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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
- Department of Physics
- A...................................................................................................................................... A- Review A: Vector Analysis
- A.1 Vectors A-
- A.1.1 Introduction A-
- A.1.2 Properties of a Vector A-
- A.1.3 Application of Vectors A-
- A.2 Dot Product A-
- A.2.1 Introduction A-
- A.2.2 Definition A-
- A.2.3 Properties of Dot Product A-
- A.2.4 Vector Decomposition and the Dot Product A-
- A.3 Cross Product A-
- A.3.1 Definition: Cross Product A-
- A.3.2 Right-hand Rule for the Direction of Cross Product A-
- A.3.3 Properties of the Cross Product A-
- A.3.4 Vector Decomposition and the Cross Product A-
Vector Analysis
A.1 Vectors
A.1.1 Introduction
Certain physical quantities such as mass or the absolute temperature at some point only have magnitude. These quantities can be represented by numbers alone, with the appropriate units, and they are called scalars. There are, however, other physical quantities which have both magnitude and direction; the magnitude can stretch or shrink, and the direction can reverse. These quantities can be added in such a way that takes into account both direction and magnitude. Force is an example of a quantity that acts in a certain direction with some magnitude that we measure in newtons. When two forces act on an object, the sum of the forces depends on both the direction and magnitude of the two forces. Position, displacement, velocity, acceleration, force, momentum and torque are all physical quantities that can be represented mathematically by vectors. We shall begin by defining precisely what we mean by a vector.
A.1.2 Properties of a Vector
A vector is a quantity that has both direction and magnitude. Let a vector be denoted by
the symbol A. The magnitude of
G
A
G
is | A | ≡ A
G
. We can represent vectors as geometric
objects using arrows. The length of the arrow corresponds to the magnitude of the vector. The arrow points in the direction of the vector (Figure A.1.1).
Figure A.1.1 Vectors as arrows.
There are two defining operations for vectors:
(1) Vector Addition: Vectors can be added.
Let A and be two vectors. We define a new vector,
G
B
G
C = A + B
G G G
, the “vector addition”
of A and , by a geometric construction. Draw the arrow that represents. Place the
G
B
G
A
G
Figure A.1.4 Associative law.
(iii) Identity Element for Vector Addition: There is a unique vector, 0 , that acts as an identity element for vector addition.
G
This means that for all vectors A ,
G
A + 0 = 0 + A = A
G G G G G
(A.1.3)
(iv) Inverse element for Vector Addition: For every vector A
G
, there is a unique inverse vector
( −^1 ) A^ ≡ − A
G G
(A.1.4)
such that
A + ( − A ) = 0
G G G
This means that the vector − A has the same magnitude as
G
A
G
, | A | | = − A | = A
JG JG
, but they
point in opposite directions (Figure A.1.5).
Figure A.1.5 additive inverse.
(2) Scalar Multiplication of Vectors: Vectors can be multiplied by real numbers.
Let be a vector. Let c be a real positive number. Then the multiplication of by c is
a new vector which we denote by the symbol
A
G
A
G
c A
G
. The magnitude of is c times the
magnitude of (Figure A.1.6a),
c A
G
A
G
cA = Ac (A.1.5)
Since c > 0 , the direction of c A is the same as the direction of
G
A
G
. However, the direction
of − c A is opposite of (Figure A.1.6b).
G
A
G
Figure A.1.6 Multiplication of vector A
G
by (a) c > 0 , and (b) − c < 0.
Scalar multiplication of vectors satisfies the following properties:
(i) Associative Law for Scalar Multiplication: The order of multiplying numbers is doesn’t matter.
Let b and c be real numbers. Then
b c ( A ) = ( bc ) A = ( cb A ) = c b ( A )
G G G G
c
(A.1.6)
(ii) Distributive Law for Vector Addition: Vector addition satisfies a distributive law for multiplication by a number.
