Chapter 3: Euclidean Space
16
SECTION B Properties of Vectors
By the end of this section you will be able to
• understand and prove the basic properties of vectors in \ n
• prove other properties of vectors in \ n
• understand what is meant by the dot product of two vectors
• understand what is meant by the terms norm and distance
• prove dot product and norm properties of vectors
Remember the space \ n is also called Euclidean n-space and is named after the Greek
mathematician Euclid.
Euclid is famously known for his work ‘The Elements’ which
is also called ‘Euclid’s Elements’. This work has been used in
mathematics teaching for over two thousand years and was
first published as a book in 1482 and only the bible has been
printed more times than ‘Euclid’s Elements’.
Up until the 1970’s school mathematics in Britain consisted
of learning various parts of Euclid’s Elements. The concept
of mathematical proof and logical reasoning is what made
this work survive for so long.
Fig 19 Euclid lived around 300BC
B1 Vector Addition and Scalar Multiplication Properties
Remember the notation \ n from the last section - \ 2 is the plane, \3 is the 3
dimensional space, etc.
Many of the properties of vectors in \ n have been proven in chapter 1, Linear
Equations and Matrices. Why?
Because a vector u in \ n is a special matrix of size n by 1 which is called a column
vector therefore all the properties of matrices also holds for a column vector u in \ n .
However we prove properties of vector addition and scalar multiplication below for
column vectors.
Proposition (3.6). Let u, v and w be vectors in \ n and k , c be real numbers (or real
scalars). We have the following results:
(i) u + v = v + u (Commutative Law)
(ii) ( u + v ) + w = u + ( v + w ) (Associative Law)
(iii) There exists a zero vector O such that u + O = u . (Neutral element)
(iv) For every vector u there is a vector −u such that
u + ( −u ) = O
(Additive inverse).
(v) k ( u + v ) = k u + k v (Distributive Law)
(vi) ( k + c ) u = k u + cu (Distributive Law)
(vii) ( kc ) u = k ( cu )
(Associative Law)
(viii) For every vector u we have 1u = u (Neutral element)
Chapter 3: Euclidean Space
17
Proof. All of these properties follow from the matrix properties that were discussed in
chapter 1. Here are some proofs and the remaining proofs are set as questions in
Exercise 3(b).
⎛ u1 ⎞
⎛ v1 ⎞
⎜ ⎟
⎜ ⎟
u
v
Since u and v are vectors in \ n therefore let u = ⎜ 2 ⎟ and v = ⎜ 2 ⎟ .
⎜# ⎟
⎜#⎟
⎜ ⎟
⎜ ⎟
⎝ un ⎠
⎝ vn ⎠
(i) We have
⎛ u1 ⎞ ⎛ v1 ⎞ ⎛ u1 + v1 ⎞
⎜ ⎟ ⎜ ⎟ ⎜
⎟
u
v
u +v
u+ v = ⎜ 2 ⎟+⎜ 2 ⎟ = ⎜ 2 2 ⎟
⎜ # ⎟ ⎜ # ⎟ ⎜ #+# ⎟
⎜ ⎟ ⎜ ⎟ ⎜
⎟
⎝ un ⎠ ⎝ vn ⎠ ⎝ un + vn ⎠
⎛ v1 + u1 ⎞ ⎛ v1 ⎞ ⎛ u1 ⎞
⎜
⎟ ⎜ ⎟ ⎜ ⎟
v2 + u2 ⎟ ⎜ v2 ⎟ ⎜ u2 ⎟
⎜
=
=
+
= v+u
⎜ # +# ⎟ ⎜ # ⎟ ⎜ # ⎟
⎜
⎟ ⎜ ⎟ ⎜ ⎟
⎝ vn + un ⎠ ⎝ vn ⎠ ⎝ un ⎠
■
(ii) See Exercise 3b.
⎛0⎞
⎜ ⎟
0
(iii) We need to prove u + O = u . Remember the zero vector is O = ⎜ ⎟ therefore we
⎜#⎟
⎜ ⎟
⎝0⎠
have
⎛ u1 ⎞ ⎛ 0 ⎞ ⎛ u1 + 0 ⎞ ⎛ u1 ⎞
⎜ ⎟ ⎜ ⎟ ⎜
⎟ ⎜ ⎟
u2 ⎟ ⎜ 0 ⎟ ⎜ u 2 + 0 ⎟ ⎜ u 2 ⎟
⎜
u+O =
+
=
=
=u
⎜ # ⎟ ⎜ # ⎟ ⎜ #+ 0 ⎟ ⎜ # ⎟
⎜ ⎟ ⎜ ⎟ ⎜
⎟ ⎜ ⎟
⎝ un ⎠ ⎝ 0 ⎠ ⎝ u n + 0 ⎠ ⎝ u n ⎠
(iv) We need to show u + ( −u ) = O . How?
