Fundamentals of Vectors and
Mathematical Models
Angie Capece
YESS Physics Instructor
California Institute of Technology
June 2010
Chapter 1
The Fundamentals of Vectors
1.1
Introduction to Scalars and Vectors
Scalars are quantities that are fully described by a value and a unit only. Examples of scalars include:
time, mass, temperature, and length. For example, we can say that the mass of the Earth is 5.9742 × 1024
kg. From this statement, we gather that the value of the mass is 5.9742 × 1024 in units of kilograms. There
is no ambiguity in this statement, and we need no additional information to fully define the mass.
Unlike scalars, vectors require a direction in order to be completely specified. In physics, we use vectors
to describe displacements, velocities, accelerations, and forces. Displacement vectors are used to indicate
a change in the position of an object. When playing chess, we can say that we plan to move the queen 3
squares. This could mean 3 squares forward, 3 squares backward, 3 squares diagonally, etc. To remove the
ambiguity, it is essential to state the direction of the displacement in addition to the value and the unit. If
an object moves from point A to point B, we can fully describe this displacement by stating the distance
moved and the direction in which the object moved. Note that the distance between the two points is the
length of the displacement vector. We call this length the magnitude of the vector, where the magnitude is
a scalar quantity.
Now consider the velocity vector. The magnitude of the velocity vector is given by the speed, which is itself
a scalar quantity. The direction of the velocity vector indicates which way the object is moving. Therefore,
we see that there is a distinction between speed and velocity. There is no spatial direction associated with
the speed. However, the velocity is described by a magnitude (the speed) and a direction.
1.2
Coordinate Systems
In order to describe the direction in which a vector points, we must first set up a coordinate system. This
acts as a reference frame. Since our world is 3-dimensional (that is, the objects around us have length, height,
and depth) we generally set up our coordinate system to also be 3-dimensional. The standard coordinate
system consists of three mutually orthogonal axes, x, y, and z. When we say that the axes are “mutually
orthogonal”, we mean that the axes are all perpendicular to each other. The point where all the axes
intersect is called the origin. The standard conventions for 2D and 3D coordinate systems are shown in Fig.
(1.1). When we set up our coordinate system, we must also indicate which direction is positive and which is
negative. Note that in Fig. (1.1) the solid lines represent the positive axes, while the dotted lines represent
the negative axes. For simplification, most of the systems we will consider in this class will be 2-dimensional.
Therefore, our axes will be represented by x and y only.
Now that we have set up our coordinate system, we can describe the location of a point with respect to the
2
Figure 1.1: 3D and 2D coordinate systems
origin. For example, consider points A, B and C in Fig. (1.2). Point A is located at the origin. Point B
is located 3 units from the origin in the positive x direction and 2 units from the origin in the positive y
direction. Point C is located 5 units in the positive x direction and 1 unit in the positive y direction. That
is: A = (0, 0), B = (3, 2), and C = (5, 1). If an object starts at A and is moved to point B, we can assign
a displacement vector to describe its change in position. We can say that our object moved 3 units in
the positive x direction and 2 units in the positive y direction. If the object now moves from point B to
point C, we can say that our object moved 2 units in the positive x direction and 1 unit in the negative y
direction.
Figure 1.2: Displacement vectors
1.3
Geometric Addition of Vectors
Suppose that our object moves from A to B, and then from B to C as in Fig. (1.2). We can describe the
object’s net displacement from A to C by geometrically adding the two displacement vectors. This allows
−−→
−−→
us to take two displacement vectors (AB and BC) and add them to obtain a single displacement vector
−→
(AC). By doing this, we will determine the net displacement without consideration of the actual path the
−−→
−−→
−→
object took. Denote AB as ~a, BC as ~b, and AC as ~c as shown in Fig. (1.3). We can now represent the net
displacement by the following vector equation:
~c = ~a + ~b
These vectors can be added geometrically by performing the following steps:
1. Sketch ~a to the proper length and angle.
2. Sketch ~b in the same manner with its tail at the head of ~a.
3
(1.3.1)
Figure 1.3: Geometric addition of two displacement vectors
3. The sum of these two vectors is the vector that extends from the tail of ~a to the head of ~b.
Vector addition is both associative and commutative. The associative property implies that the order of
operations does not affect the end result, as long as the sequence of the elements is not changed.
