Lesson 1
Chapter 1: Units, Physical Quantities, and Vectors
Course Plan
Week Date
Chapter
Topics
Units, Physical Quantities, and Vectors
1
20 February 2023
Chapter 1
2
27 February 2023
Chapter 2 and 3 Kinematics
3
6 March 2023
Chapter 4
Newton's Laws of Motion
4
13 March 2023
Chapter 5
Applying Newton's Laws
5
20 March 2023
Chapter 6
Work and Kinetic Energy
6
27 March 2023
Chapter 7
Potential Energy and Energy Conservation
7
3 April 2023
Chapter 8
Momentum, Impulse, and Collisions
8
10 April 2023
Chapter 8
Momentum, Impulse, and Collisions
9
17 April 2023
Chapter 9
Kinematics of Rotational Motion
10
24 April 2023
Chapter 10
Dynamics of Rotational Motion
11
1 May 2023
Chapter 10
Dynamics of Rotational Motion
12
8 May 2023
Chapter 12
Gravitation
13
15 May 2023
Chapter 13
Periodic Motion
14
22 May 2023
Chapter 13
Periodic Motion
2
Units, Physical Quantities, and Vectors
Nature of Physics
Physics wants to understand the nature
Understanding the nature must relay on mathematical base
Physics interprets the nature on the mathematical base
It is a basic science that reveals the causes of the consequences of natural
phenomena as a mathematical model.
Examines the results of the observations, reveals theories about the reasons,
derives mathematical expressions.
Checks the validity of these statements and makes corrections or additions if
necessary.
3
Units, Physical Quantities, and Vectors
Measurements
Measurements are important for consistency of observations.
Physical quantities should be measurable.
Examples of quantities that can be measured: length, temperature, time, etc.
We must express the magnitudes of these quantities, and we use units for this.
Different units can be used for a physical quantity. We must also define the
conversions between these units.
Meter (m) is the basic unit for length in the SI unit system (International System of
Units - Le Système International d'Unités).
Miles (mi) is the unit of length for the Royal Unit System (Imperial units).
4
Idealized Model
What we call a model in physics is a simplified version of a very
complex system to be fully understood.
The figure shows a tennis ball thrown into the air.
When we try to examine the system in all its details, we should
consider each physical event during the movement of the ball, such
as the rotation of the ball around its own axis, air resistance, and the
shape of the ball due to not being a solid object.
In such a case, the solution to the problem will be very complex.
If we start from the simplest case instead, we can proceed.
5
Idealized Model
1. We can neglect air friction in the initial approximation, assuming that air resistance is
not dominant over other effects on the case.
2. We can assume the object is a point object. We can make this assumption if the
distance the object moves is too great for the dimensions of the object.
3. The point-mas acceptance will also eliminate the rotation
of the object around its own axis, since the point-mass will
be one-dimensional, the rotation is undefined and the
change of its shape is also undefined.
Under these assumptions, the motion of the tennis ball was
modeled and turned into a solvable problem with the initial
approximation.
When analyzing the results, we should not overlook under
which approaches we obtained these results.
6
Standard Units
Measurable Physical Quantities:
Length can be measured by ruler or meter.
Time can be measured with a stopwatch.
Describable (Operational description) Physical quantities:
Average Speed: It is the measure of how long a certain distance has been covered, size
can be obtained from the recipe. For this, the distance and how long it takes to cover this
distance must be measured.
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑 =
𝐸𝑙𝑎𝑝𝑠𝑒𝑑 𝑡𝑖𝑚𝑒
The unit system has been defined so that the measurements made by different people at
different places and at different times are comparable.
Turkey and many other countries use the SI units. (International System of Units - Le
Système International d'Unités)
7
SI Units
Time:
seconds (s)
Length:
Meters (m)
There are Standard definitions for mass, length and time.
Over time, new definitions have been created with more
precise precision.
Mass:
Kilogram (kg)
1000 gram (g)
8
Huge – Tiny
We must describe physical quantities with appropriate scales.
We express the length of a straight line in our notebook in centimeters or millimeters.
We express our height in centimeters or meters.
We express our mass in kilograms and our age in years.
We indicate the distance between the Earth and the Moon in kilometers, the distance
between galaxies in light years, the distance between atoms in nanometers, and the
duration of fluorescence in nanoseconds.
9
Unit Prefixes
We need to describe physical quantities with appropriate scales.
Multiplier
Name
Symbol
Mass
Length
Time
Memory Units
1015
Peta
P
1012
Tera
T
TB
109
Giga
G
GB
106
Mega
M
MB
103
Kilo
k
kg
km
100
-
-
[gram]
[metre]
[second]
10−3
Milli
m
mg
mm
ms
10−6
Micro
m
mg
mm
ms
10−9
Nano
n
nm
ns
10−12
Pico
p
10−15
Femto
f
[byte]
fs
10
Unit Prefixes
In addition to these, there are also inter units that we use.
Multiplier
Name
Symbol
Length
Volume
103
Kilo
k
km
102
Hecto
H
Hm
101
Deca
D
Dm
100
-
-
[metre]
[Litre]
10−1
Deci
d
dm
dL
10−2
Centi
c
cm
cL
10−3
Mili
m
mm
mL
11
Unit Prefixes
Below are the approximate sizes of some objects:
12
Unit Consistency and Unit Conversions
We use symbols in expressions to correlate physical quantities. For example:
We use the symbols 𝑑 for distance, 𝑡 for time, and 𝑣 for velocity.
If 𝑑; in meters, 𝑡; in seconds:
𝑑
𝑣=
𝑡
𝑣; will be in meter/seconds.
An equation should be dimensionally consistent. The speed calculated here must be in
the unit of speed. Precisely speed is a measure of displacement, that is, how much time
has been traveled a given distance.
Its unit must be in the distance / time unit. From the SI system, this is m / s.
km / h (kilometer / hour) is also the unit of speed.
We talked about the SI and Royal Unit systems.
The Royal Unit system can use miles per hour for speed.
Example 1.1
Converting speed units: The official world land speed record is 1228 km/h, set on
October 15, 1997, by Andy Green in the jet engine car Thrust SSC (Thrust Super
Sonic Car). Express this speed in meters per second.
km
km 1000 m 1 h
𝑣 = 1228
= 1228
h
h 1 km 3600 s
1000 m
𝑣 = 1228
= 341.11 m/s
3600 s
14
Uncertainty and Significant Figures
15
Order of Magnitude Estimation
In some cases, we can make a rough estimate of the situation without detailed
calculation.
It is a situation we use very often in daily life. By rounding the available data, we can
calculate the magnitude we want with a good approximation.
When we sit in a cafe, looking around, most of us think about how well this place has
earned, and we do a rough finger count to calculate how much that amount is.
