v1chap1.pdf

Chapter 1
Units and Vectors: Tools for Physics
1.1 The Important Stuff
1.1.1 The SI System
Physics is based on measurement. Measurements are made by comparisons to well–defined
standards which define the units for our measurements.
The SI system (popularly known as the metric system) is the one used in physics. Its
unit of length is the meter, its unit of time is the second and its unit of mass is the kilogram.
Other quantities in physics are derived from these. For example the unit of energy is the
joule, defined by 1 J = 1 kg·m2
s2 .
As a convenience in using the SI system we can associate prefixes with the basic units to
represent powers of 10. The most commonly used prefixes are given here:
Factor Prefix Symbol
10−12
pico- p
10−9
nano- n
10−6
micro- µ
10−3
milli- m
10−2
centi- c
103
kilo- k
106
mega- M
109
giga- G
Other basic units commonly used in physics are:
Time : 1 minute = 60 s 1 hour = 60 min etc.
Mass : 1 atomic mass unit = 1 u = 1.6605 × 10−27
kg
1

2 CHAPTER 1. UNITS AND VECTORS: TOOLS FOR PHYSICS
1.1.2 Changing Units
In all of our mathematical operations we must always write down the units and we always
treat the unit symbols as multiplicative factors. For example, if me multiply 3.0 kg by 2.0 m
s
we get
(3.0 kg) · (2.0 m
s
) = 6.0 kg·m
s
We use the same idea in changing the units in which some physical quantity is expressed.
We can multiply the original quantity by a conversion factor, i.e. a ratio of values for
which the numerator is the same thing as the denominator. The conversion factor is then
equal to 1, and so we do not change the original quantity when we multiply by the conversion
factor.
Examples of conversion factors are:

1 min
60 s

100 cm
1 m

1 yr
365.25 day
!
1 m
3.28 ft

1.1.3 Density
A quantity which will be encountered in your study of liquids and solids is the density of a
sample. It is usually denoted by ρ and is defined as the ratio of mass to volume:
ρ =
m
V
(1.1)
The SI units of density are kg
m3 but you often see it expressed in g
cm3 .
1.1.4 Dimensional Analysis
Every equation that we use in physics must have the same type of units on both sides of the
equals sign. Our basic unit types (dimensions) are length (L), time (T) and mass (M).
When we do dimensional analysis we focus on the units of a physics equation without
worrying about the numerical values.
1.1.5 Vectors; Vector Addition
Many of the quantities we encounter in physics have both magnitude (“how much”) and
direction. These are vector quantities.
We can represent vectors graphically as arrows and then the sum of two vectors is found
(graphically) by joining the head of one to the tail of the other and then connecting head to
tail for the combination, as shown in Fig. 1.1 . The sum of two (or more) vectors is often
called the resultant.
We can add vectors in any order we want: A + B = B + A. We say that vector addition
is “commutative”.
We express vectors in component form using the unit vectors i, j and k, which each
have magnitude 1 and point along the x, y and z axes of the coordinate system, respectively.

1.1. THE IMPORTANT STUFF 3
A
B
A
B
A+B
(a) (b)
Figure 1.1: Vector addition. (a) shows the vectors A and B to be summed. (b) shows how to perform the
sum graphically.
B
C
A
Ax
By
Ay
Bx
Cy
Cx
x
y
Figure 1.2: Addition of vectors by components (in two dimensions).
Any vector can be expressed as a sum of multiples of these basic vectors; for example,
for the vector A we would write:
A = Axi + Ayj + Azk .
Here we would say that Ax is the x component of the vector A; likewise for y and z.
In Fig. 1.2 we illustrate how we get the components for a vector which is the sum of two
other vectors. If
A = Axi + Ayj + Azk and B = Bxi + Byj + Bzk
then
A + B = (Ax + Bx)i + (Ay + By)j + (Az + Bz)k (1.2)
Once we have found the (Cartesian) component of two vectors, addition is simple; just add
the corresponding components of the two vectors to get the components of the resultant
vector.
When we multiply a vector by a scalar, the scalar multiplies each component; If A is a
vector and n is a scalar, then
cA = cAxi + cAyj + cAzk (1.3)

