# v - dmami ```Vectors
MM1 module 4
Poul Nielsen
Vectors module 4 objectives
• After this module you should have a clear
understanding of:
• Coordinate systems;
• The properties of vectors;
• Vector length, scalar multiplication, addition, and
subtraction;
• The dot product and its properties;
• Resolving vector components;
• The cross product and its properties;
• How to find equations of lines;
• How to find equations of planes.
Slide number 1
Coordinate systems
• A coordinate system is a system for
specifying points in n–dimensional space
using coordinates.
• A coordinate is a set of n variables which fix a
geometric object.
• The simplest coordinate system consists of
n coordinate axes oriented perpendicularly to
each other, known as Cartesian coordinates
(Ren&eacute; Descartes 1637).
• In this module we will focus our attention on 2–
dimensional and 3–dimensional Cartesian
coordinate systems.
Slide number 2
Coordinate systems in 3D
• Right handed
coordinate
systems are
usually chosen by
convention.
z
Right
x
Slide number 3
y
• Left handed
coordinate
systems are
occasionally
encountered.
Coordinate planes in 3D
• The 3–dimensional Cartesian coordinate
system defines 3 natural coordinate planes by
taking coordinate axes in pairs.
• Illustrated below are the xy–, yz–, and zx–
coordinate planes, respectively.
z
y
x
Slide number 4
z
z
y
x
y
x
Points in 3D
• A point in a 3–dimensional coordinate system
can be characterised by a set of 3 real
numbers, representing the distances along the
x–, y–, and z–coordinate axes.
• e.g. the point p at x = 4, y = –6, and z = 5.
Note: the order of movement is not important
–6 z
z
5
p
4
4
y p
y p
5
–6
x
Slide number 5
–6
x
z
5
4
x
y
Coordinate systems in 2D
• In 2 dimensions right handed coordinate
systems are usually chosen by convention
(constructed like a 3–dimensional coordinate
system with the z–coordinate axis pointing
towards the viewer).
• A point in a 2–dimensional coordinate system
can be characterised by a set of 2 real
numbers, representing the
y
distances along the
x– and y–coordinate axes.
x
• e.g. the point p at
x = 4, y = –6
p
Slide number 6
Vector
• A vector is essentially a directed arrow.
• A vector has the properties of length and
direction, but not position.
• Two vectors are equal if they have the same
length and the same direction.
• A vector from point p to point q has its tail at p
q
v
p
tail
• A vector with its tail at the coordinate origin is
sometimes called a position vector.
Slide number 7
What are vectors used for?
• Computer graphics;
• Characterisation of force and/or flow fields in
analysis of solid and fluid systems;
• Representation of state variables to describe
complex systems.
Slide number 8
Vector component notation
• Vectors can be described in component
notation as n  1 matrices where the
elements represent the components of the
vector along each coordinate axis.
• The number of components is its dimension.
• e.g. the 2–dimensional vector, v, illustrated
below has components 4 and 3 units along the
x– and y–axes, respectively.
y
It can be represented in
3 v
component notation as
x
4
v  3
4
 
Slide number 9
Specifying vectors and components
• Vectors are generally denoted by lowercase,
non-italicised, bold characters
e.g. u, v, w will be commonly used.
• Components of vectors are usually denoted by
lowercase, italicised, non-bold characters with
an index as a subscript
e.g. u3 refers to the 3rd component of vector u.
• More generally, an n–dimensional vector, v,
can be written as
v1 
v2 
v   

vn 

Slide number 10
Length of a vector
• The length of a vector v, denoted |v|, can be
readily calculated from its components, vi
v1 
v2 
v   
if

vn 

then
v  v12  v22 

 
• e.g. for v  43
 

v  42  32  5

Slide number 11
vn2
y
3
v
4
x
Zero vector
• A zero vector, denoted 0, is a vector of length
zero.
• All the components of a zero vector are zero.
• The zero vector is the additive identity
i.e. v + 0 = v
• Listed below are zero vectors for 1–, 2–, 3–,
and n–dimensions
 