Let c be a real number. Then
c ( A + B ) = c A + B
G G G G
(A.1.7)
Figure A.1.7 illustrates this property.
mathematical objects we shall instead consider the following essential properties that enable us to represent physical quantities as vectors.
(1) Vectors can exist at any point P in space.
(2) Vectors have direction and magnitude.
(3) Vector Equality: Any two vectors that have the same direction and magnitude are equal no matter where in space they are located.
(4) Vector Decomposition: Choose a coordinate system with an origin and axes. We can decompose a vector into component vectors along each coordinate axis. In Figure A.1. we choose Cartesian coordinates for the x - y plane (we ignore the -direction for
simplicity but we can extend our results when we need to). A vector at P can be decomposed into the vector sum,
z A
G
A = A (^) x + A (^) y
G G G
(A.1.10)
where is the -component vector pointing in the positive or negative -direction,
and is the -component vector pointing in the positive or negative -direction
(Figure A.1.9).
A x
G
x x A y
G
y y
Figure A.1.9 Vector decomposition
(5) Unit vectors: The idea of multiplication by real numbers allows us to define a set of unit vectors at each point in space. We associate to each point in space, a set of three
unit vectors (
P
ˆ ˆ ˆ i j k , , ). A unit vector means that the magnitude is one: | |ˆ i = 1 , | ˆ j | = 1 , and
. We assign the direction of ˆ^ to point in the direction of the increasing -
coordinate at the point. We call ˆ^ the unit vector at pointing in the + -direction.
Unit vectors ˆ
| k ˆ | = 1 i x
P i P x j and k ˆ can be defined in a similar manner (Figure A.1.10).
Figure A.1.10 Choice of unit vectors in Cartesian coordinates.
(6) Vector Components: Once we have defined unit vectors, we can then define the - component and -component of a vector. Recall our vector decomposition,
. We can write the x -component vector,
x y A = A (^) x + A
G G G
y A x
G
, as
A (^) x = Axˆ i
G
(A.1.11)
In this expression the term Ax , (without the arrow above) is called the x -component of
the vector. The -component can be positive, zero, or negative. It is not the
magnitude of which is given by. Note the difference between the -
component, , and the -component vector,
A
G
x Ax A x
G 2 1/ 2
( Ax ) x Ax x A x
G
In a similar fashion we define the y -component, Ay , and the -component, , of the
vector
z Az A
G
y y z z A = A ˆ j^ , A = A k ˆ
G G
(A.1.12)
A vector A can be represented by its three components
G
A =( A , A , Ax y z )
G
. We can also
write the vector as
x y z A = A ˆ i^ + A ˆ j^ + A k ˆ
G
(A.1.13)
(7) Magnitude: In Figure A.1.10, we also show the vector components.
Using the Pythagorean theorem, the magnitude of the
A = ( A , A , Ax y z )
G
A
G
is,
2 2 A = Ax + Ay + A 2 z (A.1.14)
(8) Direction: Let’s consider a vector A =( A , A ,x y 0)
G
. Since the -component is zero, the
vector lies in the
z A
G
x - y plane. Let θ denote the angle that the vector A makes in the
G
A = A cos θ A ˆ i + A sin θ Aˆ j
G
(A.1.19)
B = B cos θ B ˆ i + B sin θ B ˆ j
G
(A.1.20)
In Figure A.1.13, the vector addition C = A + B
G G G
is shown. Let θ C denote the angle that
the vector C makes with the positive x -axis.
G
Figure A.1.13 Vector addition with components
Then the components of C are
G
C x = Ax + B ,x C (^) y = Ay + By
B B
(A.1.21)
In terms of magnitudes and angles, we have
cos cos cos sin sin sin
x C A y C A
C C A B
C C A B
(A.1.22)
We can write the vector C as
G
C = ( Ax + Bx ) ˆ i^ + ( Ay + By ) ˆ j^ = C (cos θ C ˆ i +sinθ C j
G
(A.1.23)
A.2 Dot Product
A.2.1 Introduction
We shall now introduce a new vector operation, called the “dot product” or “scalar product” that takes any two vectors and generates a scalar quantity (a number). We shall see that the physical concept of work can be mathematically described by the dot product between the force and the displacement vectors.