In terms of its components, the magnitude (“length”) of a vector A (which we write as
A) is given by:
A =
q
A2
x + A2
y + A2
z (1.4)
Many of our physics problems will be in two dimensions (x and y) and then we can also
represent it in polar form. If A is a two–dimensional vector and θ as the angle that A
makes with the +x axis measured counter-clockwise then we can express this vector in terms
of components Ax and Ay or in terms of its magnitude A and the angle θ. These descriptions
are related by:
Ax = A cos θ Ay = A sin θ (1.5)
A =
q
A2
x + A2
y tan θ =
Ay
Ax
(1.6)
When we use Eq. 1.6 to find θ from Ax and Ay we need to be careful because the inverse
tangent operation (as done on a calculator) might give an angle in the wrong quadrant; one
must think about the signs of Ax and Ay.
1.1.6 Multiplying Vectors
There are two ways to “multiply” two vectors together.
The scalar product (or dot product) of the vectors a and b is given by
a · b = ab cos φ (1.7)
where a is the magnitude of a, b is the magnitude of b and φ is the angle between a and b.
The scalar product is commutative: a · b = b · a. One can show that a · b is related to
the components of a and b by:
a · b = axbx + ayby + azbz (1.8)
If two vectors are perpendicular then their scalar product is zero.
The vector product (or cross product) of vectors a and b is a vector c whose mag-
nitude is given by
c = ab sin φ (1.9)
where φ is the smallest angle between a and b. The direction of c is perpendicular to the
plane containing a and b with its orientation given by the right–hand rule. One way
of using the right–hand rule is to let the fingers of the right hand bend (in their natural
direction!) from a to b; the direction of the thumb is the direction of c = a × b. This is
illustrated in Fig. 1.3.
The vector product is anti–commutative: a × b = −b × a.
Relations among the unit vectors for vector products are:
i × j = k j × k = i k × i = j (1.10)

1.2. WORKED EXAMPLES 5
A
B
C
A
B
C
(a) (b)
f
Figure 1.3: (a) Finding the direction of A × B. Fingers of the right hand sweep from A to B in the
shortest and least painful way. The extended thumb points in the direction of C. (b) Vectors A, B and C.
The magnitude of C is C = AB sinφ.
The vector product of a and b can be computed from the components of these vectors
by:
a × b = (aybz − azby)i + (azbx − axbz)j + (axby − aybx)k (1.11)
which can be abbreviated by the notation of the determinant:
a × b =
i j k
ax ay az
bx by bz
(1.12)
1.2 Worked Examples
1.2.1 Changing Units
1. The Empire State Building is 1472 ft high. Express this height in both meters
and centimeters. [FGT 1-4]
To do the first unit conversion (feet to meters), we can use the relation (see the Conversion
Factors in the back of this book):
1 m = 3.281 ft
We set up the conversion factor so that “ft” cancels and leaves meters:
1472 ft = (1472 ft)

1 m
3.281 ft

= 448.6 m .
So the height can be expressed as 448.6 m. To convert this to centimeters, use:
1 m = 100 cm

and get:
448.6 m = (448.6 m)

100 cm
1 m

= 4.486 × 104
cm
The Empire State Building is 4.486 × 104
cm high!
2. A rectangular building lot is 100.0 ft by 150.0 ft. Determine the area of this lot
in m2
. [Ser4 1-19]
The area of a rectangle is just the product of its length and width so the area of the lot
is
A = (100.0 ft)(150.0 ft) = 1.500 × 104
ft2
To convert this to units of m2
we can use the relation
1 m = 3.281 ft
but the conversion factor needs to be applied twice so as to cancel “ ft2
” and get “ m2
”. We
write:
1.500 × 104
ft2
= (1.500 × 104
ft2
) ·