How many tables are around, how long does a customer stay at the table on
average and what is the average bill they pay, how many days a week, how many
hours a day? Here are the necessary information for the endorsement. Of course
there are also expenses, ...
The difference will give you a rough overview of the earnings of this cafe. Of course,
you do not trust this rough calculation and open a cafe, you will make a detailed
feasibility study for this.
16
Example 1.4
An order-of-magnitude estimate: You are writing an adventure novel in which the
hero escapes across the border with a billion dollars' worth of gold in his suitcase. Is
this possible? Would that amount of gold fit in a suitcase? Would it be too heavy to
carry?
3
19.3
g/cm
In
real
1 ons Gold
200 ~ 600
1 ons ≈ 30 g
assuming
1 Billion $ Gold ≈ 108 g
We got
if
100 Ton
10 m3
50 Ton ~ 200 Ton
1 g/cm3
108 cm3
10 g/cm3
100 m3
Do the calculation for 100000 $ worth
5 carat (1 gram) diamonds.
Is the result more plausible?
17
Vectors
A number and with a unit may be sufficient to express some physical quantities
such as temperature, mass, density.
Since the direction is also associated with some other quantities, only number and
unit are not enough. The motion of the plane is an example for such a situation
In such cases, we must express these quantities with vectors.
Vectors in three dimensional physical space will also have three components.
𝐴Ԧ
18
Vectors
𝐴Ԧ
For example, we must use a vector to express displacement.
2
1
𝑟Ԧ
When the object is displaced from 1 to 2
over any path, the net displacement must
be expressed with a vector.
The magnitude of this vector 𝑟 Ԧ shown in blue indicates the amount of
displacement and the direction of the arrow indicates the direction of the
displacement.
As can be seen, the displacement was from 1 to 2.
Changing the direction of the arrow is sufficient to express the displacement from
2 to 1.
19
Vectors
𝐴Ԧ
𝐵
𝐴Ԧ
−𝐴Ԧ
𝐶Ԧ
Vectors showing the same direction are called
parallel, and vectors pointing opposite directions
are called anti-parallel.
Two vectors of the same length but opposite
direction are equal to each other with the (-) sign.
Ԧ indicated by 𝐴Ԧ
The magnitude of the vector is 𝐴|
Displacement cannot be interpreted as 𝐴Ԧ = 6 m. The statement should also
contain the information for direction.
20
Vector Operations
1. Vector Sum (addition, subtraction)
2. Vector Product (Scalar, Vectorial)
𝐴Ԧ
𝐵
21
Vector addition and subtraction
Addition of vector 𝐴Ԧ and 𝐵 :
𝐵
𝐴Ԧ
𝐵
𝐶Ԧ = 𝐴Ԧ + 𝐵
𝐴Ԧ
Subtraction vector 𝐵 from 𝐴Ԧ :
−𝐵
𝐴Ԧ
−𝐵
𝐵
𝐴Ԧ
𝐷 = 𝐴Ԧ − 𝐵 = 𝐴Ԧ + −𝐵
22
Vector addition and subtraction
Sum of more than one vector:
𝐶Ԧ
𝐸
𝐵
𝐷
𝐺Ԧ = 𝐴Ԧ + 𝐵 + 𝐶Ԧ + 𝐷 + 𝐸 + 𝐹Ԧ
𝐹Ԧ
𝐴Ԧ
𝐺Ԧ
23
Example 1.5
A cross-country skier skis 1 km north and then 2 km east on a horizontal snow field.
How far and in what direction is she from he starting point?
Net displacement is in North-East direction.
North
𝐵
𝐴Ԧ
East
𝐶Ԧ = 𝐴Ԧ + 𝐵
The magnitude of the displacement can be
found from the right triangle:
𝐶Ԧ
2
𝐶Ԧ
2
2
= 𝐴Ԧ + 𝐵
2
= (1 km)2 +(2 km)2 = 5 km2
𝐶Ԧ = 5 km = 2.24 km
For the direction, we can determine the angle the
plane makes with the north:
2 km
tan 𝜃 =
=2
1 km
𝜃 = 63.4°
24
Vector Components
The sum of two or more vectors is
again a vector.
We can write a vector as the sum of
other vectors.
Specifically, we can choose these two
vectors to be perpendicular to each other.
𝐴Ԧ
𝐴Ԧ2
𝐴Ԧ
𝐴Ԧ2
𝐴Ԧ1
𝐴Ԧ1
𝐴Ԧ = 𝐴Ԧ1 + 𝐴Ԧ2
𝐴Ԧ = 𝐴Ԧ1 + 𝐴Ԧ2
If vector 𝐴Ԧ1 is the component of 𝐴Ԧ in
the 𝑥-direction and vector 𝐴Ԧ2 is the
component of 𝐴Ԧ in the 𝑦-direction, ;
𝐴Ԧ = 𝐴Ԧ𝑥 + 𝐴Ԧ𝑦
𝐴Ԧ1 and 𝐴Ԧ2 vectors are the components
Ԧ
of the vector 𝐴.
25
Vector Components
y
𝐴Ԧ
𝐴Ԧ𝑦
𝜃
0
We can define the magnitudes of the components of the vector 𝐴Ԧ in the
direction of 𝑥 and 𝑦 and interpret the vector 𝐴Ԧ as its components:
𝐴Ԧ𝑥
x
𝑎𝑥 = 𝐴Ԧ𝑥 = 𝐴Ԧ cos(𝜃)
𝐴Ԧ𝑥 / 𝐴Ԧ𝑥
Gives the unit vector 𝑖Ƹ that is in the direction of 𝑥.
𝐴Ԧ𝑦 / 𝐴Ԧ𝑦
Gives the unit vector 𝑗Ƹ that is in the direction of 𝑦.
In terms of the unit vector 𝑖Ƹ and 𝑗Ƹ
vector a can be written as:
𝐴Ԧ = 𝑎𝑥 𝑖 Ƹ + 𝑎𝑦 𝑗 Ƹ
𝑎𝑦 = 𝐴Ԧ𝑦 = 𝐴Ԧ sin(𝜃)
With the additional component in the third direction in three
dimensional space;
𝐴Ԧ = 𝑎𝑥 𝑖 Ƹ + 𝑎𝑦 𝑗 Ƹ + 𝑎𝑧 𝑘መ
26
Example 1.6
Finding components: (a) What are the 𝑥 and 𝑦-components of vector 𝐷 in Fig. 1.19 a? The
magnitude of the vector is 𝐷 = 3 m and the angle 𝛼 = 45°. (b) What are the 𝑥 and 𝑦components of vector 𝐸 in Fig. 1.19 b? The magnitude of the vector is 𝐷 = 4.5 m and the angle
𝛽 = 37°.