Let and B be two vectors. Since any two non-collinear vectors form a plane, we
define the angle
A
G G
θ to be the angle between the vectors A
G
and B
G
as shown in Figure
A.2.1. Note that θ can vary from 0 to π.
Figure A.2.1 Dot product geometry.
A.2.2 Definition
The dot product A B ⋅ of the vectors
G G
A
G
and B
G
is defined to be product of the magnitude
of the vectors A and with the cosine of the angle
G
B
G
θ between the two vectors:
A B ⋅ = AB cos θ
JG JG
(A.2.1)
Where A =| A | and represent the magnitude of
G
B =| B |
G
A
G
and B
G
respectively. The dot
product can be positive, zero, or negative, depending on the value of cos θ. The dot
product is always a scalar quantity.
We can give a geometric interpretation to the dot product by writing the definition as
A B ⋅ = ( A cos θ) B
G G
(A.2.2)
In this formulation, the term A cos θ is the projection of the vector A
G
in the direction of
the vector. This projection is shown in Figure A.2.2a. So the dot product is the product
of the projection of the length of
B
G
A
G
in the direction of B
G
with the length of B
G
. Note that we could also write the dot product as
A B ⋅ = A B ( cos θ)
G G
(A.2.3)
Now the term (^) B cos θ is the projection of the vector B
G
in the direction of the vector A
G
as shown in Figure A.2.2b.From this perspective, the dot product is the product of the
projection of the length of B in the direction of
G
A
G
with the length of A
G
vector B pointing along the positive -axis with positive -component , i.e.,
G
x x Bx B = B x ˆ i
JG
The vector A can be written as
G
A = A x i + Ay j + Az k
JG
(A.2.9)
We first calculate that the dot product of the unit vector ˆ i^ with itself is unity:
ˆ ˆ i i ⋅ =| ˆ i (^) || ˆ i | cos(0) = 1 (A.2.10)
since the unit vector has magnitude | |ˆ i = 1 and cos(0) = 1. We note that the same rule
applies for the unit vectors in the y and z directions:
ˆ ˆ j j ⋅ = k k ˆ (^) ⋅ ˆ (^) = 1 (A.2.11)
The dot product of the unit vector ˆ i^ with the unit vector ˆ j is zero because the two unit
vectors are perpendicular to each other:
ˆ ˆ i j ⋅ = | ||ˆ i ˆ j | cos( π/2) = 0 (A.2.12)
Similarly, the dot product of the unit vector ˆ i^ with the unit vector k ˆ , and the unit vector ˆ j with the unit vector k ˆ are also zero:
ˆ i k (^) ⋅ ˆ (^) = ˆ j k ⋅ ˆ = 0 (A.2.13)
The dot product of the two vectors now becomes
(A.2.14)
( ˆ^ ˆ^ ˆ) ˆ
ˆ ˆ ˆ ˆ ˆ ˆ property (2a)
(ˆ ˆ ) (ˆ ˆ ) ( ˆ ˆ) property (1a) and (1b)
x y z x
x x y x z x
x x y x z x x x
A A A B
A B A B A B
A B A B A B
A B
A B i j k i
i i j i k i
i i j i k i
G G
This third step is the crucial one because it shows that it is only the unit vectors that undergo the dot product operation.
Since we assumed that the vector B
G
points along the positive -axis with positive - component , our answer can be zero, positive, or negative depending on the -
component of the vector A
G
. In Figure A.2.3, we show the three different cases.
x x Bx x
Figure A.2.3 Dot product that is (a) positive, (b) zero or (c) negative.