0  0 0  00
 
0
0
0  0 0  0
 
0
0
Slide number 12
Unit vector
• A unit vector is a vector of length 1,
sometimes called a direction vector.
• An arbitrary vector, v, can be converted to a
unit vector, vˆ, by dividing each of its
components by its length
v1 v 
v v2 v 
vˆ   

v 

vn v 

4 5 
• e.g.   is a vector of unit length with the
3 5 
same direction as the vector
4

3

Slide number 13
Base vectors…
• A coordinate system provides a natural set of
base vectors of unit length with direction in
each of the coordinate axes.
• In a 2–dimensional coordinate system with
coordinate axes x and y, the base vectors are
denoted xˆ and yˆ , respectively.
• In a 3–dimensional coordinate system with
coordinate axes
y
z
x, y, and

 z, the
yˆ ˆ
zˆ
y
x x
base vectors
are denoted
xˆ
yˆ
xˆ , yˆ , and zˆ
x

 
Slide number 14


…Base vectors

• Any vector can be expressed in terms of its
coordinate base vectors
4 ˆ
ˆ

4
x

3
y


3
4 
6 4 xˆ  6 yˆ  5 zˆ
5 
• For an n–dimensional coordinate system with
coordinates x1, x2, … xn, and base vectors
xˆ 1, xˆ 2 , xˆ n
v1 
v2 
v    v1xˆ 1  v2 xˆ 2  vn xˆ n

v 

n
Slide number 15
Vector scalar multiplication
• Multiplying an n–dimensional vector by a
scalar results in an n–dimensional vector with
each of its elements multiplied by the scalar.
• e.g.
4  8 
26 12 
5  10
v1  cv1 
v2  cv2 
2v
cv  c   
v
–0.5v

vn 
 
cvn 

• Note: the length scales by c but the direction
remains the same (or is reversed if c &lt; 0).

Slide number 16
Vector negation
• Multiplying an n–dimensional vector by –1
results in a vector of the same length but
reversed direction.
v
Slide number 17
–v
or more vectors together into a vector sum.
• For vectors u and v, the vector sum u + v is
obtained by placing them head to tail and
drawing the vector from the free tail to the free
• Note that u + v = v + u
• Vectors being added must have the same
dimension
v
v
u
u
u+v
v+u
u+v
u
Slide number 18
v
u
v
• In Cartesian coordinates, vector addition can
be performed simply by adding the
corresponding components of the vectors
u1  v1  u1  v1 
u2  v2  u2  v2 
u  v      


un 
 
vn 
 
un  vn 

• e.g.

4  4 8 
 3 1

4
 
   
4 xˆ  4 yˆ  4 xˆ  3yˆ   8xˆ  yˆ 

Slide number 19
Vector subtraction…
• A vector difference is the result of subtracting
one vector from another.
• A vector difference is denoted using the
normal minus sign, i.e., the vector difference of
vectors u and v is written u – v.
• A vector difference is equivalent to a vector
sum with the orientation of the second vector
reversed, i.e. u – v = v + (–u)
u
v
u–v
–v
Slide number 20
…Vector subtraction
• In Cartesian coordinates, vector subtraction
can be performed simply by subtracting the
corresponding components of the vectors
u1  v1  u1  v1 
u2  v2  u2  v2 
u  v      


un 
 
vn 
 
un  vn 

• e.g.


Slide number 21
4  4 0 
 3 7


4
     
4 xˆ  4 yˆ   4 xˆ  3yˆ   0 xˆ  7 yˆ 
Vector between two points
• If p and q are two points with position vectors
p and q, respectively, then the vector joining p
to q is q – p.
• The finite line segment starting at p and
terminating at q is denoted pq
• Remember as final – initial.
y

x
q
q
q–p
Slide number 22
p
p
–p
q–p
q
Dot product…
• The dot product can be defined for two n–
dimensional vectors u and v by
u•v = |u| |v| cos()
where  is the angle between the vectors and
|u| and |v| are the lengths of u and v,
respectively.
• Note that the dot product of two vectors results
in a scalar.
Slide number 23
…Dot product…