𝐷𝑥 = 𝐷 cos 𝛼 = 3 m cos −45° = +2.41 m
𝐷𝑦 = 𝐷 sin 𝛼 = 3 m sin −45° = −2.41 m
𝐸𝑥 = 𝐸 sin 𝛽 = 4.5 m sin 37° = +2.71 m
𝐸𝑦 = 𝐸 cos 𝛽 = 4.5 m cos 37° = +3.59 m
27
Vector Operations with Components Method
y
Ԧ we can calculate the direction and
From the components of 𝐴,
magnitude ( 𝐴Ԧ ) of the vector 𝐴Ԧ
𝐴Ԧ
𝐴Ԧ𝑦
𝜃
0
𝐴Ԧ𝑥
𝐴Ԧ𝑦
tan 𝜃 =
𝐴Ԧ𝑥
The direction of the vector 𝐴Ԧ can be stated with respect to the 𝑥-axis.
x
𝐴Ԧ
and
2
= 𝐴Ԧ𝑥
𝜃 = 𝑎𝑟𝑐𝑡𝑎𝑛
2
+ 𝐴Ԧ𝑦
𝐴Ԧ𝑦
𝐴Ԧ𝑥
2
= 𝑡𝑎𝑛−1
𝐴Ԧ𝑦
𝐴Ԧ𝑥
28
Vector Product
Vector multiplied by a scalar :
𝐷 = 𝑐 𝐴Ԧ
Multiplying vector 𝐴Ԧ with a scalar like 𝑐 give a new vector like 𝐷.
𝐴Ԧ
𝐷 = 𝑐 𝐴Ԧ
Ԧ If 𝑐 is positive (𝑐 ≥ 0) 𝐷 is also in the same direction,
𝐷 has the same alignment with 𝐴.
Ԧ
and the magnitude of 𝐷 is 𝑐 times the magnitude of 𝐴.
Multiplying a vector by a scalar is equivalent to multiplying its components
by that scalar.
𝐷𝑥 = 𝑐 𝐴Ԧ𝑥
𝐷 = 𝑐 𝐴Ԧ = 𝑐 𝑎𝑥 𝑖 Ƹ + 𝑎𝑦 𝑗 Ƹ = 𝑐𝑎𝑥 𝑖 Ƹ + 𝑐𝑎𝑦 𝑗 Ƹ = 𝑐 𝐴Ԧ𝑥 + 𝑐 𝐴Ԧ𝑦
𝐷𝑦 = 𝑐 𝐴Ԧ𝑦
29
Vector Sum by Components Method
y
𝐶Ԧ = 𝐴Ԧ + 𝐵
𝐶Ԧ
𝐵𝑦
𝐵
𝐴Ԧ = 𝑎𝑥 𝑖Ƹ + 𝑎𝑦 𝑗Ƹ
𝐴Ԧ𝑥 = 𝑎𝑥 𝑖Ƹ
𝐴Ԧ𝑦 = 𝑎𝑦 𝑗Ƹ
𝐵 = 𝑏𝑥 𝑖Ƹ + 𝑏𝑦 𝑗Ƹ
𝐵𝑥 = 𝑏𝑥 𝑖Ƹ
𝐵𝑦 = 𝑏𝑦 𝑗Ƹ
𝜃𝐵
𝐴Ԧ𝑦
0
𝜃
Ԧ
𝜃𝐴 𝐴
𝐴Ԧ𝑥
x
𝐵𝑥
The lengths of vectors in the same direction can be summed directly.
𝑎𝑥 𝑖Ƹ + 𝑏𝑥 𝑖Ƹ = 𝑐𝑥 𝑖Ƹ
𝑎𝑦 𝑗Ƹ + 𝑏𝑦 𝑗Ƹ = 𝑐𝑦 𝑗Ƹ
𝐶Ԧ = 𝑐𝑥 𝑖Ƹ + 𝑐𝑦 𝑗Ƹ
The sum of 𝐴Ԧ and 𝐵 is again a vector.
The magnitude of this vector is
The alignment of this vector with
respect to the 𝑥 axis is
𝐶Ԧ
2
= 𝐶Ԧ𝑥
2
+ 𝐶Ԧ𝑦
𝜃 = 𝑡𝑎𝑛−1
2
𝐶Ԧ𝑦
𝐶Ԧ𝑥
30
Vector Sum by Components Method
Addition of more than one vector:
𝑅 = 𝐴Ԧ + 𝐵 + 𝐶Ԧ + 𝐷 + ⋯ + 𝑍Ԧ
𝑅𝑥 𝑖Ƹ = 𝑎𝑥 𝑖Ƹ + 𝑏𝑥 𝑖Ƹ + 𝑐𝑥 𝑖Ƹ + 𝑑𝑥 𝑖Ƹ + ⋯ + 𝑧𝑥 𝑖Ƹ
and
𝑅𝑦 𝑗Ƹ = 𝑎𝑦 𝑗Ƹ + 𝑏𝑦 𝑗Ƹ + 𝑐𝑦 𝑗Ƹ + 𝑑𝑦 𝑗Ƹ + ⋯ + 𝑧𝑦 𝑗Ƹ
𝑅 = 𝑟𝑥 𝑖Ƹ + 𝑟𝑦 𝑗Ƹ
If the vectors are three-dimensional:
𝑅𝑧 𝑘෠ = 𝑎𝑧 𝑘෠ + 𝑏𝑧 𝑘෠ + 𝑐𝑧 𝑘෠ + 𝑑𝑧 𝑘෠ + ⋯ + 𝑧𝑧 𝑘෠
𝑅 = 𝑟𝑥 𝑖Ƹ + 𝑟𝑦 𝑗Ƹ + 𝑟𝑧 𝑘෠
𝑛
A 𝑛-dimensional vector
𝑅 = ෍ 𝑎𝑖 𝑖 Ƹ
𝑖=1
𝑎𝑖 is the magnitude of the component of the
vector in the direction of 𝑖 and 𝑖Ƹ is the unit
vector in this direction.
31
Example 1.7 (1/2)
Adding vectors with components: Three players on a reality TV show are brought to the center of a large, flat
field. Each is given a meter stick, a compass, a calculator, a shovel, and (in a different order for each
contestant) the following three displacements:
72.4 m, 32.0" east of north, 57.3 m, 36.0" south of west, 17.8 m straight south
The three displacements lead to the point where the keys to a new Porsche are buried. Two players start
measuring immediately, but the winner first calculates where to go. What does she calculate?