1 m
3.281 ft
2
= 1.393 × 103
m2
The area of the lot is 1.393 × 103
m2
.
3. The Earth is approximately a sphere of radius 6.37 × 106
m. (a) What is its
circumference in kilometers? (b) What is its surface area in square kilometers?
(c) What is its volume in cubic kilometers? [HRW5 1-6]
(a) The circumference of the sphere of radius R, i.e. the distance around any “great circle”
is C = 2πR. Using the given value of R we find:
C = 2πR = 2π(6.37 × 106
m) = 4.00 × 107
m .
To convert this to kilometers, use the relation 1 km = 103
m in a conversion factor:
C = 4.00 × 107
m = (4.00 × 107
m) ·
1 km
103 m
!
= 4.00 × 104
km
The circumference of the Earth is 4.00 × 104
km.
(b) The surface area of a sphere of radius R is A = 4πR2
. So we get
A = 4πR2
= 4π(6.37 × 106
m)2
= 5.10 × 1014
m2
Again, use 1 km = 103
m but to cancel out the units “ m2
” and replace them with “ km2
” it
must be applied twice:
A = 5.10 × 1014
m2
= (5.10 × 1014
m2
) ·
1 km
103 m
!2
= 5.10 × 108
km2

The surface area of the Earth is 5.10 × 108
km2
.
(c) The volume of a sphere of radius R is V = 4
3
πR3
. So we get
V = 4
3
πR3
= 4
3
π(6.37 × 106
m)3
= 1.08 × 1021
m3
Again, use 1 km = 103
m but to cancel out the units “m3
” and replace them with “ km3
” it
must be applied three times:
V = 1.08 × 1021
m3
= (1.08 × 1021
m3
) ·
1 km
103 m
!3
= 1.08 × 1012
km3
The volume of the Earth is 1.08 × 1012
km3
.
4. Calculate the number of kilometers in 20.0 mi using only the following conver-
sion factors: 1 mi = 5280 ft, 1 ft = 12 in, 1 in = 2.54 cm, 1 m = 100 cm, 1 km = 1000 m.
[HRW5 1-7]
Set up the “factors of 1” as follows:
20.0 mi = (20.0 mi) ·
5280 ft
1 mi
!
·

12 in
1 ft

·

2.54 cm
1 in

·

1 m
100 cm

·
1 km
1000 m
!
= 32.2 km
Setting up the “factors of 1” in this way, all of the unit symbols cancel except for km
(kilometers) which we keep as the units of the answer.
5. One gallon of paint (volume = 3.78 × 10−3
m3
) covers an area of 25.0 m3
. What
is the thickness of the paint on the wall? [Ser4 1-31]
We will assume that the volume which the paint occupies while it’s covering the wall is
the same as it has when it is in the can. (There are reasons why this may not be true, but
let’s just do this and proceed.)
The paint on the wall covers an area A and has a thickness τ; the volume occupied is the
area time the thickness:
V = Aτ .
We have V and A; we just need to solve for τ:
τ =
V
A
=
3.78 × 10−3
m3
25.0 m2
= 1.51 × 10−4
m .
The thickness is 1.51 × 10−4
m. This quantity can also be expressed as 0.151 mm.
6. A certain brand of house paint claims a coverage of 460 ft2
gal
. (a) Express this
quantity in square meters per liter. (b) Express this quantity in SI base units. (c)

What is the inverse of the original quantity, and what is its physical significance?
[HRW5 1-15]
(a) Use the following relations in forming the conversion factors: 1 m = 3.28 ft and 1000 liter =
264 gal. To get proper cancellation of the units we set it up as:
460 ft2
gal
= (460 ft2
gal
) ·

1 m
3.28 ft
2
·
264 gal
1000 L
!
= 11.3 m2
L
(b) Even though the units of the answer to part (a) are based on the metric system, they
are not made from the base units of the SI system, which are m, s, and kg. To make the
complete conversion to SI units we need to use the relation 1 m3
= 1000 L. Then we get:
11.3 m2
L
= (11.3 m2
L
) ·

1000 L
1 m3

= 1.13 × 104
m−1
So the coverage can also be expressed (not so meaningfully, perhaps) as 1.13 × 104
m−1
.
(c) The inverse (reciprocal) of the quantity as it was originally expressed is