• In Cartesian coordinates, the dot product can
be performed by summing the products of the
corresponding components of the vectors
u1  v1 
u2  v2 
u  v      u1v1  u2 v2  un vn

un 
 
vn 

• e.g.
4  4
 3 4  4 4 3 4

4
 
 
4  3 
6 2  4  3 6 2 5  1  5
5  1
Slide number 24
…Dot product
• The dot product, u•v, has the geometric
interpretation as the length of v times the
length of the projection of u onto v when the
two vectors are placed so that their tails
coincide.
projection of
u on v

u
Slide number 25
v
Dot product properties
• The dot product is commutative, u•v = v•u.
This follows from the definitions:
u•v = |u| |v| cos() = |v| |u| cos() = v•u
u•v = u1v1 + u2v2 + …unvn = v•u
• The dot product is distributive,
u•(v + w) = u•v + u•w
• The associative property, (u•v)•w = v•(u•w), is
meaningless as (u•v)•w is not defined since
(u•v) is a scalar and cannot be dot multiplied.
• From u•u = u12 + u22 + …un2 and
u  u12 u22  un2
it follows that |u|2 = u•u
Slide number 26
Perpendicular vectors…
• Two vectors are perpendicular if the angle
between them is a right angle
• Since u•v = |u| |v| cos() and cos(/2) = 0, it
follows that two non–zero vectors u and v are
perpendicular if, and only if, u•v = 0.
• e.g.
2 
1
u  1
,
v


 

2

2  1
u  v  1
 2 2 1  1  2  0

   
So u and v are perpendicular.

Slide number 27
Dot product of base vectors


• Recall that in a Cartesian coordinate system
the base vectors are mutually perpendicular
and of unit length.
• Thus in 2–dimensions:
xˆ  xˆ 1 xˆ  yˆ  0
yˆ  xˆ  0 yˆ  yˆ 1
in 3–dimensions:
xˆ  xˆ  1 xˆ  yˆ  0 xˆ  zˆ  0
yˆ  xˆ  0 yˆ  yˆ  1 yˆ  zˆ  0
zˆ  xˆ  0 zˆ  yˆ  0 zˆ  zˆ  1
in n–dimensions:
xˆ 1  xˆ 1  1 xˆ 1  xˆ 2  0
xˆ 1  xˆ n  0
xˆ 2  xˆ 1  0 xˆ 2  xˆ 2  1
xˆ 2  xˆ n  0
xˆ n  xˆ 1  0 xˆ n  xˆ 2  0
Slide number 28
xˆ n  xˆ n  1
Finding angles between vectors…
• Since u•v = |u| |v| cos()
and u•v = u1v1 + u2v2 + …unvn
the dot product can be used to find the angle,
, between vectors u and v
u  v u1v1  u2 v2  un vn
cos  

uv
uv
4 
4
,
v

• e.g. u  4

 

3

u  4 2  4  32, v  4 2  32  5

2
Slide number 29
4  4
u  v  4 3 4  4  4 3  4
   
uv
4
cos  

 0.14142    81.87
uv
32  5
…Finding angles between vectors
• e.g.
4 
3 
u  6, v  2 
5 
1
u  4 2  6  5 2  75
2
v  32  2 2  1  14
2
4  3 
u  v  6 2  4  3  6  2 5  1  5
5  1
uv
5
cos  

 0.15430    98.88
uv
75 14
Slide number 30
Projection of vector on vector…
• Given a vector u, find its projection p in the
direction of vector v.
p  u cos 
 u vˆ cos , since vˆ  1
 u  vˆ
 v v u  v 
p  p vˆ  u  vˆ vˆ  u   
v
2
v
 v v
u  v 
2
v
p  
v, since v  v  v
p
v  v 

u

Slide number 31
…Projection of vector on vector
• e.g. find the projection of u on v, where
 3 
1 
u  2, v  1 
4 
1
u  v  31  2 1 4  1  3
v  v  11  11  1  1  3
u  v  3 1  1
p  
v   1  1
v  v   3 1 1 

Slide number 32
Resolving vector components…
• The projection formula can be used to resolve
a vector, u, parallel, u||, and perpendicular, u,
to a direction, v.
v
u  v 
u 
u||  
v
v  v 
u

u   u  u||
u
• e.g. resolve u into vectors
parallel and perpendicular to v, where
0
1 
u