𝑅 = 𝐴Ԧ + 𝐵 + 𝐶Ԧ
North
36°
𝐵
𝐴Ԧ
𝐶Ԧ
32°
West
𝑅
𝐴Ԧ = 𝑎𝑥 𝑖Ƹ + 𝑎𝑦 𝑗Ƹ = 72.4 sin(32°) 𝑖Ƹ + cos(32°) 𝑗Ƹ
0
South
𝐵 = 𝑏𝑥 𝑖Ƹ + 𝑏𝑦 𝑗Ƹ = 57.3 cos(−36°) 𝑖Ƹ + sin(−36°) 𝑗Ƹ
East
𝐶Ԧ = 𝑐𝑥 𝑖Ƹ + 𝑐𝑦 𝑗Ƹ = 72.4 cos(−90°) 𝑖Ƹ + sin(−90°) 𝑗Ƹ
𝑅 = 𝑎𝑥 + 𝑏𝑥 + 𝑐𝑥 𝑖Ƹ + 𝑎𝑦 + 𝑏𝑦 + 𝑐𝑦 𝑗Ƹ
32
Example 1.7 (2/2)
North
36°
𝐵
32°
𝑅
𝐵 = −46.36 m 𝑖Ƹ − 33.68 m 𝑗Ƹ
𝐴Ԧ
𝐶Ԧ
West
𝐴Ԧ = 38.37m 𝑖Ƹ + 61.40 m 𝑗Ƹ
0
Eat
𝐶Ԧ = 0 m 𝑖Ƹ − 17.80 m 𝑗Ƹ
𝑅 = 𝑎𝑥 + 𝑏𝑥 + 𝑐𝑥 𝑖Ƹ + 𝑎𝑦 + 𝑏𝑦 + 𝑐𝑦 𝑗Ƹ
South
𝑅 = −7.99 m 𝑖Ƹ + 9.92 m 𝑗Ƹ
𝑅 =
−7.99 m 2 + 9.92 m 2 = 12.7 m
𝜃 = 𝑡𝑎𝑛−1
9.92 m
= 129°
−7.99 m
33
Determination of Axes
To determine the orthogonal coordinate axis set in three-dimensional space:
𝑧
𝑦
𝑥
We can choose the first axis in any direction. Let's call this axis the x-axis.
The second axis should be perpendicular to the first axis.
We can choose an infinite number of axes perpendicular to the first axis.
All of these axes remain in the plane perpendicular to the x axis.
We can specify any one and call it the y-axis.
As the third axis, we must determine the axis that will be perpendicular to
these first two axes.
The axis perpendicular to both axes must be in a direction perpendicular to
the 𝑥 − 𝑦 plane formed by these two axes.
𝑍
We will decide which of the two possible axes is the axis according to the
right hand rule.
When we stretch our four fingers of our right hand in the direction of the 𝑥
axis and rotate them shortly towards 𝑦 axis, our thumb will point in the
direction of the 𝑧 axis.
34
Unit Vectors
𝐴Ԧ = 𝐴Ԧ𝑥 + 𝐴Ԧ𝑦 + 𝐴Ԧ𝑧
z
Ԧ 𝑦, 𝑧) or
𝐴(𝑥,
Ԧ 𝜃, ∅)
𝐴(𝑟,
𝐴Ԧ𝑥 = 𝐴Ԧ sin 𝜃 cos ∅
𝐴Ԧ𝑧
𝐴Ԧ𝑥 0
𝜃
𝐴Ԧ
𝐴Ԧ𝑦
∅
x
𝐴Ԧ = 𝑎𝑥 𝑖Ƹ + 𝑎𝑦 𝑗Ƹ + 𝑎𝑧 𝑘෠
𝐴Ԧ𝑦 = 𝐴Ԧ sin 𝜃 sin ∅
y
𝐴Ԧ𝑧 = 𝐴Ԧ cos 𝜃
and
𝐴Ԧ
2
= 𝐴Ԧ𝑥
2
+ 𝐴Ԧ𝑦
2
+ 𝐴Ԧ𝑧
2
𝑖Ƹ = 𝐴Ԧ𝑥 / 𝐴Ԧ𝑥
𝑗Ƹ = 𝐴Ԧ𝑦 / 𝐴Ԧ𝑦
unit vectors in related directions.
𝑘෠ = 𝐴Ԧ𝑧 / 𝐴Ԧ𝑧
35
Example 1.9
Given the two displacements 𝐷 = 6𝑖Ƹ + 3𝑗Ƹ − 𝑘෠ m and 𝐸 = 4𝑖Ƹ − 5𝑗Ƹ + 8𝑘෠ m find the magnitude of
the displacements 2𝐷 − 𝐸.
𝐹Ԧ = 2𝐷 − 𝐸
𝐹Ԧ = 2 6𝑖Ƹ + 3𝑗Ƹ − 𝑘෠ m − 4𝑖Ƹ − 5𝑗Ƹ + 8𝑘෠ m
𝐹Ԧ = 12 − 4 m 𝑖Ƹ + 6 + 5 m 𝑗Ƹ + −2 − 8 m 𝑘෠
𝐹Ԧ = 8 𝑖Ƹ + 11 𝑗Ƹ − 10 𝑘෠ m
𝐹Ԧ =
8 m 2 + 11 m 2 + −10 m 2
𝐹Ԧ = 17 m
36
Vector Multiplication
Scalar Product
The scalar (dot) product of 𝐴Ԧ and 𝐵 is given with 𝐴Ԧ ∙ 𝐵.
The scalar (dot) product of two vectors gives a scalar quantity.
This magnitude gives the projection of one vector on the other.
𝐴Ԧ
𝜃
𝐴Ԧ
𝐷 = 𝐵 cos 𝜃
𝐵
𝐶Ԧ = 𝐴Ԧ cos 𝜃
𝜃
𝐵
𝐴Ԧ ∙ 𝐵 = 𝐵 ∙ 𝐴Ԧ = 𝐴Ԧ 𝐵 cos 𝜃
37
Vector Multiplication
Scalar Product
The scalar (dot) product of a vector by itself gives the square of its magnitude.
𝐴Ԧ ∙ 𝐴Ԧ = 𝐴Ԧ 𝐴Ԧ cos 0 = 𝐴Ԧ
2
z
The scalar (dot) product of 𝐴Ԧ and 𝐵
𝑘෠
𝐴Ԧ = 𝑎𝑥 𝑖Ƹ + 𝑎𝑦 𝑗Ƹ + 𝑎𝑧 𝑘෠
𝐵 = 𝑏𝑥 𝑖Ƹ + 𝑏𝑦 𝑗Ƹ + 𝑏𝑧 𝑘෠
𝑗Ƹ
0
x
y
𝑖Ƹ
This magnitude gives the projection of one vector on the other.
𝐴Ԧ ∙ 𝐵 = 𝐵 ∙ 𝐴Ԧ = 𝐴Ԧ 𝐵 cos 𝜃
Remember that the unit vectors are perpendicular to
each other.