460 ft2
gal
−1
= 2.17 × 10−3 gal
ft2 .
Of course when we take the reciprocal the units in the numerator and denominator also
switch places!
Now, the first expression of the quantity tells us that 460 ft2
are associated with every
gallon, that is, each gallon will provide 460 ft2
of coverage. The new expression tells us that
2.17×10−3
gal are associated with every ft2
, that is, to cover one square foot of surface with
paint, one needs 2.17 × 10−3
gallons of it.
7. Express the speed of light, 3.0 × 108 m
s
in (a) feet per nanosecond and (b)
millimeters per picosecond. [HRW5 1-19]
(a) For this conversion we can use the following facts:
1 m = 3.28 ft and 1 ns = 10−9
s
to get:
3.0 × 108 m
s
= (3.0 × 108 m
s
) ·
3.28 ft
1 m
!
·
10−9
s
1 ns
!
= 0.98 ft
ns
In these new units, the speed of light is 0.98 ft
ns
.
(b) For this conversion we can use:
1 mm = 10−3
m and 1 ps = 10−12
s

and set up the factors as follows:
3.0 × 108 m
s
= (3.0 × 108 m
s
) ·

1 mm
10−3 m

·
10−12
s
1 ps
!
= 3.0 × 10−1 mm
ps
In these new units, the speed of light is 3.0 × 10−1 mm
ps
.
8. One molecule of water (H2O) contains two atoms of hydrogen and one atom
of oxygen. A hydrogen atom has a mass of 1.0 u and an atom of oxygen has a
mass of 16 u, approximately. (a) What is the mass in kilograms of one molecule
of water? (b) How many molecules of water are in the world’s oceans, which
have an estimated total mass of 1.4 × 1021
kg? [HRW5 1-33]
(a) We are given the masses of the atoms of H and O in atomic mass units; using these
values, one molecule of H2O has a mass of
mH2O = 2(1.0 u) + 16 u = 18 u
Use the relation between u (atomic mass units) and kilograms to convert this to kg:
mH2O = (18 u)
1.6605 × 10−27
kg
1 u
!
= 3.0 × 10−26
kg
One water molecule has a mass of 3.0 × 10−26
kg.
(b) To get the number of molecules in all the oceans, divide the mass of all the oceans’
water by the mass of one molecule:
N =
1.4 × 1021
kg
3.0 × 10−26 kg
= 4.7 × 1046
.
. . . a large number of molecules!
1.2.2 Density
9. Calculate the density of a solid cube that measures 5.00 cm on each side and
has a mass of 350 g. [Ser4 1-1]
The volume of this cube is
V = (5.00 cm) · (5.00 cm) · (5.00 cm) = 125 cm3
So from Eq. 1.1 the density of the cube is
ρ =
m
V
=
350 g
125 cm3
= 2.80 g
cm3

r2
r1
Figure 1.4: Cross–section of copper shell in Example 11.
10. The mass of the planet Saturn is 5.64 × 1026
kg and its radius is 6.00 × 107
m.
Calculate its density. [Ser4 1-2]
The planet Saturn is roughly a sphere. (But only roughly! Actually its shape is rather
distorted.) Using the formula for the volume of a sphere, we find the volume of Saturn:
V = 4
3
πR3
= 4
3
π(6.00 × 107
m)3
= 9.05 × 1023
m3
Now using the definition of density we find:
ρ =
m
V
=
5.64 × 1026
kg
9.05 × 1023 m3
= 6.23 × 102 kg
m3
While this answer is correct, it is useful to express the result in units of g
cm3 . Using our
conversion factors in the usual way, we get:
6.23 × 102 kg
m3 = (6.23 × 102 kg
m3 ) ·
103
g
1 kg
!
·

1 m
100 cm
3
= 0.623 g
cm3
The average density of Saturn is 0.623 g
cm3 . Interestingly, this is less than the density of
water.
11. How many grams of copper are required to make a hollow spherical shell
with an inner radius of 5.70 cm and an outer radius of 5.75 cm? The density of
copper is 8.93 g/ cm3
. [Ser4 1-3]
A cross–section of the copper sphere is shown in Fig. 1.4. The outer and inner radii are
noted as r2 and r1, respectively. We must find the volume of space occupied by the copper
metal; this volume is the difference in the volumes of the two spherical surfaces:
Vcopper = V2 − V1 = 4
3
πr3
2 − 4
3
πr3
1 = 4
3
π(r3
2 − r3
1)
With the given values of the radii, we find:
Vcopper = 4
3
π((5.75 cm)3
− (5.70 cm)3
) = 20.6 cm3