,
v




4

1

u  v  4 1  2
u||  
 2 
v  

1


 
v  v 
2
  2 2
u  u  u||  0
 2  2

4
     
Slide number 33
…Resolving vector components
• e.g. resolve the vector, u, parallel, u||, and
perpendicular, u, to the vector, v, where
2 
1 
u  6, v  1
2 
1
u  v  6 1  2 
u||  
v  1 2
v  v  3 1 2
2  2  0 
u  u  u||  6 2 4
2  2 4 

Slide number 34
Cross product…
• For 3–dimensional vectors u and v, the cross
product is defined as
uv = |u| |v| sin() nˆ
where  is the angle between the vectors, |u|
and |v| are the lengths of u and v, and nˆ
is the unit vector
normal to the plane

containing u and v.
• The cross product uses the 
uv
right hand rule to determine
the direction of the result.
Right
v
i.e. u, v, and uv form a

right handed system.
u
Slide number 35
…Cross product
• Ensure that you understand how to orient uv
relative to u and v using the right hand rule.
uv
u
v
u
uv
v
v
Slide number 36
uv
uv
u
u
uv
v
u
v
u
uv
v
Cross product properties
• The cross product is anti–commutative,
uv = –vu.
This is a consequence of the right hand rule.
• The cross product is distributive,
u(v + w) = uv + uw
• For a scalar c, (cu)v = c(uv)
• Note that the cross product operates on two
3–dimensional vectors to produce a 3–
dimensional vector perpendicular to the first
two.
• Question: What do you get when you cross a
mountain-climber with a mosquito?
Answer: Nothing: you can't cross a scaler with a vector.
Slide number 37
Aside: determinant of 33 matrix…
• Determinants are very useful in the analysis
and solution of systems of linear equations. A
nonhomogeneous system of linear equations
has a unique solution iff the determinant of
the system's matrix is nonzero (i.e., the matrix
is nonsingular)
• Recall that if A is a 2  2 matrix, then the
determinant of A is denoted by
det(A) = |A| = a11 a22 – a21 a12
• e.g.


2
5
5  2  3  1 5  11

det 1 3 2
  

 1 3 
Slide number 38
…Aside: determinant of 33 matrix…
• If A is a 3  3 matrix, then the determinant of A
may be calculated using the formula
a11 a12 a13 
A  deta21 a22 a23 


a
a
a
 31
33 
32
a21 a23 
a22 a23 
 a11 det

 a12 det
a
a
a
a
 31
 32
33 
33 
a21 a22 
 a13 det

a31 a32 
 a11a22 a33  a11a23 a32  a12 a21a33  a12 a23 a31
 a13 a21a32  a21a22 a31
Slide number 39
…Aside: determinant of 33 matrix…
• Remember using the following pattern
a11 a12 a13 
A  deta21 a22 a23 


a31 a32 a33 
a
a
a23 
 a11 det 22
 a12 det 21
a32 a33 
a31
a
a23 
 a13 det 21
a33 
a31
a22 

a32 
a11 a12 a13 
A  deta21 a22 a23 


a31 a32 a33 
a
a
a23 
 a11 det 22
 a12 det 21
a32 a33 
a31
a
a23 
 a13 det 21
a33 
a31
a22 

a32 
a11 a12 a13 
A  deta21 a22 a23 


a31 a32 a33 
a
a
a23 
 a11 det 22
 a12 det 21
a32 a33 
a31
a
a23 
 a13 det 21
a33 
a31
a22 

a32 
Slide number 40
…Aside: determinant of 33 matrix…
• e.g.
3 2 9
A  1 4 2 
2 8 4 
 3 2 9
A  det
1 4 2 

2
8
4


 4 2 
 2 det
1 2 
 9det
1 4 

 3det8
4 

2 4 
2 8
 316 16 24  4  98  8
 9616 0
 112

Slide number 41
…Aside: determinant of 33 matrix
• e.g.
1 2 1
A  3 4 3 
1 2 1
1 2 1
A  det
 3 4 3 

