Download Lecture 9: 3.2 Norm of a Vector

Lecture 9: 3.2
Norm of a Vector
Wei-Ta Chu
2008/10/17
Translation of Axes




In the figure we have translated the axes of an
xy-coordinate system to obtain an x
y
coordinate system whose O
is at point (x, y) =
(k,l).
A point P in 2-space now has both (x, y)
coordinates and (x
, y
) coordinates.
x
= x –k, y
= y –l, these formulas are called
the translation equations.
In 3-space the translation equations are
x
= x –k, y
= y –l, z
= z –m,
where (k, l, m) are the xyz-coordinates of the
x
y
z
-origin.
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Example


Suppose that an xy-coordinate system is translated to obtain an
x
’
y
’
-coordinate system whose origin has xy-coordinate (k, l) =
(4, 1)
The point with xy-coordinates P(2,0) has



x
’= x –4, y
’= y –1.
The x
’
y
’
-coordinates of P are (-2,-1)
The point with x
’
y
’
-coordinates Q(-1,5)


x=x
’
+4, y = y
’
+1
The xy-coordinates of Q are (3, 6)
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Theorem 3.2.1 (Properties of Vector
Arithmetic)

If u, v and w are vectors in 2- or 3-space and k and l are
scalars, then the following relationships hold.








u+v=v+u
(u + v) + w = u + (v + w)
u+0=0+u=u
u + (-u) = 0
k(lu) = (kl)u
k(u + v) = ku + kv
(k + l)u = ku + lu
1u = u
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Proof of part (b) (geometric)
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Norm of a Vector


The length of a vector u is often called the norm (範數) of u
and is denoted by ||u||.
It follows from the Theorem of Pythagoras that the norm of
a vector u = (u1,u2,u3) in 3-space is
u  u12 u22 u32



A vector of norm 1 is called a unit vector. (單位向量)
The distance between two points is the norm of the vector.
The length of the vector ku = ||ku|| = |k| ||u||.
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Distance

If P1(x1, y1, z1) and P2(x2, y2, z2) are two points in 3-space,
then the distance d between them is the norm of the
vector

Euclidean distance (歐幾里德距離, 歐式距離)
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Lecture 9: 3.3
Dot Product and Projection
Wei-Ta Chu
2008/10/17
Definitions


Let u and v be two nonzero vectors in 2-space or 3-space,
and assume these vectors have been positioned so their
initial points coincided. By the angle between u and v, we
shall mean the angle determined by u and v that satisfies 0
.
If u and v are vectors in 2-space or 3-space and is the
angle between u and v, then the dot product (點積) or
Euclidean inner product (內積) u · v is defined by
u v cos  if u 0 and v 0
u
v 
0
if u 0 or v 0

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Example

If the angle between the vectors u = (0,0,1) and v =
(0,2,2) is 45
, then
1 
u
v u v cos  0 0 1 0 4 4 
 2
2
u
v (u1 , u2 , u3 ) 
(v1 , v2 , v3 ) u1v1 u2 v2 u3v3 2
u
v
2
1
cos 


u v
0 0 1 0 4 4
2
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Component Form of Dot Product


Let u=(u1,u2,u3) and v=(v1,v2,v3) be two nonzero vectors.
According to the law of cosine
P
Q
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Theorems

Theorem 3.3.1
 Let u and v be vectors in 2- or 3-space.


v · v = ||v||2; that is, ||v|| = (v · v)½
If the vectors u and v are nonzero and  is the angle between them, then




u v cos if u 0 and v 0
u
v 
0
if u 0 or v 0

 is acute (銳角)
 is obtuse (鈍角)
= /2 (直角)
if and only if u · v > 0
if and only if u · v < 0
if and only if u · v = 0
Theorem 3.3.2 (Properties of the Dot Product)
 If u, v and w are vectors in 2- or 3-space, and k is a scalar, then