Substituting the above u into this
⎛ u1 ⎞ ⎛ −u1 ⎞ ⎛ u1 − u1 ⎞ ⎛ 0 ⎞
⎜ ⎟ ⎜
⎟ ⎜
⎟
u2 ⎟ ⎜ −u2 ⎟ ⎜ u2 − u2 ⎟ ⎜⎜ 0 ⎟⎟
⎜
u + ( −u ) =
+
=
=
=O
⎜ # ⎟ ⎜ # ⎟ ⎜ # ⎟ ⎜#⎟
⎜ ⎟ ⎜
⎟ ⎜
⎟ ⎜ ⎟
⎝ u n ⎠ ⎝ −u n ⎠ ⎝ u n − u n ⎠ ⎝ 0 ⎠
■
(v) See Exercise 3(b).
(vi) We need to prove ( k + c ) u = k u + cu . Substituting the above u gives:
Chapter 3: Euclidean Space
18
⎛ u1 ⎞ ⎛ ( k + c ) u1 ⎞
⎟
⎜ ⎟ ⎜
u2 ⎟ ⎜ ( k + c ) u 2 ⎟
⎜
(k + c) u = (k + c) ⎜ ⎟ = ⎜
⎟
#
#
⎟⎟
⎜ ⎟ ⎜⎜
⎝ un ⎠ ⎝ ( k + c ) u n ⎠
[Scalar Multiplication ]
⎛ ku1 + cu1 ⎞
⎜
⎟
ku2 + cu2 ⎟
⎜
=
[ Expanding Brackets ]
⎜
⎟
#
⎜
⎟
⎝ kun + cun ⎠
⎛ ku1 ⎞ ⎛ cu1 ⎞
⎛ u1 ⎞ ⎛ u1 ⎞
⎜
⎟ ⎜
⎟
⎜ ⎟ ⎜ ⎟
ku2 ⎟ ⎜ cu2 ⎟
u
u
⎜
=
+
= k ⎜ 2 ⎟ + c ⎜ 2 ⎟ = k u + cu
⎜ # ⎟ ⎜ # ⎟
⎜# ⎟ ⎜# ⎟
⎜
⎟ ⎜
⎟
⎜ ⎟ ⎜ ⎟
⎝ kun ⎠ ⎝ cun ⎠
⎝ un ⎠ ⎝ un ⎠
■
(vii) See Exercise 3(b).
(viii) Need to prove 1u = u . (The scalar k is one not the letter l).
⎛ u1 ⎞
⎛ 1× u1 ⎞ ⎛ u1 ⎞
⎜ ⎟
⎜
⎟ ⎜ ⎟
u2 ⎟
1 × u2 ⎟ ⎜ u 2 ⎟
⎜
⎜
1u = 1
=
=
=u
N
⎜ # ⎟ Scalar Multiplication
⎜ # ⎟ ⎜# ⎟
⎜ ⎟
⎜
⎟ ⎜ ⎟
⎝ un ⎠
⎝ 1 × un ⎠ ⎝ u n ⎠
■
We can use this proposition (3.6) to prove other properties of vectors in \ n .
Proposition (3.7). Let u be a vector in \ n . Then the vector −u which satisfies property
(iv) in the above Proposition (3.6) is unique:
u + ( −u ) = O
Proof. Let v be a vector in \ n such that
u+v =O
(*)
What do we need to show?
Show that v = −u . Adding −u to both sides of (*) gives
−u + ( u + v ) = ( − u ) + O
( −u + u ) + v = ( − u ) + O
( −u
) ) + v = −u
(u+
=O
O + v = −u
v = −u
■
What does Proposition (3.7) mean?
It means that for every vector u in \ n there is only one vector −u in \ n such that
u + ( −u ) = O
Chapter 3: Euclidean Space
19
Proposition (3.8). Let u be a vector in \ n then ( −1) u = −u .