1
2
1



3 
 2 det
 3 3 
 1det
 3 4 

 1det4
2 1
1 1
1 2 
 14  6 23  3 16  4
 10 0 10
 20

Slide number 42
Cross product calculation…
• In Cartesian coordinates, the cross product
can be calculated using the following method
u1 
v1 
u  u2 , v  v2 


u3 

v3 

xˆ yˆ zˆ 
u  v  detu1 u2 u3 


v1 v2 v3 
 xˆ u2 v3  u3v2   yˆ u1v3  u3v1   zˆ u1v2  u2 v1 
u2 v3  u3v2 
 u3v1  u1v3 

u1v2  u2 v1 

Slide number 43
…Cross product calculation…
• Remember using the following pattern
xˆ yˆ zˆ 
u  v  detu1 u2 u3 


v
v
v
 1 2
3 
 xˆ u2 v3  u3v2   yˆ u1v3  u3v1   zˆ u1v2  u2 v1 
xˆ yˆ zˆ 
u  v  detu1 u2 u3 


v
v
v
 1 2
3 
 xˆ u2 v3  u3v2   yˆ u1v3  u3v1   zˆ u1v2  u2 v1 
xˆ yˆ zˆ 
u  v  detu1 u2 u3 


v
v
v
 1 2
3 
 xˆ u2 v3  u3v2   yˆ u1v3  u3v1   zˆ u1v2  u2 v1 
Slide number 44
…Cross product calculation…
• e.g.
3 
1
u  2 , v  3
3
2
xˆ yˆ zˆ 
u  v  det3 2 3
1 3 2 


 xˆ 4  9 yˆ 6  3 zˆ 9  2
 13xˆ  9 yˆ  7 zˆ
13
 9
7 
Slide number 45
…Cross product calculation
• e.g.
1 
2
u  2 , v  1
1
1
xˆ yˆ zˆ 
u  v  det1 2 1
2 1 1 


 xˆ 2  1 yˆ 1 2 zˆ 1 4 
 3xˆ  3yˆ  3zˆ
3 
 3
3
Slide number 46
Area of a parallelogram
• If a parallelogram is defined by the vectors u
and v then the area of the parallelogram is
given by |u  v|
• i.e.
area  base  height
 u v sin 
 uv
v


u
Slide number 47
|u|sin()
Lines
• A line is a straight one-dimensional figure
having no thickness and extending infinitely in
both directions. A line is sometimes called a
straight line.
• A line is uniquely determined by either:
• two points; or
• a point and a direction vector.
Slide number 48
Line defined by two points…
• The line passing through points p and q is
denoted pq
• Recall that if p and q are two points with
position vectors p and q, respectively, then the
vector joining p to q is q – p.

• We can thus define the line that passes
through points p and q parametrically as
x = p + (q – p)t
where t is a parameter that defines the
position on the line (i.e. x = p at t = 0, and
x = q at t = 1).
p
q
Slide number 49
…Line defined by two points
• e.g. find an equation that defines the line that
passes through points p and q where
1
2
p  3, q  5 
4
9 
x  p  q  pt
1 2 1
 3 
t
5  3


4 9  4
1 3
 3 2 t
4 5 
x  1 3t, y  3 2t, z  4  5t
p
q
Slide number 50
Line defined by a point and vector…
• We can also define the line that passes
through a point p in the direction of a vector u
parametrically as
x = p + ut
where t is a parameter that defines the
position on the line (i.e. x = p at t = 0, and
x = p + u at t = 1).
• Note that this is the same equation as that
obtained for the line defined by two points, p
and q, where u = q – p.
p u
Slide number 51
…Line defined by a point and vector
• e.g. find an equation that defines the line that
passes through the point p in the direction of a
vector u where
1
3
p  3, u  2 
4
5 
x  p  ut
1 3
 3 2 t
4 5 
x  1 3t, y  3 2t, z  4  5t
p u