38
Vector Multiplication
Scalar Product
𝐴Ԧ ∙ 𝐴Ԧ = 𝐴Ԧ 𝐴Ԧ cos 0 = 𝐴Ԧ
2
The scalar (dot) product of unit vectors
𝑗Ƹ
z
𝑘෠
0
x
𝑖Ƹ
𝑘෠
𝑖Ƹ
𝑗Ƹ
y
𝑘෠
𝑗Ƹ
𝑖Ƹ
There is no projection
onto each other.
𝑖Ƹ 𝑖Ƹ
The projection onto itself
other is one.
𝑖Ƹ ∙ 𝑗Ƹ = 𝑗Ƹ ∙ 𝑘෠ = 𝑘෠ ∙ 𝑖Ƹ = 0
𝑗Ƹ
𝑗Ƹ
𝑘෠
𝑘෠
𝑖Ƹ ∙ 𝑖Ƹ = 𝑗Ƹ ∙ 𝑗Ƹ = 𝑘෠ ∙ 𝑘෠ = 1
39
Vector Multiplication
Scalar Product
𝐴Ԧ ∙ 𝐵 = 𝐵 ∙ 𝐴Ԧ = 𝐴Ԧ 𝐵 cos 𝜃
𝑖Ƹ ∙ 𝑖Ƹ = 𝑖Ƹ 𝑖Ƹ cos 0 = 1
𝐴Ԧ ∙ 𝐵 = 𝑎𝑥 𝑖Ƹ + 𝑎𝑦 𝑗Ƹ + 𝑎𝑧 𝑘෠ ∙ 𝑏𝑥 𝑖Ƹ + 𝑏𝑦 𝑗Ƹ + 𝑏𝑧 𝑘෠
𝑖Ƹ ∙ 𝑗Ƹ = 𝑖Ƹ 𝑗Ƹ cos 90 = 0
𝐴Ԧ ∙ 𝐵 = 𝑎𝑥 𝑖Ƹ ∙ 𝑏𝑥 𝑖Ƹ + 𝑎𝑦 𝑗Ƹ ∙ 𝑏𝑥 𝑖Ƹ + 𝑎𝑧 𝑘෠ ∙ 𝑏𝑥 𝑖Ƹ
𝑖Ƹ ∙ 𝑘෠ = 𝑖Ƹ 𝑘෠ cos 90 = 0
+ 𝑎𝑥 𝑖Ƹ ∙ 𝑏𝑦 𝑗Ƹ + 𝑎𝑦 𝑗Ƹ ∙ 𝑏𝑦 𝑗Ƹ + 𝑎𝑧 𝑘෠ ∙ 𝑏𝑦 𝑗Ƹ
+ 𝑎𝑥 𝑖Ƹ ∙ 𝑏𝑧 𝑘෠ + 𝑎𝑦 𝑗Ƹ ∙ 𝑏𝑧 𝑘෠ + 𝑎𝑧 𝑘෠ ∙ 𝑏𝑧 𝑘෠
𝐴Ԧ ∙ 𝐵 = 𝑎𝑥 𝑏𝑥 + 𝑎𝑦 𝑏𝑦 + 𝑎𝑧 𝑏𝑧
and
𝐴Ԧ ∙ 𝐴Ԧ = 𝑎𝑥 𝑎𝑥 + 𝑎𝑦 𝑎𝑦 + 𝑎𝑧 𝑎𝑧 = 𝑎𝑥 2 + 𝑎𝑦 2 + 𝑎𝑧 2
Gives the 3-D Pythagorean theorem.
40
Example 1.10
Find the scalar product 𝐴Ԧ ∙ 𝐵 of the two vectors in figure. The magnitudes of the vectors are 𝐴Ԧ =
4 and 𝐵 = 5.
𝐴Ԧ ∙ 𝐵 = 𝐴Ԧ 𝐵 cos 𝜃
y
𝐴Ԧ ∙ 𝐵 = 4 ∙ 5 cos 130° − 53° = 20 cos 77° = 4.5
𝐵
130°
We can get the same result by finding the components.
𝐴Ԧ
𝑗Ƹ
53°
𝑖Ƹ
x
𝑎𝑥 = 4 cos 53° = 2.407
𝑏𝑥 = 5 cos 130° = −3.214
𝑎𝑦 = 4 sin 53° = 3.195
𝑏𝑦 = 5 sin 130° = 3.830
𝑎𝑧 = 0
𝑏𝑧 = 0
𝐴Ԧ ∙ 𝐵 = 𝑎𝑥 𝑏𝑥 + 𝑎𝑦 𝑏𝑦 + 𝑎𝑧 𝑏𝑧 = 2.407 −3.214 + 3.195 3.830 + 0 = 4.5
41
3D Vectors
z
𝐴Ԧ = 2𝑖Ƹ + 3𝑗Ƹ + 3𝑘෠
z
𝐴Ԧ𝑥 = 2𝑖Ƹ
𝐵𝑦
𝐴Ԧ𝑦 = 3𝑗Ƹ
𝐴Ԧ𝑧 = 3𝑘෠
𝑘෠
෠
𝐵 = −3𝑖Ƹ + 2𝑗Ƹ − 𝑘.
y
𝑗Ƹ
y
𝑖Ƹ
𝐴Ԧ𝑥
𝐵𝑦 = 2𝑗Ƹ
𝐵𝑧 = −𝑘෠
𝑂
𝑖Ƹ
𝐵𝑥 = −3𝑖Ƹ
𝐵
𝑘෠
𝑗Ƹ
𝑂
𝐵𝑧
𝐵𝑥
𝐴Ԧ𝑧
𝐴Ԧ
x
𝐴Ԧ𝑦
x
42
Example 1.11
෠
Find the angle between the two vectors 𝐴Ԧ = 2𝑖Ƹ + 3𝑗Ƹ + 𝑘෠ and 𝐵 = −4𝑖Ƹ + 2𝑗Ƹ − 𝑘.
𝐴Ԧ ∙ 𝐵 = 𝐴Ԧ 𝐵 cos 𝜃
𝐴Ԧ ∙ 𝐵 = 2 −4 + 3 2 + (1)(−1)
𝐴Ԧ ∙ 𝐵 = 𝑎𝑥 𝑏𝑥 + 𝑎𝑦 𝑏𝑦 + 𝑎𝑧 𝑏𝑧
𝐴Ԧ
2
= 22 + 32 + 12
𝐴Ԧ = 14
𝐵
2
= (−4)2 +22 + (−1)2
𝐵 = 21
𝐴Ԧ ∙ 𝐵 = −8 + 6 − 1 = −3
𝐴Ԧ ∙ 𝐵
cos 𝜃 =
𝐴Ԧ 𝐵
cos 𝜃 =
−3
14 21
=
−3
7 6
𝜃 = 100°
43
Vector Multiplication
Vector Product
The vector (cross) product of 𝐴Ԧ with 𝐵 :
𝐴Ԧ × 𝐵
The cross product of two vectors gives a
vector quantity.