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u·v=v·u
u · (v + w) = u · v + u · w
k(u · v) = (ku) · v = u · (kv)
v · v > 0 if v 0, and v · v = 0 if v = 0
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Proof of Theorem 3.3.1

(a) Since the angle

(b) Since satisfies
, it follows that is acute if
and only if
, that is obtuse if and only if
, and that
if and only if
. But
has the
same sign as
since
,
. Thus, the result follows.
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between
and
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is 0, we have
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Proof of Theorem 3.3.2
k(u · v) = (ku) · v = u · (kv)

Let u=(u1,u2,u3) and v=(v1,v2,v3)
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Orthogonal Vectors

Definition



Perpendicular vectors are also called orthogonal (正交)
vectors.
Two nonzero vectors are orthogonal if and only if their dot
product is zero.
To indicate that u and v are orthogonal vectors we write
u⊥v.
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Example
Show that in 2-space the nonzero vector n = (a,b) is perpendicular
to the line ax + by + c = 0.

Let P1(x1,y1) and P2(x2,y2) be distinct points on the line, so that

Since the vector
runs along the line,
we need only show that n and
are perpendicular.
But on subtracting the equations, we obtain
y

ax + by + c = 0
P1
P2
x
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An Orthogonal Projection




To "decompose" a vector u into a sum of two terms, one parallel to
a specified nonzero vector a and the other perpendicular to a.
We have w2 = u –w1 and w1 + w2 = w1 + (u –w1) = u
The vector w1 is called the orthogonal projection (正交投影) of u
on a or sometimes the vector component (分向量) of u along a, and
denoted by projau
The vector w2 is called the vector component of u orthogonal to a,
and denoted by w2 = u –projau
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w1 proja u
Theorem 3.3.3

w 2 u proja u
If u and a are vectors in 2-space or 3-space and if a 0, then
u
a
proja u  2 a
a
(vector component of u along a)
u
a
u proja u u  2 a
a
(vector component of u orthogonal to a)
u
a
proja u 
u cos 
a
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Proof of Theorem 3.3.3

Let
and
. Since w1 is parallel
to a, it must be a scalar multiple of a, so it can be written in the
form w1 = ka. Thus

Taking the dot product

But

Since w1 = ka, we obtain
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Example
u
a
proja u  2 a
a
u
a
u proja u u  2 a
a
Let u ( 2,1,3) and a ( 4,1,2). Find the vector component of u along a
and the vector component of u orthogonal to a.

Solution:
u
a ( 2)( 4) ( 1)( 1) (3)( 2) 15
a
2
4 2 ( 1) 2 2 2 21
Thus, the vector component of u along a is
u
a
proj a u  2 a 15
( 4,1,2) ( 207 ,75 , 107 )
21
a
and the vector component of u orthogonal to a is
u proj a u ( 2, 1,3) ( 207 ,75 , 107 ) ( 67 ,72 , 117 )
Verify tha t the vector u proj a u and a are perpendicu lar by showing that
their dot product is zero.
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Length of Orthogonal Projection
scalar
Formula (5) of Section 3.2
Since
If
denotes the angle between u and a, then
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Length of Orthogonal Projection
u
a
u
a
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Distance between
a Point and a Line
y
ax + by + c = 0
n = (a,b)
D
Q(x1,y1)
P0(x0,y0)
x
Find a formula for the distance D between point P0 ( x0 , y0 ) and the line ax by c 0.
Solution:
Let Q( x1 , y1 ) be any point on the line and position the vector n (a, b) so that its initial
point is at Q.
By virtueof Example5, the vector n is perpendicular to the line (Fig 3.3.8).
As indicated in the figure, the distance D is equal to the length of the orthogonal

projection of QP0 on n; thus,
QP0 
n
D  projn QP0 
n
But
QP0 ( x0 x1 , y0 y1 ),
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QP0 
n a( x0 x1 ) b( y0 y1 ),
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n  a 2 b 2
23
Distance between a Point and a Line

