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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
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Vector Geometry
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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.
?
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Vector Geometry
?
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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?
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etc.
Need a convenient tool: vector geometry.
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Vector Geometry
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Vector Geometry
Vector Geometry
Vector geometry studies plane geometry using vector algebra.
Can be 2-D, 3-D or m-D.
Basic geometrical elements
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0-D: point
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1-D: line
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2-D: plane
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m-D: hyperplane
A hyperplane is a m-D linear structure in (m + 1)-D vector space.
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In 2-D space, it is a 1-D line.
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In 3-D space, it is a 2-D plane.
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Vector Geometry
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Vector Geometry
Geometry studies many interesting properties:
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Normal of a plane.
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Distance of a point to a plane.
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Projection of a point on a plane.
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Intersection of two planes.
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Distance of a point to a line.
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Projection of a point on a line.
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Intersection of two lines.
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Vector Geometry
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Vector Geometry
Vector Space
Vector Space
A vector space is a structure in which vector addition and scalar
multiplication are defined:
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commutative: u + v = v + u
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associative: (u + v) + w = u + (v + w)
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zero vector: v + 0 = v
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negatives: v + (−v) = 0
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distributive: s(u + v) = su + sv
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associative: r(sv) = (rs)v
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distributive: (r + s)v = rv + sv
◮
scalar-identity: 1v = v
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Vector Geometry
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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)
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Vector Geometry
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Vector Geometry
◮
◮

length or norm: kvk = 
m
X
j=1
Vector Space
1/2
vj2 
1/2
m
X
distance: d(u, v) = ku − vk =  (uj − vj )2 

j=1
m
X
uj v j .
◮
dot product: u · v =
◮
In matrix notation, we denote a vector as a column matrix
u = [u1 · · · um ]⊤ , and vector dot product becomes matrix inner
product u⊤ v.
◮
j=1
v · v = kvk2
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Vector Geometry
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Vector Geometry
◮
Vector Space
u · v = kukkvk cos θ, where θ is the angle between u and v.
u
θ
◮
◮
v
If v is a unit vector, then u · v = kuk cos θ, which is the
component of u along v.
In 3-D vector space, cross product u×v is orthogonal to u and v.
i
j k u×v = u1 u2 u3 v1 v2 v3 (1)
= (u2 v3 − u3 v2 )i + (u3 v1 − u1 v3 )j + (u1 v2 − u2 v1 )k.
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Vector Geometry
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Vector Geometry
◮
◮
◮
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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: kvk ≥ 0
◮
Vector Space
non-degenerate: kvk = 0 iff v = 0
◮
multiplicative: ksvk = |s| kvk
◮
positive: d(u, v) ≥ 0
◮
non-degenerate: d(u, v) = 0 iff u = v
◮
symmetric: d(u, v) = d(v, u)
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Vector Geometry
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Plane Geometry
Normal of Plane
Normal of Plane
A hyperplane π in m-D space is given by the linear equation
a1 x1 + a2 x2 + · · · + am xm + am+1 = 0.
◮
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.
(2)
Denote vectors a = (a1 , . . . , am ), x = (x1 , . . . , xm ).
Then, equation of π can be written as
a · x + am+1 = 0.
◮
a and am+1 represent the plane π.
◮
x is a point on π.
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Vector Geometry
(3)
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Plane Geometry
Normal of Plane
Plane π has a unit normal vector n.
n
p1
p1 − p2
p2
π
Let p1 and p2 be two points on π. Then,
a · p1 + am+1 = 0,
(4)
a · p2 + am+1 = 0.
(5)
a · (p1 − p2 ) = 0.
(6)
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Subtracting Eq. 4 and 5 yields
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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=
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a
.
kak
Vector Geometry
(7)
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Plane Geometry
Distance to Plane
Distance to Plane
Now, consider a point q not on π.
q
n
d
x
π
O
Let x denote the (perpendicular) projection of q on π along n.