(~a + ~b) + ~c = ~a + (~b + ~c) (associative law )
(1.3.2)
The commutative property implies that the order of the elements can be inverted without affecting the end
result.
~a + ~b = ~b + ~a (commutative law )
(1.3.3)
Vectors can also be subtracted. Note that −~b has the same magnitude as ~b, but is in the opposite direction.
Therefore, we can write:
~b + (−~b) = 0
(1.3.4)
Note that subtracting ~b is equivalent to adding −~b. For example:
~a − ~b = ~a + (−~b)
1.4
(1.3.5)
Vector Components
Since adding vectors geometrically can be very cumbersome, we can instead add them by components. A
vector component is the projection of the vector onto an axis. In a 2-dimensional coordinate system, we
have two axes: the x-axis and the y-axis. Therefore, each vector in 2D space has two components: an x
component and a y component as shown in Fig. (1.4).
We denote the x and y components of ~a by ax and ay , respectively. We can “resolve” the vector into its
components using the following trigonometric relations:
ax = |~a| cos θ
and
ay = |~a| sin θ
(1.4.1)
where |~a| is the magnitude of ~a. (Note: If ~a is a displacement vector, then |~a| gives the length or distance.
If ~a is a velocity vector, then |~a| gives the speed.) There are no vector signs over ax and ay because these
are scalar values.
Using the Pythagorean Theorem, we can write the magnitude of a vector as follows:
4
Figure 1.4: The x and y components of a vector, ~a
2
|~a| = a2x + a2y
(1.4.2)
Equation (1.4.2) is consistent with the relations for ax and ay :
2
|~a| = a2x + a2y = |~a|2 cos2 θ + |~a|2 sin2 θ = |~a|2 (cos2 θ + sin2 θ) = |~a|2
(1.4.3)
Note that the magnitude of a three-dimensional vector is given by:
2
|~a| = a2x + a2y + a2z
1.5
(1.4.4)
Unit Vectors
A unit vector is a vector with a magnitude of 1 that points in a specific direction. In a 3D coordinate
system, the unit vectors in the x, y, and z directions are: î, ĵ, and k̂, respectively. In a 2D coordinate
system, we use only î and ĵ.
Figure 1.5: 3D coordinate system showing the unit vectors, î, ĵ, and k̂
We can use unit vectors to help describe other vectors. For example, we can write ~a as:
~a = ax î + ay ĵ + az k̂.
5
(1.5.1)
Equation (1.5.1) above means that ~a has a component, ax , in the x direction, a component, ay , in the y
direction, and a component, az , in the z direction. Using the example given in Fig. (1.2), we can express
−−→
the vector, AB as:
−−→
AB = 3î + 2ĵ.
(1.5.2)
Note that ax , ay , and az are scalar quantities.
1.6
Vector Notation
We can express vectors in different forms. Here are some examples:
~a = ax î + ay ĵ + az k̂
~a = (ax , ay , az )


ax
~a =  ay 
az
(1.6.1)
(1.6.2)
(1.6.3)
In Section 1.5, we learned that unit vectors have magnitude equal to 1 and point in a particular direction.
We can write our unit vectors as:
î = (1, 0, 0)
ĵ = (0, 1, 0)
k̂ = (0, 0, 1)
(1.6.4)
This representation tells us that î is a vector that points only in the +x direction, ĵ is a vector that points
only in the +y direction, and k̂ is a vector that points only in the +z direction.
In this class, we will use the following vector notation:
~a = ax î + ay ĵ + az k̂
(1.6.5)
It is important to remember that the components, ax , ay , az , are scalar quantities that are multiplied by
the unit vectors, î, ĵ, k̂, to give them a direction.