The direction of this vector is
perpendicular to the plane formed by the
vector 𝐴Ԧ and 𝐵, and its direction is
determined by the right-hand rule.
The magnitude of this vector ise given by
𝐴Ԧ 𝐵 sin 𝜃
𝐶Ԧ = 𝐴Ԧ × 𝐵 = −𝐵 × 𝐴Ԧ
𝐶Ԧ = 𝐴Ԧ 𝐵 sin 𝜃
𝐶Ԧ = 𝐴Ԧ × 𝐵
𝐵
𝜃
𝐴Ԧ
44
Vector Product
𝐶Ԧ = 𝐴Ԧ 𝐵 sin 𝜃
The vector (cross) product of 𝐴Ԧ with 𝐵 : 𝐴Ԧ × 𝐵
𝐵 sin 𝜃
𝐶Ԧ = 𝐴Ԧ × 𝐵
𝐶Ԧ = 𝐴Ԧ × 𝐵
𝐵
𝐵
𝜃
𝜃
𝐴Ԧ
𝐴Ԧ
𝐴Ԧ sin 𝜃
−𝐶Ԧ = 𝐵 × 𝐴Ԧ
−𝐶Ԧ = 𝐵 × 𝐴Ԧ
Magnitude of the shaded area: 𝐴Ԧ 𝐵 sin 𝜃
Magnitude of the shaded area: 𝐴Ԧ 𝐵 sin 𝜃
𝐶Ԧ = 𝐴Ԧ × 𝐵 = −𝐵 × 𝐴Ԧ
The magnitude of the vector (cross) product 𝐴Ԧ × 𝐵 is the area formed by this two vectors, and the direction of the
resultant vector can be found by the right hand rule.
45
Vector Multiplication
Vector Product
𝐶Ԧ = 𝐴Ԧ 𝐵 sin 𝜃
𝑖Ƹ × 𝑖Ƹ = 0
𝐴Ԧ × 𝐵 = 𝑎𝑥 𝑖Ƹ + 𝑎𝑦 𝑗Ƹ + 𝑎𝑧 𝑘෠ × 𝑏𝑥 𝑖Ƹ + 𝑏𝑦 𝑗Ƹ + 𝑏𝑧 𝑘෠
𝑖Ƹ × 𝑗Ƹ = 𝑘෠
𝐴Ԧ × 𝐵 = 𝑎𝑥 𝑖Ƹ × 𝑏𝑥 𝑖Ƹ + 𝑎𝑦 𝑗Ƹ × 𝑏𝑥 𝑖Ƹ + 𝑎𝑧 𝑘෠ × 𝑏𝑥 𝑖Ƹ
𝑖Ƹ × 𝑘෠ = −𝑗Ƹ
+ 𝑎𝑥 𝑖Ƹ × 𝑏𝑦 𝑗Ƹ + 𝑎𝑦 𝑗Ƹ × 𝑏𝑦 𝑗Ƹ + 𝑎𝑧 𝑘෠ × 𝑏𝑦 𝑗Ƹ
+ 𝑎𝑥 𝑖Ƹ × 𝑏𝑧 𝑘෠ + 𝑎𝑦 𝑗Ƹ × 𝑏𝑧 𝑘෠ + 𝑎𝑧 𝑘෠ × 𝑏𝑧 𝑘෠
𝐴Ԧ × 𝐵 = 𝑎𝑦 𝑏𝑧 − 𝑎𝑧 𝑏𝑦 𝑖Ƹ + 𝑎𝑧 𝑏𝑥 − 𝑎𝑥 𝑏𝑧 𝑗Ƹ + 𝑎𝑥 𝑏𝑦 − 𝑎𝑦 𝑏𝑥 𝑘෠ = 𝑐𝑥 𝑖Ƹ + 𝑐𝑦 𝑗Ƹ + 𝑐𝑧 𝑘෠
𝐶Ԧ = 𝐴Ԧ × 𝐵
46
Vector Multiplication
Vector Product
The vector (cross) product of 𝐴Ԧ and 𝐵 is given by
𝑖Ƹ
𝐴Ԧ × 𝐵 = 𝑎𝑥
𝑏𝑥
𝑗Ƹ
𝑎𝑦
𝑏𝑦
𝑘෠
𝑎𝑦
1+1
𝑎𝑧 = −1
𝑏𝑦
𝑏𝑧
𝑖Ƹ
𝐴Ԧ × 𝐵 = 𝑎𝑥
𝑏𝑥
𝑎𝑧
𝑎𝑥
1+2
𝑏𝑧 𝑖Ƹ + −1
𝑏𝑥
𝑗Ƹ
𝑎𝑦
𝑏𝑦
𝑘෠
𝑎𝑧
𝑏𝑧
𝑎𝑥
𝑎𝑧
1+3
𝑏𝑧 𝑗Ƹ + −1
𝑏𝑥
𝑎𝑦
෠
𝑘
𝑏𝑦
𝐴Ԧ × 𝐵 = 𝑎𝑦 𝑏𝑧 − 𝑎𝑧 𝑏𝑦 𝑖Ƹ + 𝑎𝑧 𝑏𝑥 − 𝑎𝑥 𝑏𝑧 𝑗Ƹ + 𝑎𝑥 𝑏𝑦 − 𝑎𝑦 𝑏𝑥 𝑘෠
47
Example 1.12
Vector 𝐴Ԧ has magnitude 6 units and is in the direction of the +𝑥 − axis. Vector 𝐵 has
magnitude 4 units and lies in the 𝑥𝑦 − plane, making an angle of 30° with the +𝑥 − axis.
Find the vector product 𝐴Ԧ × 𝐵.
𝑦
𝐴Ԧ × 𝐵 = 6𝑖Ƹ × 4 cos 30° 𝑖Ƹ + 4 sin(30°)𝑗Ƹ
𝐵
𝐴Ԧ × 𝐵 = 24 sin(30°)𝑘෠
𝜃 = 30°
𝐴Ԧ
𝑧
𝑥
𝐴Ԧ × 𝐵 = 12𝑘෠
𝐶Ԧ = 𝐴Ԧ × 𝐵
48
Problems
Q 1.49
(a) Write each vector in Fig. 1.37 in terms of the unit vectors 𝑖Ƹ and 𝑗Ƹ . (b) Use unit
Ԧ where 𝐶Ԧ = 3𝐴Ԧ − 4𝐵. (c) Find the magnitude and
vectors to express the vector𝐶,
Ԧ
direction of 𝐶.
