Consider a point q not on l that projects to x on l.The distance between p and x is
q · u− p · u = (q− p) · u. (38)
So,x = p+ (q− p) · uu. (39)
Leow Wee Kheng (NUS) Vector Geometry 32 / 41
Line Geometry Distance to Line
Distance to Line
O
p
q
ux l
The (perpendicular) distance d from q to its projection x on line l is
d = ‖q− x‖ = ‖(q− p)− (q− p) · uu‖. (40)
Leow Wee Kheng (NUS) Vector Geometry 33 / 41
Line Geometry Line Intersection
Line Intersection
In general, two non-parallel lines l1 and l2 in m-D space with m ≥ 3do not intersect.
They can intersect only if they are coplanar, i.e., lie on a 2-D plane.
In this case, suppose they are given by the implicit equations
a1x+ a2y + a3 = 0,
b1x+ b2y + b3 = 0.(41)
Then, their intersection is (Homework)
x =a2b3 − a3b2a1b2 − a2b1
,
y =a3b1 − a1b3a1b2 − a2b1
.
(42)
Leow Wee Kheng (NUS) Vector Geometry 34 / 41
Line Geometry Line Intersection
Example
l2l1
x
y
O 1
1
x
The line equations are:
l1 : x+ y − 1 = 0
l2 : x− y − 0 = 0(43)
So,
x =0− 1
−1− 1=
1
2, y =
−1− 0
−1− 1=
1
2.
Leow Wee Kheng (NUS) Vector Geometry 35 / 41
Line Geometry Line Intersection
In general, can only solve for a point that is closest to the lines.
l2
l1
d2
d1
q
Point q projects perpendicularly to x1 on line l1 and x2 on line l2.Point q is mid-way between x1 and x2.
Point q minimizes its distance to l1 and l2.
Leow Wee Kheng (NUS) Vector Geometry 36 / 41
Line Geometry Line Intersection
Given n lines of the form x = pi + sui, the closest point q to the linesminimizes the sum squared distance D
D =n∑
i=1
‖(q− pi)− (q− pi) · ui ui‖2. (44)
There is no closed-form solution for this problem.
In practice, have to apply optimization algorithm instead.
Leow Wee Kheng (NUS) Vector Geometry 37 / 41
Summary
Summary
◮ Vector geometry studies plane geometry using vector algebra.
◮ Studies geometrical properties such as normal, distance,projection, intersection.
◮ Two representations for planes:implicit equation vs. point-normal form.
◮ Lines in m-D space with m ≥ 3 are represented by point-directionform.
Leow Wee Kheng (NUS) Vector Geometry 38 / 41
Probing Questions
Probing Questions
◮ A point is on a plane when its distance to the plane is 0. Inprogramming, we may not get exactly 0 due to floating-pointrounding error. How to handle this problem?
◮ The distance between a point to a plane has a sign. Does thedistance between a point to a line has a sign?
◮ What is the intersection of 3 planes?How to get the intersection of 3 planes?
Leow Wee Kheng (NUS) Vector Geometry 39 / 41
Homework
Homework
1. What are the key concepts that you have learned?
2. Consider a plane π represented by a known point p and a unitnormal n on π. Let q denote a point whose projected point on π isnot p. Show that the perpendicular distance d of q to π is givenby d = (q− p) · n.
3. Solve Eq. 32 to obtain Eq. 33 for the intersecting line of twoplanes.
4. Show that for the n1 and n2 in Eq. 34, n1 · n2 = 0, andn1×n2 = (0, 0,−1).
5. Solve Eq. 41 to obtain Eq. 42 for the intersection point of twocoplanar lines.
Leow Wee Kheng (NUS) Vector Geometry 40 / 41
References
References
1. J. E. Marsden and M. J. Hoffman, Elementary Classical Analysis, 2nded., W. H. Freeman, 1993.
2. B. Noble and J. W. Daniel, Applied Linear Algebra, 3rd ed.,Prentice-Hall, 1988.
Leow Wee Kheng (NUS) Vector Geometry 41 / 41