1.7
1.7.1
Vector Algebra
Vector Addition and Subtraction
As we saw in Section 1.3, vector addition is both commutative and associative. Vectors can be added by
equating their components. For example, consider the following relation:
~c = ~a + ~b
(1.7.1)
In order for this statement to be true, each component of the left hand side must be equivalent to the
corresponding component on the right hand side. We can write each term in component form:
6
cx î + cy ĵ + cz k̂ = (ax î + ay ĵ + az k̂) + (bx î + by ĵ + bz k̂).
(1.7.2)
cx î + cy ĵ + cz k̂ = (ax + bx )î + (ay + by )ĵ + (az + bz )k̂.
(1.7.3)
Regroup terms to obtain:
Finally, equate the components to obtain a set of three equations:
cx = ax + bx
cy = ay + by
cz = az + bz .
(1.7.4)
It is important that we only add quantities that are dimensionally equivalent. For example, we would not
add a quantity that has units of length to a quantity that has units of velocity. In the same manner, we
cannot add a vector to a scalar quantity. We can only add vectors to other vectors, and scalars to other
scalars.
1.7.2
Scalar Multiplication
Vectors can be multiplied by scalars to get new vectors. This is called scalar multiplication. Physically, it is
equivalent to stretching or shrinking the vector. If the scalar is positive, only the magnitude of the vector
will change. The direction will remain the same. However, if the scalar is negative, the magnitude of the
vector will change and the direction of the vector will flip. Scalar multiplication is distributive, and so we
can write the following:
s~a = s(ax î + ay ĵ) = sax î + say ĵ
(1.7.5)
We can also divide a vector by a scalar, s, by multiplying the vector by 1/s.
1.7.3
Dot Product
The dot product is a special type of multiplication between two vectors, which returns a scalar quantity.
The definition of the dot product is given below.
~a · ~b = ax bx + ay by + az bz
(1.7.6)
~a · ~b = ~b · ~a.
(1.7.7)
The dot product is commutative:
A physical way of understanding the dot product is to think of it in terms of a projection of one vector onto
another. When we dot two vectors, ~a and ~b, we are multiplying the length of the projection of ~a onto ~b by
the length of ~b.
From Fig. (1.6), we see that the quantity, |~a| cos α is the length of the projection of ~a onto ~b. Therefore, we
can express the dot product as:
~a · ~b = |~a||~b| cos α
7
(1.7.8)
Figure 1.6: The projection of ~a onto ~b.
where α is the angle between ~a and ~b.
If the two vectors, ~a and ~b, are perpendicular to each other, there will be no component of ~a along ~b and the
dot product will be equal to zero. We can arrive at this result by considering Eq. (1.7.8). If the two vectors
are perpendicular, then α = 90◦ . Since the cosine of 90◦ is zero, the dot product is zero.
Now suppose that ~b is the unit vector, î. Using our definition of the dot product, we obtain:
~a · ~b = ~a · î = (ax î + ay ĵ + az k̂) · (î + 0ĵ + 0k̂) = ax · 1 + ay · 0 + az · 0 = ax
(1.7.9)
We therefore see that by dotting a vector ~a with a unit vector, we obtain the component of ~a along the unit
vector. Note that the magnitude of a vector can be described in terms of the dot product:
2
~a · ~a = a2x + a2y + a2z = |~a|
(1.7.10)
We therefore see that the square of the magnitude of a vector is equivalent to the dot product of that vector
with itself:
|~a|2 = ~a · ~a
1.7.4
(1.7.11)
Cross Product
The cross product is an operation between two vectors, ~a and ~b, and is denoted as:
~a × ~b = ~c.