⎛ u1 ⎞
⎜ ⎟
u
Proof. Let u = ⎜ 2 ⎟ . We first show that u + ( −1) u = O and then use the above
⎜# ⎟
⎜ ⎟
⎝ un ⎠
Proposition (3.7) which says −u is unique to deduce ( −1) u = −u :
⎛ u1 ⎞
⎛ u1 ⎞ ⎛ u1 ⎞ ⎛ −u1 ⎞
⎜ ⎟
⎜ ⎟ ⎜ ⎟ ⎜
⎟
u2 ⎟
u2 ⎟ ⎜ u2 ⎟ ⎜ −u2 ⎟
⎜
⎜
u + ( −1) u =
+ ( −1)
=
+
[Scalar Multiplication ]
⎜# ⎟
⎜# ⎟ ⎜# ⎟ ⎜ # ⎟
⎜ ⎟
⎜ ⎟ ⎜ ⎟ ⎜
⎟
⎝ un ⎠
⎝ un ⎠ ⎝ un ⎠ ⎝ −un ⎠
⎛ u1 − u1 ⎞ ⎛ 0 ⎞
⎜
⎟
u2 − u2 ⎟ ⎜⎜ 0 ⎟⎟
⎜
=
=
=O
⎜ # # ⎟ ⎜#⎟
⎜
⎟ ⎜ ⎟
⎝ un − u n ⎠ ⎝ 0 ⎠
We have u + ( −1) u = O and by Proposition (3.7) the vector −u which satisfies
u + ( −u ) = O is unique, hence
( −1) u = −u
■
So far we have looked at vector addition and scalar multiplication of vectors in \ n but
we have not yet examined multiplication of vectors. Next we define the dot product.
B2 Introduction to the Dot (Inner) Product
⎛ u1 ⎞
⎛ v1 ⎞
⎜ ⎟
⎜ ⎟
u2 ⎟
v
⎜
Let u =
and v = ⎜ 2 ⎟ be vectors in \ n then the dot product of u and v which is
⎜# ⎟
⎜#⎟
⎜ ⎟
⎜ ⎟
⎝ un ⎠
⎝ vn ⎠
denoted by u ⋅ v is defined as the matrix multiplication of the transpose of u multiplied
by the column vector v:
T
⎛ u1 ⎞ ⎛ v1 ⎞
⎜ ⎟ ⎜ ⎟
u
v
u ⋅ v = uT v = ⎜ 2 ⎟ ⎜ 2 ⎟
⎜# ⎟ ⎜#⎟
⎜ ⎟ ⎜ ⎟
⎝ un ⎠ ⎝ vn ⎠
⎛ v1 ⎞
⎜ ⎟
v
= ( u1 u2 " un ) ⎜ 2 ⎟ = u1v1 + u2 v2 + u3v3 + " + un vn
⎜#⎟
⎜ ⎟
⎝ vn ⎠
Rather than carry out this matrix multiplication we say the dot product or inner
product of the vectors u and v is given by
Chapter 3: Euclidean Space
20
u ⋅ v = u1v1 + u2 v2 + u3v3 + " + un vn
(3.9)
What do you notice about your answer?
The result of the dot (inner) product of two vectors in \ n is a real number, that is
carrying out the dot product operation on two vectors gives us a scalar not a vector.
Also note that the dot product of two vectors u and v is obtained by multiplying each
component u j with it’s corresponding component v j and adding the result.
Example 5
⎛ −3 ⎞
⎛9⎞
⎜ ⎟
⎜ ⎟
1⎟
2
⎜
and v = ⎜ ⎟ . Find u ⋅ v .
Let u =
⎜7⎟
⎜ −4 ⎟
⎜ ⎟
⎜ ⎟
⎝ −5 ⎠
⎝1⎠
Solution.
Applying the above formula (3.9) gives
⎛ −3 ⎞ ⎛ 9 ⎞
⎜ ⎟ ⎜ ⎟
1
2
u ⋅ v = ⎜ ⎟ ⋅ ⎜ ⎟ = ( −3 × 9 ) + (1× 2 ) + ( 7 × ( −4 ) ) + ( ( −5 ) × 1)
⎜ 7 ⎟ ⎜ −4 ⎟
⎜ ⎟ ⎜ ⎟
⎝ −5 ⎠ ⎝ 1 ⎠
= −27 + 2 − 28 − 5 = −58
Hence u ⋅ v = −58 .
You can check this answer, u ⋅ v = −58 , by using the numerical software MATLAB on
your computer. The MATLAB command for the dot product of two vectors u and v is
dot ( u, v ) .