The result for the dot product can be generalized easily for arbitrary vectors
A = A x ˆ i^ + Ay ˆ j^ + Az k ˆ
G
(A.2.15)
and
B = B x ˆ i^ + By ˆ j^ + Bzˆ k
G
(A.2.16)
to yield
A B ⋅ = A B x x + A By y + A Bz z
G G
(A.2.17)
A.3 Cross Product
We shall now introduce our second vector operation, called the “cross product” that takes any two vectors and generates a new vector. The cross product is a type of “multiplication” law that turns our vector space (law for addition of vectors) into a vector algebra (laws for addition and multiplication of vectors). The first application of the cross product will be the physical concept of torque about a point which can be described mathematically by the cross product of a vector from
P
P to where the force acts, and the force vector.
A.3.1 Definition: Cross Product
G Let A and B be two vectors. Since any two vectors form a plane, we define the angle
G
θ
to be the angle between the vectors A
G
and B
G
as shown in Figure A.3.2.1. The magnitude
of the cross product A × B of the vectors
G G
A
G
and B
G
is defined to be product of the
magnitude of the vectors A and B
G G
with the sine of the angle θ between the two vectors,
A × B = AB sin θ
G G
(A.3.1)
A × B = A B ( sin θ)
G G
(A.3.2)
The vectors and form a parallelogram. The area of the parallelogram equals the height times the base, which is the magnitude of the cross product. In Figure A.3.3, two different representations of the height and base of a parallelogram are illustrated. As
depicted in Figure A.3.3(a), the term
A
G
B
G
B sin θ is the projection of the vector in the
direction perpendicular to the vector
B
G
A
G
. We could also write the magnitude of the cross product as
A × B = ( A sin θ) B
G G
(A.3.3)
Now the term A sin θ is the projection of the vector A
G
in the direction perpendicular to
the vector B as shown in Figure A.3.3(b).
G
Figure A.3.3 Projection of vectors and the cross product
The cross product of two vectors that are parallel (or anti-parallel) to each other is zero since the angle between the vectors is 0 (or π ) and sin(0) = 0 (or sin( π ) = 0 ).
Geometrically, two parallel vectors do not have any component perpendicular to their common direction.
A.3.3 Properties of the Cross Product
(1) The cross product is anti-commutative since changing the order of the vectors cross product changes the direction of the cross product vector by the right hand rule:
A × B = − B × A
G G G G
(A.3.4)
(2) The cross product between a vector c A
G
where c is a scalar and a vector B is
G
c A × B = c ( A × B )
G G G G
(A.3.5)
Similarly,
A × c B = c ( A × B )
G G G G
(A.3.6)
(3) The cross product between the sum of two vectors A
G
and B
G
with a vector C is
G
( A + B ) × C = A × C + B × C
G G G G G G G
(A.3.7)
Similarly,
A × ( B + C ) = A × B + A × C
G G G G G G G
(A.3.8)
A.3.4 Vector Decomposition and the Cross Product
We first calculate that the magnitude of cross product of the unit vector ˆ i^ with ˆ j :
| ˆ^ ˆ^ | | ˆ^ || ˆ| sin 2
× = ⎜ ⎟ 1
i j i j = (A.3.9)
since the unit vector has magnitude | ˆ i^ | |= ˆ j | = 1 and sin( π / 2) = 1. By the right hand rule,
the direction of ˆ i^ × ˆ j is in the + k ˆ as shown in Figure A.3.4. Thus ˆ i^ × ˆ j = ˆ k.
Figure A.3.4 Cross product of ˆ i^ × ˆ j
We note that the same rule applies for the unit vectors in the y and z directions,
ˆ j (^) × k ˆ (^) = ˆ i (^) , ˆ k × ˆ i = ˆ j (A.3.10)
Note that by the anti-commutatively property (1) of the cross product,
ˆ j (^) × ˆ i (^) = − k ˆ (^) , ˆ i (^) × k ˆ = − ˆ j (A.3.11)
The cross product of the unit vector ˆ^ with itself is zero because the two unit vectors are parallel to each other, ( sin( ),
i
- = 0