(1.7.12)
The cross product returns a third vector, ~c, that is perpendicular to both ~a and ~b. The cross product is
defined in three dimensions, and is given by:
~a × ~b = (ay bz − by az )î + (az bx − bz ax )ĵ + (ax by − bx ay )k̂
(1.7.13)
The cross product can also be expressed as:
~c = |~a||~b| sin α n̂
8
(1.7.14)
where α is the angle between ~a and ~b, and n̂ is the unit vector perpendicular to the plane containing ~a and
~b as shown in Fig. (1.7).
Figure 1.7: Geometric depiction of the cross product, ~c = ~a × ~b
Using the above relations, we can see that the cross product between two parallel vectors is zero. Recall that
the angle between two parallel vectors is either 0◦ or 180◦ . The sine of 0◦ and 180◦ is zero, and therefore
the cross product is zero.
Note the following equalities:
(î × ĵ) = k̂
(ĵ × k̂) = î
(k̂ × î) = ĵ
(1.7.15)
The cross product is not commutative:
~a × ~b = −(~b × ~a)
1.8
(1.7.16)
Example Problems
1. Point A is located 25 meters from the origin in the negative x direction and 40 meters from the origin in
the positive y direction. Let ~a be the displacement vector that extends from the origin to point A. a) What
is the magnitude of ~a? b) What is the angle between the direction of ~a and the positive x direction?
SOLUTION:
In vector notation, we have:
~a = −25 m î + 40 m ĵ.
(1.8.1)
p
(−25)2 + 402 m = 47.17 m
~
b) ax = A
cos θ
Substitute to obtain: −25 m = (47.17 cos θ) m. We therefore see that θ = 122◦ .
a) |~a| =
2. A ship sets out to sail to a point 120 km due north. An unexpected storm blows the ship to a point
100 km due east of its starting point. How far, and in what direction must it now sail to reach its original
destination?
SOLUTION: See the figure below.
Let east correspond to î, west correspond to −î, north correspond to ĵ, and south correspond to
−ĵ. To reach its destination, the ship must travel 100 km in the −î direction and 120 km in the
9
ĵ direction. Its displacement vector is therefore: ~c = −100 km î + 120 km ĵ.
p
(−100)2 + 1202 km = 156.2 km
cy
120 km
Direction: tan θ =
=
, therefore we see that θ = 50.2◦ .
cx
−100 km
Magnitude: |~c| =
3. Consider the rotated coordinate system in the figure below. Determine the components, a1 and a2 , of the
vector, ~a.
SOLUTION:
a1 = |~a| sin θ
a2 = |~a| cos θ
~ = 5î + 2ĵ + 1k̂ and B
~ = 8î + −3ĵ + 2k̂
4. Find the dot product of the following two vectors: A
~·B
~ = (5 · 8) + 2 · (−3) + (2 · 1) = 36
SOLUTION: A
10
Chapter 2
Mathematical Models
Mathematical models are used to describe trends in data. If there is no theoretical law to describe a certain
phenomenon, an empirical model is often formulated from curve-fits of experimental data. Several types
of curves often appropriately describe experimental data including: lines, polynomials, power functions,
exponential functions, and logarithmic functions. When a given curve is used to fit a given set of data, we
refer to the fit as a model of the data. For example, when data is fitted to a line we call this a linear model.
Each of the aforementioned models will be defined in the following sections. A short discussion on how to
use Excel to curve-fit experimental data to one of these models is also presented.
2.1
Lines
Points (x, y) that lie on a line satisfy the relationship:
y = f (x) = mx + b,
(2.1.1)
where m is the slope of the line and b is the y-intercept. An example of a linear function is shown below in
Fig. (2.1).
Figure 2.1: Graph of f (x) = mx + b
2.2
Polynomials
A function, f (x), is considered a polynomial if it has the form
f (x) = an xn + an−1 xn−1 + . . . + a2 x2 + a1 x + a0 ,
11
(2.2.1)
where n is a non-negative integer and a1 , a2 , . . . , an are constant coefficients. Note the following cases:
1. A linear function is a polynomial is of degree n = 1:
f (x) = a1 x + a0
(2.2.2)
2. A quadratic function is a polynomial of degree n = 2:
f (x) = a2 x2 + a1 x + a0
(2.2.3)
3. A cubic function is a polynomial of degree n = 3:
f (x) = a3 x3 + a2 x2 + a1 x + a0
(2.2.4)
Examples of polynomials are shown below in Fig. (2.2).