Then, the signed (perpendicular, shortest) distance d of q to π is
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d = q · n − x · n.
(8)
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Plane Geometry
Distance to Plane
d =q·n−x·n
=
q·a−x·a
kak
(9)
From Eq. 3, for any point x on π,
a · x = −am+1 .
So,
d=
a · q + am+1
.
kak
◮
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.
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Vector Geometry
(10)
(11)
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Plane Geometry
Projection on Plane
Projection on Plane
q
n
d
x
π
O
The projection x of q on π along n is given by
x = q − d n.
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Vector Geometry
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Plane Geometry
So,
x=q−
Projection on Plane
a · q + am+1
a.
kak2
(13)
For q at the origin, q = 0, and its projection is
x=−
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am+1
a.
kak2
Vector Geometry
(14)
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Plane Geometry
Representations
Representations
A hyperplane can be represented in several ways.
The implicit equation represents a hyperplane as
a1 x1 + a2 x2 + · · · + am xm + 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 p
and unit normal n on the plane, with
n · (x − p) = 0,
(16)
where x is a point on the plane.
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Vector Geometry
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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,
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x is a point on the line.
◮
p is a known point on the line.
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u is the unit vector along the line.
◮
s is a scalar parameter.
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Vector Geometry
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Plane Geometry
Representations
Consider a point p = (x1 , . . . , xm ) in m-D vector space.
b2
p
x2
b1
O
x1
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 = c 1 b1 + · · · + c m bm .
(18)
The vectors bj form a basis.
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Vector Geometry
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Plane Geometry
Representations
In particular, if bj are orthogonal to each other, they form an
orthonormal basis.
b2
p
x2
b1
O
x1
In this case,
p = p · b1 b1 + · · · + p · bm bm .
(19)
The coordinates of p in this basis is (p · b1 , . . . , p · bm ).
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Vector Geometry
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Plane Geometry
Conversion of Representations
Conversion of Representations
Consider a plane π represented by
a1 x1 + · · · + am xm + am+1 = 0.
(20)
As shown previously, π can be represented in point-normal form by
a
◮ its unit normal n =
,
kak
am+1
◮ the projected point p of the origin on π, where p = −
a.
kak2
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Vector Geometry
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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
n1 x1 + · · · + nm xm − n · p = 0.
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Vector Geometry
(22)
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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)
y
1
n
p
x
O
1
In this case, a1 = 1, a2 = 1, a3 = −1.
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Vector Geometry
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Plane Geometry
Conversion of Representations
Line normal n is
(1, 1)
(a1 , a2 )
= √ =
n=
k(a1 , a2 )k
2
1 1
√ ,√
2 2
(24)
The projected point p of the origin on the line is
1
a3
(a1 , a2 ) = (1, 1) =
p=−
2
k(a1 , a2 )k
2
1 1
,
2 2
.
(25)
The signed distance d of the origin to the line is
d=
a · 0 + a3
1
= −√ .
k(a1 , a2 )k
2
(26)
That is, the origin is on the negative side of the line.
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Vector Geometry
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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?
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Vector Geometry
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Plane Geometry
Plane Intersection
Now, let us consider 2-D planes in 3-D space.
n1
n2
n1 x n2
π2
π1
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.
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Plane Geometry
Plane Intersection
So, any point p can be represented as
p = c1 n1 + c2 n2 + c3 n1 ×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 = c1 n1 + c2 n2 + sn1 ×n2
(31)
where s is a scalar parameter.
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Vector Geometry
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Plane Geometry
Plane Intersection
Let’s denote h1 = n1 · p1 and h2 = n2 · p2 .
Substituting Eq. 31 into Eq. 29 gives
c1 + c2 n1 · n2 = h1 ,
c1 n1 · n2 + c2 = h2 .