Slide number 52
Planes…
• A plane is a 2–dimensional surface in a 3–
dimensional space.
• More generally, a hyperplane is an (n–1)–
dimensional vector subspace in an n–
dimensional vector space.
• In this module we will be dealing only with
planes in 3D.
• A plane is uniquely determined by either:
•
•
•
•
Slide number 53
Three points; or
Two points and a direction vector; or
One point and two direction vectors; or
One point and a normal vector.
…Planes…
• One way to think of a plane is as a flat surface
with a handle perpendicular to the surface.
• The handle, or normal vector, determines the
orientation of the plane.
• The magnitude of the (non–zero) normal
vector does not affect the orientation of the
plane.
Slide number 54
…Planes
• Specifying the normal vector does not uniquely
determine the plane.
• We need to choose a point to fix an individual
plane.
Slide number 55
Plane defined by one point and a
normal vector…
• The equation of a plane passing through the
point p with a non–zero normal vector n can
be defined by
n • (x – p) = 0
or, equivalently, n • x = n • p
where x is any point in the plane.
• Note that this equation states that any vector
starting from the point p and ending on a point
in the plane must be perpendicular to the
n
normal vector n.
x
p
Slide number 56
…Plane defined by one point and a
normal vector…
• e.g. find the equation of the plane passing
through the point p with normal vector n,
where
1
1
p  3, n  3 
2
4 
nx  np
1 x 1 1
3  y 3  3
4  
z 
 4  2
x  3y  4z  1 9  8
 16

Slide number 57
…Plane defined by one point and a
normal vector…
• e.g. find the equation of the plane passing
through the point p with normal vector n,
where
4
5
p  5 , n  4
2
1 
nx  np
5 x 5 4
4 y 4 5 
1  
z 
 1  2
5x  4y  z  20  20  2
 2

Slide number 58
…Plane defined by one point and a
normal vector

• Note that the coefficients of x, y, and z in the
equation for a plane are simply the
components of the normal vector
• i.e. a normal vector to the plane defined by
ax + by + cz = d
is
a
n  b
c
• e.g. a normal vector to the plane defined by
–5x – 4y + z = –2
is
5
n  4
1 
Slide number 59
Plane intercepts coordinate axes…

• The plane defined by the equation
ax + by + cz = d
crosses the x–, y–, and z–axes at the points
x = d/a, y = d/b, and z = d/c, respectively.
• This plane lies at a distance from the origin of
d
D
a2  b2  c2
• A plane crossing the x–, y–, and z–axes at the
points x = a’, y = b’, and z = c’, respectively,
can be defined by the equation
x y z
  1
a b c
Slide number 60
…Plane intercepts coordinate axes


• e.g. a plane defined by the equation 3x –
2y + z = 6
crosses the x–, y–, and z–axes at the points
2 0 
0
0, 3, and 0
0 0 
6
lies at a distance
6
6
D

2
2
2
14
3  2 1
from the origin,
and can be described also by the equation
x y z
  1
2 3 6
Slide number 61
Plane defined by three points…
• Consider a plane passing through three points
p, q, and r.
• First, choose point p and calculate the vectors
joining p to q, (q – p), and p to r, (r – p).
• Next find the normal to the plane by taking the
cross product n = (q – p)  (r – p).
• Finally, use the formula for defining a plane
from a point, p, and normal vector, n.
n
x
r
p
q
Slide number 62
…Plane defined by three points…


• e.g. find the plane that passes through the
points
2 
5
3
p  1, q  2, and r  1
6 
2
8
• First, choose point p and calculate the vectors
joining p to q, (q – p), and p to r, (r – p)
5 2   3 
3 2  1
q  p  2 1  3 , r  p  1 1 2
2 6  4
8 6  2
n
x
p
r
q
Slide number 63
…Plane defined by three points…

• Next find the normal to the plane by taking the
cross product n = (q – p)  (r – p)
n  q  p  r  p
xˆ yˆ zˆ 
 det3 3 4 
1 2 2 


 xˆ 6  8 yˆ 6  4 zˆ 6  3
14 
 10
 3 
n
x
p
r
q
Slide number 64
…Plane defined by three points

• Finally, use the formula for defining a plane
from a point, p, and normal vector, n.
2 
14 
p  1, n  10
6 
 3 
nx  np
14  x 14  2 
10 y 10 1
 3  
z 
  3  6 
14x 10y  3z  281018
n
 56
x
p
r
q
Slide number 65
Vectors module 4 objectives
• After this module you should have a clear
understanding of:
• Coordinate systems;
• The properties of vectors;
• Vector length, scalar multiplication, addition, and
subtraction;
• The dot product and its properties;
• Resolving vector components;
• The cross product and its properties;
• How to find equations of lines;
• How to find equations of planes.
Slide number 66
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