Figure 2.2: Examples of polynomials
2.3
Power Functions
A function, f (x), is considered a power function if it has the form
f (x) = xa ,
(2.3.1)
where a is a constant. Examples of various power functions are shown below in Fig. (2.3).
Figure 2.3: Graphs of f (x) = xn
2.4
Exponential Functions
A function, f (x), is considered an exponential function if it has the form
f (x) = ax ,
12
(2.4.1)
where a is a positive constant. Graphs of various exponential functions are shown in Fig. (2.4).
Figure 2.4: Graph of f (x) = ax
2.5
Logarithmic Functions
A function, f (x), is considered a logarithmic function if it has the form
f (x) = loga x,
(2.5.1)
where base a is some positive constant. The logarithmic function is the inverse of the exponential function.
Graphs of logarithmic functions with different bases are shown in Fig. (2.5) below.
Figure 2.5: Graph of f (x) = loga x
2.6
Creating Empirical Models
In this section we will use Excel to plot data and determine an appropriate model (or curve-fit) that describes
the thermal expansion of a copper bar. The table below lists data for the length of the copper bar at various
temperatures. From the data we can see that the bar expands as the temperature is increased. We would
like to find a model to describe how the change in length of the bar varies with the change in temperature. In
other words, we would like to find the change in length of the bar as a function of the change in temperature.
13
Temperature (◦ C)
20
30
40
50
60
70
80
90
100
Length (m)
1.00000
1.00018
1.00033
1.00051
1.00070
1.00083
1.00100
1.00120
1.00138
Table 2.1: Thermal Expansion Data
In order to do this we will first create a new data set for ∆L and ∆T . We will then plot the data in Excel
and add a trendline. Follow the instructions given below.
1. Modify the current data set:
Let ∆L = L − L0 , where L0 is the reference length. Let ∆T = T − T0 , where T0 is the reference temperature.
Take the first data point in the table to be the reference, i.e. T0 = 20◦ C and L0 = 1.00000 m.
2. Plot the data in Excel:
Input the data for ∆L and ∆T into Excel. The first point in your modified data set should be (0,0). It
is important to include this as it is a valid data point in this particular example and will improve the fit.
Put the x-data in column A and the y-data in column B. (Note that in this example, the x-data is ∆T and
the y-data is ∆L since we are looking for the change in length of the bar as a function of the change in
temperature.) Select the two columns containing the data you would like to plot then go to Insert and click
on Chart. On the Standard Types tab select XY (Scatter). Click Next. Click Next again. On the Titles tab,
enter a chart title and labels for the x and y axes. Click Finish.
3. Add a trendline to the data:
Left click on any data point in the graph to highlight the data series. Right click and select Add Trendline.
On the Type tab, click the type of regression trendline you want. (Note: If you select Polynomial, enter
in the Order box the highest power for the independent variable.) On the Options tab, check the boxes
to Display equation on chart and Display R-squared value on chart. Note that if (0,0) is included in the
data set, Excel will limit the regression lines you can choose from to only linear and polynomial. This is an
oddity of Excel; other curve-fitting programs will do their best to fit data to a given curve, no matter the
data. This odd feature of Excel will not pose a problem in this class because we will mostly be fitting to
polynomial curves when the point (0,0) is included in our data set.
3. Evaluate the curve-fit:
Excel will fit a curve to the data set based on your selection of regression trendline. Try fitting your data
with various curves to see which model best describes the data. The R2 value is a statistical measure of how
well the regression line approximates the data points. Therefore, it is an important parameter to consider
when evaluating the fit of your data to the mathematical model. Values of R2 that are close to 1 indicate
a good fit. Note that R2 = 1 indicates that the line perfectly fits the data. What model best describes
the data given in the table above? What is the resulting equation? What is the R2 value? To increase the
number of decimal places given in the equation and R2 value printed on the chart in Excel, do the following:
click on the equation then go to Format on the Menu Bar, click Selected Data Labels..., on the Number tab
use the up and down arrows to adjust the decimal places, click OK.