Since the point Q(x1,y1) lies on the lines, it coordinates
satisfy the equation of the line, so


ax1 + by1 + c = 0 or c = -ax1 –by1
Example: the distance D from the point (1,-2) to the line
3x+4y-6 = 0 is
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Cross Product (叉積, 外積)

Definition

If u = (u1, u2, u3) and v = (v1, v2, v3) are vectors in 3-space, then
the cross product u v is the vector defined by
u v (u2 v3 u3v2 , u3v1 u1v3 , u1v2 u2 v1 )
or in determinant notation
u2
u v 
v
2
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u3 u1 u3 u1 u2 

,
,
v3 v1 v3 v1 v2 

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Example
Find u v, where u = (1, 2, -2) and v = (3, 0, 1).
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Cross Product


Form the
matrix
whose first row contains
the components of u and whose second row contains the
components of v.
To find the first component of
, delete the first
column and take the determinant; to find the second
component, delete the second column and take the
negative of the determinant; and to find the third
component, delete the third column and take the
determinant.
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Theorems

Theorem 3.4.1 (Relationships Involving Cross Product and
Dot Product)
 If u, v and w are vectors in 3-space, then





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u · (u v) = 0
v · (u v) = 0
|| u v ||2 = ||u||2||v||2 –(u · v)2
u (v
 w)
 = (u · w) v –(u · v) w
product)
(u v) w = (u · w) v –(v · w) u
product)
(u v is orthogonal to u)
(u v is orthogonal to v)
(
La
gr
a
nge
’
si
de
nt
i
t
y)
(relationship between cross & dot
(relationship between cross & dot
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Proof of Theorem 3.4.1(a)
Let u=(u1,u2,u3) and v=(v1,v2,v3)

Example: u=(1, 2, -2) and v=(3, 0, 1)
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Proof of Theorem 3.4.1(c)
u v (u2 v3 u3v2 , u3v1 u1v3 , u1v2 u2 v1 )
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Theorems

Theorem 3.4.2 (Properties of Cross Product)
 If u, v and w are any vectors in 3-space and k is any scalar,
then







u v = - (v u)
u (v + w) = u v + u w
(u + v) w = u w + v w
k(u v) = (ku) v
 = u (kv)

u 0 = 0 u = 0
u u = 0
Proof of (a)

Interchanging u and v interchanges the rows of the three
determinants and hence changes the sign of each component in
the cross product. Thus u v = - (v u).
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Standard Unit Vectors


z
(0,0,1)
x
The vectors
(0,1,0) y
(1,0,0)
i = (1,0,0), j = (0,1,0), k = (0,0,1)
have length 1 and lie along the coordinate axes. They are called the
standard unit vectors in 3-space.
Every vector v = (v1, v2, v3) in 3-space is expressible in terms of i, j,
k since we can write
v = (v1, v2, v3) = v1(1,0,0) + v2 (0,1,0) + v3 (0,0,1) = v1i + v2j + v3k


For example, (2, -3, 4) = 2i –3j +4k
Note that
i i = 0, j j = 0, k k = 0
i j = k, j k = i, k i = j
j i = -k, k j = -i, i k = -j
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Cross Product

A cross product can be represented symbolically in the
form of 33 determinant:
i
j
u v u1 u2
v1

v2
k
u2
u3 
v2
v3
u3
u1 u3
u1 u2
i
j
k
v3
v1 v3
v1 v2
Example: if u=(1,2,-2) and v=(3,0,1)
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Cross Product

I
t
’
snott
r
uei
nge
ne
r
a
lt
ha
t
For example:

Right-hand rule


If the fingers of the right hand are cupped
so they point in the direction of rotation,
then the thumb indicates the direction of
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u
v
34
Exercises





Sec. 3.1: 6, 12, 21
Sec. 3.2: 2, 9, 12
Sec. 3.3: 3, 4, 8, 16, 24
Sec. 3.4: 2, 8, 13, 17, 36
Sec. 3.5: 3, 6, 9, 15, 20, 27
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