Solving Eq. 32 yields
(Homework)
c1 =
h1 − h2 n1 · n2
,
1 − (n1 · n2 )2
h2 − h1 n1 · n2
.
c2 =
1 − (n1 · n2 )2
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(32)
Vector Geometry
(33)
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Plane Geometry
Plane Intersection
Example
Consider these two 2-D planes 3-D space:
π1 : x + y − 1 = 0,
z
π2 : x − y = 0.
π1
π2
n1
Want to find the intersection
of the planes.
y
n2
1
O
1
x
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Vector Geometry
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Plane Geometry
Plane Intersection
Convert to point-normal form with the projected point of the origin.
1 1
1 1
p1 =
plane π1 : n1 = √ , √ , 0 ,
, ,0 ,
2 2
2 2
(34)
1
1
plane π2 : n2 = √ , − √ , 0 , p2 = (0, 0, 0).
2
2
So,
1
h1 = n1 · p1 = √ ,
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 =
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1 1
, , 0 + s(0, 0, −1).
2 2
Vector Geometry
(36)
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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)
q
p
x
u
l
O
Consider a point q not on l that projects to x on l.
The distance between p and x is
So,
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q · u − p · u = (q − p) · u.
(38)
x = p + (q − p) · u u.
(39)
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Line Geometry
Distance to Line
Distance to Line
q
p
x
u
l
O
The (perpendicular) distance d from q to its projection x on line l is
d = kq − xk = k(q − p) − (q − p) · u uk.
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Vector Geometry
(40)
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Line Geometry
Line Intersection
Line Intersection
In general, two non-parallel lines l1 and l2 in m-D space with m ≥ 3
do 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
a1 x + a2 y + a3 = 0,
b1 x + b2 y + b3
Then, their intersection is
(41)
(Homework)
x =
y =
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= 0.
a 2 b3 − a 3 b2
,
a 1 b2 − a 2 b1
a 3 b1 − a 1 b3
.
a 1 b2 − a 2 b1
Vector Geometry
(42)
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Line Geometry
Line Intersection
Example
y
l1
l2
1
x
x
O
1
The line equations are:
l1 : x + y − 1 = 0
l2 : x − y − 0 = 0
So,
x=
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0−1
1
= ,
−1 − 1
2
y=
Vector Geometry
(43)
−1 − 0
1
= .
−1 − 1
2
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Line Geometry
Line Intersection
In general, can only solve for a point that is closest to the lines.
l1
d1
q
l2
d2
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 .
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Vector Geometry
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Line Geometry
Line Intersection
Given n lines of the form x = pi + sui , the closest point q to the lines
minimizes the sum squared distance D
D=
n
X
i=1
k(q − pi ) − (q − pi ) · ui ui k2 .
(44)
There is no closed-form solution for this problem.
In practice, have to apply optimization algorithm instead.
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Vector Geometry
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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-direction
form.
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Vector Geometry
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Probing Questions
Probing Questions
◮
A point is on a plane when its distance to the plane is 0. In
programming, we may not get exactly 0 due to floating-point
rounding error. How to handle this problem?
◮
The distance between a point to a plane has a sign. Does the
distance between a point to a line has a sign?
◮
What is the intersection of 3 planes?
How to get the intersection of 3 planes?
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Vector Geometry
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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 unit
normal n on π. Let q denote a point whose projected point on π is
not p. Show that the perpendicular distance d of q to π is given
by d = (q − p) · n.
3. Solve Eq. 32 to obtain Eq. 33 for the intersecting line of two
planes.
4. Show that for the n1 and n2 in Eq. 34, n1 · n2 = 0, and
n1 ×n2 = (0, 0, −1).
5. Solve Eq. 41 to obtain Eq. 42 for the intersection point of two
coplanar lines.
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Vector Geometry
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References
References
1. J. E. Marsden and M. J. Hoffman, Elementary Classical Analysis, 2nd
ed., W. H. Freeman, 1993.
2. B. Noble and J. W. Daniel, Applied Linear Algebra, 3rd ed.,
Prentice-Hall, 1988.
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Vector Geometry
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