14
Chapter 3
Homework Problems
Please work out the following homework problems to be handed in when you arrive at Caltech. Be sure to
use vector notation.
1. Two vectors are given by: ~a = 4î − 3ĵ + 1k̂ and ~b = −1î + 1ĵ + 4k̂. Determine the following quantities:
~a + ~b, ~a − ~b, and a third vector, ~c, such that ~a − ~b + ~c = 0.
2. Two vectors are given by: ~a = 3î + 5ĵ + 2k̂ and ~b = 2î + 4ĵ + 6k̂. Determine the following quantities:
~a × ~b, (~a + ~b) · ~b, and ~a · ~b.
3. If v~1 + v~2 = 5 v~3 , v~1 − v~2 = 3 v~3 , and v~3 = 2î + 4ĵ, find v~1 and v~2 .
4. a) Calculate (î × ĵ). What angle does this new vector make with î and ĵ? b) Now calculate (ĵ × î).
Compare your answer to the one you obtained in part a. Is the cross product commutative? c) Calculate
(k̂ × k̂). What do you get when you cross two vectors that are parallel to each other? d) Calculate (k̂ · k̂).
What do you get when you dot two vectors that are parallel to each other? e) Calculate (ĵ · k̂). What do
you get when you dot two vectors that are perpendicular to each other?
5. Calculate the angle between the following two vectors: ~a = 3î + 3ĵ + 3k̂ and ~b = 2î + 1ĵ + 3k̂.
Hint: Consider the definition for the dot product.
6. Two vectors are given by: f~ = 4î − 3ĵ and ~g = 6î + 8ĵ. a) What is the magnitude of f~ ? b) What is the
angle of f~ relative to î? c) What is the magnitude of ~g ? d) What is the angle of ~g relative to î? e) What
is the magnitude of f~ + ~g ? f) What is the angle of f~ + ~g relative to î? g) What is the angle between the
directions of f~ − ~g and ~g − f~ ?
7. A room has dimensions 3 m (height) x 3.7 m x 4.3 m. A fly starting at one corner flies around, ending up
at the diagonally opposite corner. a) What is the magnitude of its displacement? b) Could the length of the
fly’s path ever be less than this magnitude? If so, when? c) Could the length of the fly’s path ever be greater
than this magnitude? If so, when? d) Could the length of the fly’s path ever be equal to this magnitude? If
so, when? e) Choose a suitable coordinate system and express the components of the displacement vector
in that system. f) If the fly walks instead of flying, what is the length of the shortest path? Hint: The room
is like a box. Unfold its walls to flatten them into a plane.
8. In the following figure, a cube of edge length L sits with one corner at the origin of an xyz coordinate
15
system. A body diagonal is a line that extends from one corner to another through the center.
In vector notation, what is the body diagonal that extends from the corner at the following coordinates: a)
(0, 0, 0), b) (L, 0, 0), c) (0, L, 0), d) (L, L, 0), e) Determine the angles that the body diagonals make with the
two adjacent edges. f) Determine the length of the body diagonals in terms of a.
9. A rock is dropped from the top of a cliff at a height of 500 meters above the ground. The rock’s height
above the ground is recorded at various times during its descent. This data is presented in the table below.
Plot the data in Excel and find a model to describe how the height changes with time. Print out a graph
of your data and the trendline that best fits the data. List the equation and R2 value. Use your model to
predict the time at which the ball hits the ground.
Time (seconds)
0
1
2
3
4
5
6
7
8
8.5
9
9.5
Height (meters)
500
495
480
456
422
377
323
260
186
146
103
57
Table 3.1: Height and Time Data for a Falling Object
16