The dot product does signify a geometric property for \ 2 and \3 which we will discuss
in the next section.
Two vectors u and v in \ n are said to be orthogonal or perpendicular if
u⋅v = 0
(3.10)
Example 6
⎛ 2⎞
⎛ 5⎞
Let u = ⎜ ⎟ and v = ⎜ ⎟ . Plot these vectors u and v in \ 2 and show that these
⎝5⎠
⎝ −2 ⎠
vectors are orthogonal.
Solution.
Plotting the vectors u and v in \ 2 gives
y
5
u=
4
⎛ 2⎞
⎜ ⎟
⎝ 5⎠
3
2
1
1
2
3
-1
Fig 20
-2
v=
⎛5⎞
⎜ ⎟
⎝ -2 ⎠
4
5
x
Chapter 3: Euclidean Space
21
We show that the vectors u and v are perpendicular or orthogonal by evaluating the dot
or inner product:
⎛ 2⎞ ⎛ 5 ⎞
u ⋅ v = ⎜ ⎟ ⋅ ⎜ ⎟ = ( 2 × 5 ) + ( 5 × ( −2 ) ) = 10 − 10 = 0
⎝ 5 ⎠ ⎝ −2 ⎠
Since the dot product u ⋅ v is zero therefore the given vectors u and v are orthogonal
(perpendicular).
Example 7
⎛a⎞
⎛ b ⎞
Let u = ⎜ ⎟ and v = ⎜ ⎟ be vectors in \ 2 . Prove that these vectors are orthogonal.
⎝b⎠
⎝ −a ⎠
Proof.
Carrying out the dot product of the given two vectors by applying the above formula
(3.9):
⎛a⎞ ⎛ b ⎞
u ⋅ v = ⎜ ⎟ ⋅ ⎜ ⎟ = ( a × b ) + ( b × ( − a ) ) = ab − ba = 0
⎝ b ⎠ ⎝ −a ⎠
Since the dot product u ⋅ v is zero therefore by formula (3.10) we conclude that the
vectors u and v are orthogonal (perpendicular).
B3 Properties of Dot (Inner) Product
Next we state some basic properties of the dot product in \ n . Remember \ n is the set
of all n dimensional vectors. \ 2 is the x − y plane, \3 is the 3 dimensional space with
axes x, y and z etc.
Proposition (3.11). Let u, v and w be vectors in \ n and k be a real scalar (real number).
We have the following:
(i) ( u + v ) ⋅ w = u ⋅ w + v ⋅ w [Distributive Law]
(ii) u ⋅ v = v ⋅ u [Commutative Law]
(iii) ( k u ) ⋅ v = k ( u ⋅ v )
(iv) u ⋅ u ≥ 0 and we have u ⋅ u = 0 if and only if u = O
Proof.
⎛ u1 ⎞
⎛ v1 ⎞
⎜ ⎟
⎜ ⎟
u2 ⎟
v
⎜
Let u =
and v = ⎜ 2 ⎟ be vectors in \ n .
⎜# ⎟
⎜#⎟
⎜ ⎟
⎜ ⎟
⎝ un ⎠
⎝ vn ⎠
(i) See Exercise 3(b).
(ii) Required to prove u ⋅ v = v ⋅ u :
u ⋅ v = u1v1 + u2 v2 + u3v3 + " + un vn [ Applying (3.9)]
= v1u1 + v2u2 + v3u3 + " + vn un
⎡ Because u j and v j are real numbers ⎤
⎢
⎥
⎢⎣ therefore u j v j = v j u j
⎥⎦
= v ⋅u
(iii) How do we prove ( k u ) ⋅ v = k ( u ⋅ v ) ?
By expanding the Left Hand Side and showing it is equal to the Right.
■
Chapter 3: Euclidean Space
22
⎡ ⎛ u1 ⎞ ⎤ ⎛ v1 ⎞ ⎛ ku1 ⎞ ⎛ v1 ⎞
⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎜
⎟ ⎜ ⎟
u2 ⎟ ⎥ ⎜ v2 ⎟ ⎜ ku2 ⎟ ⎜ v2 ⎟
⎢
⎜
( k u ) ⋅ v = ⎢k ⎜ ⎟⎥ ⋅ ⎜ ⎟ = ⎜ ⎟ ⋅ ⎜ ⎟
#
#
#
#
⎢ ⎜ ⎟⎥ ⎜ ⎟ ⎜
⎟ ⎜ ⎟
⎢⎣ ⎝ un ⎠ ⎥⎦ ⎝ vn ⎠ ⎝ kun ⎠ ⎝ vn ⎠
⎡Scalar Multiplication ⎤
⎢ of the vector u by k ⎥
⎣
⎦
= ku1v1 + ku2 v2 + ku3v3 + " + kun vn
= k ( u1v1 + u2 v2 + u3v3 + " + un vn )
[ Applying (3.9)]
[ Factorizing ]
= k (u ⋅ v )
Hence we have shown our result.