𝐴Ԧ = 𝑎𝑥 𝑖Ƹ + 𝑎𝑦 𝑗Ƹ + 𝑎𝑧 𝑘෠
𝐴Ԧ = 3.6 cos 70° m 𝑖Ƹ + 3.6 sin 70° m 𝑗Ƹ
𝐵 = 𝑏𝑥 𝑖Ƹ + 𝑏𝑦 𝑗Ƹ + 𝑏𝑧 𝑘෠
𝐵 = 2.4 cos 210° m 𝑖Ƹ + 2.4 sin 210° m 𝑗Ƹ
𝐶Ԧ = 3𝐴Ԧ − 4𝐵
𝐶Ԧ = 3 3.6 cos 70° m 𝑖Ƹ + 3.6 sin 70° m 𝑗Ƹ
− 4 2.4 cos 210° m 𝑖Ƹ + 2.4 sin 210° m 𝑗Ƹ
𝐶Ԧ = 3 1.23 m 𝑖Ƹ + 3.38 m 𝑗Ƹ
− 4 −2.08 m 𝑖Ƹ − 1.2 m 𝑗Ƹ
𝐶Ԧ = 12.01 m 𝑖Ƹ + 14.94 m 𝑗Ƹ
and
𝜃 = 𝑡𝑎𝑛−1
𝐶Ԧ ∙ 𝐶Ԧ = 𝐶Ԧ
2
𝐶Ԧ =
12.01 m 2 + 14.94 2 = 19.17 m
14.94 m
= 51.2°
12.01 m
50
Q 1.51
෠ a unit vector? Justify your answer. (b) Can a unit vector have any
(a) Is the vector (𝑖Ƹ + 𝑗Ƹ + 𝑘)
components with magnitude greater than unity? Can it have any negative components? In each
case, justify your answer. (c) If 𝐴Ԧ = 𝑎(3𝑖Ƹ + 4𝑗),
Ƹ where 𝑎 is a constant, determine the value of 𝑎 that
makes 𝐴Ԧ a unit vector.
𝐴Ԧ = 𝑖Ƹ + 𝑗Ƹ + 𝑘෠
𝐴Ԧ ∙ 𝐴Ԧ = 𝐴Ԧ
2
𝐴Ԧ =
12 + 12 + 12 = 3 ≠ 1
Is not a unit vector
Any component of a unit vector cannot be greater than 1 or less than −1
𝐴Ԧ = 𝑎(3𝑖Ƹ + 4𝑗)Ƹ
𝐴Ԧ =
𝑎𝑥 𝑎𝑥 + 𝑎𝑦 𝑎𝑦 + 𝑎𝑧 𝑎𝑧 > 1
𝐴Ԧ =
3𝑎 2 + 4𝑎 2 =
𝐴Ԧ = ±5𝑎 = 1
25𝑎2
𝑎 = ±1/5
51
Q 1.86 (1/2)
For the two vectors 𝐴Ԧ and 𝐵 in Fig. 1.37, (a) find the scalar product 𝐴Ԧ ∙ 𝐵,
and (b) find the magnitude and direction of the vector product 𝐴Ԧ × 𝐵.
𝐴Ԧ = 3.6 cos 70° m 𝑖Ƹ + 3.6 sin 70° m 𝑗Ƹ = 1.23 m 𝑖Ƹ + 3.38 m 𝑗Ƹ
𝐵 = 2.4 cos 210° m 𝑖Ƹ + 2.4 sin 210° m 𝑗Ƹ = −2.08 m 𝑖Ƹ − 1.2 m 𝑗Ƹ
𝐴Ԧ ∙ 𝐵 = 𝐴Ԧ 𝐵 cos 𝜃
𝐴Ԧ =
1.23 m 2 + 3.38 m 2 = 3.6 m
𝐵 =
−2.08 m 2 + −1.2 m 2 = 2.4 m
𝐴Ԧ ∙ 𝐵 = 3.6 m 2.4 m cos 210° − 70° = 6.62 m2
𝐴Ԧ ∙ 𝐵 = 𝑎𝑥 𝑏𝑥 + 𝑎𝑦 𝑏𝑦 + 𝑎𝑧 𝑏𝑧 = 1.23 m −2.08 m + 3.38 m −1.2 m + 0 = 6.62 m2
52
Q 1.86 (2/2)
For the two vectors 𝐴Ԧ and 𝐵 in Fig. 1.37, (a) find the scalar product 𝐴Ԧ ∙ 𝐵,
and (b) find the magnitude and direction of the vector product 𝐴Ԧ × 𝐵.
𝐴Ԧ = 1.23 m 𝑖Ƹ + 3.38 m 𝑗Ƹ + 0 m 𝑘෠
𝐴Ԧ = 3.6 m
𝐵 = −2.08 m 𝑖Ƹ − 1.2 m 𝑗Ƹ + 0 m 𝑘෠
𝐵 = 2.4 m
𝑖Ƹ
𝐴Ԧ × 𝐵 = 𝑎𝑥
𝑏𝑥
𝑗Ƹ
𝑎𝑦
𝑏𝑦
𝑘෠
𝑎𝑧
𝑏𝑧
𝐶Ԧ = 𝐴Ԧ 𝐵 sin 𝜃 = 3.6 m 2.4 m sin 140° = 5.5 m
𝐴Ԧ × 𝐵 = 𝑎𝑦 𝑏𝑧 − 𝑎𝑧 𝑏𝑦 𝑖Ƹ + 𝑎𝑧 𝑏𝑥 − 𝑎𝑥 𝑏𝑧 𝑗Ƹ + 𝑎𝑥 𝑏𝑦 − 𝑎𝑦 𝑏𝑥 𝑘෠
= 𝑎𝑥 𝑏𝑦 − 𝑎𝑦 𝑏𝑥 𝑘෠ =
=
1.23 m −1.2 m − 3.38 m −2.08 m 𝑘෠
1.23 m −1.2 m − 3.38 m −2.08 m 𝑘෠ = −1.48 + 7.03 𝑘෠ = 5.5𝑘෠
53
Q 1.89
෠ do the following. (a) Find the magnitude
Given two vectors 𝐴Ԧ = −2𝑖Ƹ + 3𝑗Ƹ + 4𝑘෠ and 𝐵 = 3𝑖Ƹ + 1𝑗Ƹ − 3𝑘,
of each vector. (b) Write an expression for the vector difference 𝐴Ԧ − 𝐵, using unit vectors. (c) Find
Ԧ Explain.
the magnitude of the vector difference 𝐴Ԧ − 𝐵. Is this the same as the magnitude of 𝐵 − 𝐴?
