vector geometry - nus computing - homecs5240/lecture/vector-geometry.pdf · vector geometry cs5240...

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.

? ?


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.


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.


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.


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



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)



◮ 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



◮ 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)



◮ 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)


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.



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)



Since p1 − p2 is on π, a must be normal to π.Then, the unit normal n of π is

n =a

‖a‖ . (7)


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)


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.


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)


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)


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.



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.



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.



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).


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.



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)



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.



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.


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?



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.



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.



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)



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



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)


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)


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)


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)



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.



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.



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.


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.


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?


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.


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.


vector geometry - nus computing - homecs5240/lecture/vector-geometry.pdf · vector geometry cs5240...

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