■
(iv) First we prove u ⋅ u ≥ 0 . How?
⎛ u1 ⎞
⎜ ⎟
u
Substitute u = ⎜ 2 ⎟ and then evaluate u ⋅ u :
⎜# ⎟
⎜ ⎟
⎝ un ⎠
⎛ u1 ⎞ ⎛ u1 ⎞
⎜ ⎟ ⎜ ⎟
u
u
u ⋅ u = ⎜ 2 ⎟ ⋅ ⎜ 2 ⎟ = u1u1 + u2u2 + u3u3 + " + unun
⎜# ⎟ ⎜# ⎟
⎜ ⎟ ⎜ ⎟
⎝ un ⎠ ⎝ u n ⎠
[ Applying (3.9)]
= ( u1 ) + ( u2 ) + ( u3 ) + " + ( un ) ≥ 0
2
2
2
2
Since all the u’s are real numbers therefore each ( u j ) ≥ 0 . Hence we have shown
2
u ⋅u ≥ 0 .
How do we show the last part, that is u ⋅ u = 0 if and only if u = O ?
Remember if and only if means that two statements are equivalent, that is if P and Q
are statements then P if and only if Q means that P implies Q ( P ⇒ Q ) and
Q implies P ( Q ⇒ P ) . In this case we need to show that u ⋅ u = 0 implies u = O and
the other way that u = O implies u ⋅ u = 0 . In mathematical notation this is written as
u ⋅ u = 0 ⇔ u = O . The implication symbol goes both ways.
Since from above we have u ⋅ u = ( u1 ) + ( u2 ) + ( u3 ) + " + ( un ) therefore
2
2
2
2
u ⋅ u = 0 ⇔ ( u1 ) + ( u2 ) + ( u3 ) + " + ( un ) = 0
2
2
2
2
⇔ u1 = u2 = u3 = " = un = 0
Remember from the definition of the zero vector u1 = u2 = u3 = " = un = 0 means that
u = O.
■
B4 The Norm or Length of a Vector
Let u be a vector in \ n . The norm or length of a vector u is written as u and is
defined by
(3.12)
u = u ⋅u
[Positive Root]
Chapter 3: Euclidean Space
23
The norm of a vector u is a real number which gives the size of the vector u.
This u = u ⋅ u is sometimes called the Euclidean norm.
2
Generally to find the norm u it is easier to determine u = u ⋅ u and then take the
square root of your result.
Proposition (3.13). Let u be a vector in \ n then
u =
( u1 ) + ( u2 ) + ( u3 )
2
2
2
+ " + ( un )
2
It is easier to use this to find the norm or length of a vector u rather than the formula
given in (3.12).
Proof. Note that since u is in \ n therefore by the dot product formula (3.9) given
earlier we have
⎛ u1 ⎞ ⎛ u1 ⎞
⎜ ⎟ ⎜ ⎟
u
u
u ⋅ u = ⎜ 2 ⎟ ⋅ ⎜ 2 ⎟ = u1u1 + u2u2 + u3u3 + " + unun
⎜# ⎟ ⎜# ⎟
⎜ ⎟ ⎜ ⎟
⎝ un ⎠ ⎝ un ⎠
= ( u1 ) + ( u2 ) + ( u3 ) + " + ( un )
2
2
2
2
By the above definition (3.12) we have u = u ⋅ u therefore taking the square root of
the above result gives
u = u ⋅u =
( u1 ) + ( u2 ) + ( u3 )
2
2
2
+ " + ( un )
2
This is our required result.
■
2
3
We can apply this norm formula (3.13) to particular n – spaces such as \ and \ as
shown below.