𝐴Ԧ = −2𝑖Ƹ + 3𝑗Ƹ + 4𝑘෠
𝐴Ԧ ∙ 𝐴Ԧ = 𝐴Ԧ
𝐵 = 3𝑖Ƹ + 1𝑗Ƹ − 3𝑘෠
2
𝐴Ԧ =
−2 2 + 32 + 42 = 29 ≅ 5.38
𝐵 =
32 + 12 + −3 2 = 19 ≅ 4.36
𝐴Ԧ − 𝐵 = −2𝑖Ƹ + 3𝑗Ƹ + 4𝑘෠ − 3𝑖Ƹ + 1𝑗Ƹ − 3𝑘෠ = −5𝑖Ƹ + 2𝑗Ƹ + 7𝑘෠
𝐴Ԧ − 𝐵 =
−5 2 + 22 + 72 = 25 + 4 + 49 = 78 ≅ 8.83
𝐴Ԧ − 𝐵 = 𝐵 − 𝐴Ԧ = 78 ≅ 8.83
and
54
Q 1.93
A cube is placed so that one comer is at the origin and three edges are along the
𝑥, y, and 𝑧-axes of a coordinate system (Fig. 1.43). Use vectors to compute (a) the
angle between the edge along the z-axis (line ab) and the diagonal from the origin
to the opposite comer (line ad), and (b) the angle between line ac (the diagonal of
a face) and line ad.
𝐴Ԧ =
02 + 02 + 12 = 1
𝐵 =
12 + 12 + 12 = 3
Vector from 𝑎 to 𝑑
𝐶Ԧ =
02 + 12 + 12 = 2
𝐵 = 1𝑖Ƹ + 1𝑗Ƹ + 1𝑘෠
𝐴Ԧ ∙ 𝐵 = 𝐴Ԧ 𝐵 cos 𝜃
Vector from 𝑎 to 𝑐
𝐴Ԧ ∙ 𝐵 = 𝑎𝑥 𝑏𝑥 + 𝑎𝑦 𝑏𝑦 + 𝑎𝑧 𝑏𝑧
𝐶Ԧ = 0𝑖Ƹ + 1𝑗Ƹ + 1𝑘෠
𝐴Ԧ ∙ 𝐵 = 0 1 + 0 1 + 1 1 = 1
Vector from 𝑎 to 𝑏
𝐴Ԧ = 0𝑖Ƹ + 0𝑗Ƹ + 1𝑘෠
cos 𝜃 =
1
3
𝜃 = 54.7°
𝐵 ∙ 𝐶Ԧ = 1 0 + 1 1 + 1 1 = 2
𝐵 ∙ 𝐶Ԧ = 𝐵 𝐶Ԧ cos 𝛽
cos 𝛽 =
2
3 2
𝛽 = 35.3°
Q 1.95
You are given vectors 𝐴Ԧ = 5𝑖Ƹ − 6.5𝑗Ƹ and 𝐵 = −3.5 + 7𝑗.Ƹ A third vector 𝐶Ԧ lies in the xy-pIane. Vector
Ԧ and the scalar product of 𝐶Ԧ with 𝐵 is 15. From this information,
𝐶Ԧ is perpendicular to vector 𝐴,
Ԧ
find the components of vector 𝐶.
𝐴Ԧ = 5𝑖Ƹ − 6.5𝑗Ƹ
𝐵 = −3.5𝑖Ƹ + 7𝑗Ƹ
If vector 𝐶Ԧ is in
the 𝑥𝑦 plane;
𝐶Ԧ = 𝑐𝑥 𝑖Ƹ + 𝑐𝑦 𝑗Ƹ
If 𝐴Ԧ is perpendicular to 𝐶Ԧ
5
𝑐𝑦 =
𝑐
6.5 𝑥
𝐴Ԧ ∙ 𝐶Ԧ = 𝑎𝑥 𝑐𝑥 + 𝑎𝑦 𝑐𝑦 + 𝑎𝑧 𝑐𝑧 = 0
5
−3.5𝑐𝑥 + 7
𝑐 = 15
6.5 𝑥
𝐴Ԧ ∙ 𝐶Ԧ = 5𝑐𝑥 − 6.5𝑐𝑦 = 0
70
−3.5 +
𝑐 = 15
13 𝑥
𝐵 ∙ 𝐶Ԧ = 𝑏𝑥 𝑐𝑥 + 𝑏𝑦 𝑐𝑦 + 𝑏𝑧 𝑐𝑧 = 15
𝐵 ∙ 𝐶Ԧ = −3.5𝑐𝑥 + 7𝑐𝑦 = 15
15
𝑐𝑥 =
≅8
1.89
5
5
𝑐𝑦 =
𝑐𝑥 ≅
8 = 6.1
6.5
6.5
56
Q 1.96
Two vectors A and B have magnitude 𝐴 = 3 and 𝐵 = 3. Their vector product is 𝐴Ԧ × 𝐵 = −5𝑘෠ + 2𝑖.Ƹ
What is the angle between 𝐴Ԧ and 𝐵?
𝐶Ԧ = 𝐴Ԧ × 𝐵 = −5𝑘෠ + 2𝑖.Ƹ
𝐶Ԧ ∙ 𝐶Ԧ = 𝐶Ԧ
𝐶Ԧ = 𝐴Ԧ 𝐵 sin 𝜃
2
𝐶Ԧ =
−5 2 + 22 = 29
𝐶Ԧ
29
sin 𝜃 =
=
Ԧ
3 ∙3
𝐴 𝐵
𝜃 = 36.8°
57
Q 1.102
෠ called the position vector, points from the origin (0,0,0) to an arbitrary
The vector 𝑟Ԧ = 𝑥 𝑖Ƹ + 𝑦𝑗Ƹ + 𝑧𝑘,
point in space with coordinates (𝑥, 𝑦, 𝑧). Use what you know about vectors to prove the following:
All points (𝑥, 𝑦, 𝑧) that satisfy the equation 𝐴𝑥 + 𝐵𝑦 + 𝐶𝑧 = 0, where 𝐴,𝐵, and 𝐶 are constants, lie in
෠ Sketch
a plane that passes through the origin and that is perpendicular to the vector 𝐴𝑖Ƹ + 𝐵 𝑗Ƹ + 𝐶 𝑘.
this vector and the plane.
𝑟Ԧ ∙ 𝑆Ԧ = 𝐴𝑥 + 𝐵𝑦 + 𝐶𝑧 = 0
𝑟Ԧ = 𝑥𝑖Ƹ + 𝑦𝑗Ƹ + 𝑧𝑘෠
𝑆Ԧ = 𝐴𝑖Ƹ + 𝐵𝑗Ƹ + 𝐶 𝑘෠
If vector 𝑟Ԧ is perpendicular
to vector 𝑆Ԧ
𝑟Ԧ ∙ 𝑆Ԧ = 𝑟𝑥 𝑆𝑥 + 𝑟𝑦 𝑆𝑦 + 𝑟𝑧 𝑆𝑧 = 0
so;
𝐴𝑥 + 𝐵𝑦 + 𝐶𝑧 = 0
than
𝑟Ԧ ∙ 𝑆Ԧ = 0
The projection of vector 𝑟Ԧ onto vector 𝑆Ԧ is
zero
so, 𝑟Ԧ ⊥ 𝑆Ԧ
58