Consider a vector u in \ 2 :
u
y
x
Fig 21
⎛ x⎞
The norm or length of a vector u = ⎜ ⎟ in \ 2 is given by Pythagoras’s Theorem:
⎝ y⎠
u = x2 + y2
Consider a vector v in \3 :
z
v
⎛ x⎞
v = ⎜⎜ y ⎟⎟
⎜z⎟
⎝ ⎠
y
x
Chapter 3: Euclidean Space
24
⎛ x⎞
The norm or length of a vector v = ⎜⎜ y ⎟⎟ in \3 is given by
Fig 22
⎜z ⎟
⎝ ⎠
v = x2 + y2 + z 2
⎛ v1 ⎞
⎜ ⎟
v2
The norm (length) of a vector v = ⎜ ⎟ in \ n is given by
⎜#⎟
⎜ ⎟
⎝ vn ⎠
v =
( v1 ) + ( v2 )
2
2
+ " + ( vn )
2
Example 8
⎛ −7 ⎞
⎛8⎞
Let u = ⎜ ⎟ and v = ⎜ ⎟ . Plot these vectors u and v in \ 2 and evaluate the norms
⎝ −2 ⎠
⎝ 3⎠
u and v . Also label the norm (length) of each vector in \ 2 .
Solution.
We first calculate the norm or length of the vectors u and v. How?
By using the above Proposition (3.13), which means that you add the square of each
component and then take the square root of the result:
u =
( −7 ) + ( −2 )
2
2
= 49 + 4 = 53 = 7.280 [3 dp]
v = 82 + 32 = 64 + 9 = 72 = 8.485 [3 dp]
⎛ −7 ⎞
⎛8⎞
Plotting the vectors u = ⎜ ⎟ and v = ⎜ ⎟ and labelling u and v in \ 2 gives:
⎝ −2 ⎠
⎝ 3⎠
y
3
2
u=
||u|| = 8.485
⎛ 8⎞
⎜ ⎟
⎝ 3⎠
1
-6
-4
||v|| = 7.280
v=
Fig 23
⎛ -7 ⎞
⎜ ⎟
⎝ -2 ⎠
-2
2
4
6
8
x
-1
-2
B5 Properties of the Norm of a Vector
Next we state certain properties of the norm of a vector.
Proposition (3.14).
Let u be a vector in \ n and k be a real scalar. We have the following:
Chapter 3: Euclidean Space
25
(i) u ≥ 0 [Positive] and u = 0 if and only if u = O .
(ii) k u = k u
[Note that for a scalar k we have
k 2 = k where k is the modulus of k].
Proof.
⎛ u1 ⎞
⎜ ⎟
u
n
Since u is a vector in \ therefore we can write this as u = ⎜ 2 ⎟ .
⎜# ⎟
⎜ ⎟
⎝ un ⎠
(i) By the above Proposition (3.13) we have
( u1 ) + ( u2 ) + ( u3 )
2
u =
2
2
+ " + ( un )
2
Since the square root is the positive root therefore u ≥ 0 .
Next we prove the equality, that is u = 0 if and only if u = O . We have u = 0 which
means that
u = u ⋅u = 0 ⇔
u ⋅u = 0
By Proposition (3.11) part (iv) we have u ⋅ u = 0 if and only if u = O . We have proven
our required equality.
■
(ii) Expanding the Left Hand Side by applying Proposition (3.13) gives
k u = ku ⋅ ku = k 2 u ⋅ u
= k2 u ⋅u
⎡ Because k 2 and u ⋅ u are real numbers therefore ⎤
⎢
⎥
⎣ we can take the square root of each part
⎦
=k u
■
Normally to obtain the length (norm) of a given vector v you will find it easier to find
2
v = v ⋅ v and then take the square root of your result to find v .
SUMMARY
Vectors in \ n have certain properties with regards to vector addition such as it is
commutative, associative etc.
Scalar multiplication of vectors obeys the distributive law.
For every vector u there is a unique vector −u such that u + ( −u ) = O and ( −1) u = −u .
Let u and v be vectors in \ n then the dot product u ⋅ v is given by
(3.9)
u ⋅ v = u1v1 + u2 v2 + u3v3 + " + un vn
If
u⋅v = 0
(3.10)
then the vectors u and v are orthogonal or perpendicular.
The dot product is commutative, distributive etc
The norm or length of a vector u in \ n is written as u and defined as
u = u ⋅u =
( u1 ) + ( u2 ) + ( u3 )
2
2
2
+ " + ( un )
2
Chapter 3: Euclidean Space
The norm of a vector is positive or zero.
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