Now use the definition of density to find the mass of the copper contained in the shell:
ρ =
mcopper
Vcopper
=⇒ mcopper = ρVcopper =

8.93 g
cm3

(20.6 cm3
) = 184 g
184 grams of copper are required to make the spherical shell of the given dimensions.
12. One cubic meter (1.00 m3
) of aluminum has a mass of 2.70 × 103
kg, and 1.00 m3
of iron has a mass of 7.86 × 103
kg. Find the radius of a solid aluminum sphere
that will balance a solid iron sphere of radius 2.00 cm on an equal–arm balance.
[Ser4 1-39]
In the statement of the problem, we are given the densities of aluminum and iron:
ρAl = 2.70 × 103 kg
m3 and ρFe = 7.86 × 103 kg
m3 .
A solid iron sphere of radius R = 2.00 cm = 2.00 × 10−2
m has a volume
VFe = 4
3
πR3
= 4
3
π(2.00 × 10−2
m)3
= 3.35 × 10−5
m3
so that from MFe = ρFeVFe we find the mass of the iron sphere:
MFe = ρFeVFe =

7.86 × 103 kg
m3

(3.35 × 10−5
m3
) = 2.63 × 10−1
kg
If this sphere balances one made from aluminum in an “equal–arm balance”, then they
have the same mass. So MAl = 2.63 × 10−1
kg is the mass of the aluminum sphere. From
MAl = ρAlVAl we can find its volume:
VAl =
MAl
ρAl
=
2.63 × 10−1
kg
2.70 × 103 kg
m3
= 9.76 × 10−5
m3
Having the volume of the sphere, we can find its radius:
VAl = 4
3
πR3
=⇒ R =

3VAl
4π
1
3
This gives:
R =
3(9.76 × 10−5
m3
)
4π
!1
3
= 2.86 × 10−2
m = 2.86 cm
The aluminum sphere must have a radius of 2.86 cm to balance the iron sphere.

1.2.3 Dimensional Analysis
13. The period T of a simple pendulum is measured in time units and is
T = 2π
s
`
g
.
where ` is the length of the pendulum and g is the free–fall acceleration in units
of length divided by the square of time. Show that this equation is dimensionally
correct. [Ser4 1-14]
The period (T) of a pendulum is the amount of time it takes to makes one full swing
back and forth. It is measured in units of time so its dimensions are represented by T.
On the right side of the equation we have the length `, whose dimensions are represented
by L. We are told that g is a length divided by the square of a time so its dimensions must
be L/T2
. There is a factor of 2π on the right side, but this is a pure number and has no
units. So the dimensions of the right side are:
v
u
u
t
L

L
T 2
=
√
T2 = T
so that the right hand side must also have units of time. Both sides of the equation agree in
their units, which must be true for it to be a valid equation!
14. The volume of an object as a function of time is calculated by V = At3
+ B/t,
where t is time measured in seconds and V is in cubic meters. Determine the
dimension of the constants A and B. [Ser4 1-15]
Both sides of the equation for volume must have the same dimensions, and those must
be the dimensions of volume where are L3
(SI units of m3
). Since we can only add terms
with the same dimensions, each of the terms on right side of the equation (At3
and B/t)
must have the same dimensions, namely L3
.
Suppose we denote the units of A by [A]. Then our comment about the dimensions of
the first term gives us:
[A]T3
= L3
=⇒ [A] =
L3
T3
so A has dimensions L3
/T3
. In the SI system, it would have units of m3
/ s3
.
Suppose we denote the units of B by [B]. Then our comment about the dimensions of
the second term gives us:
[B]
T
= L3
=⇒ [B] = L3
T
so B has dimensions L3
T. In the SI system, it would have units of m3
s.

15. Newton’s law of universal gravitation is
F = G
Mm
r2
Here F is the force of gravity, M and m are masses, and r is a length. Force has
the SI units of kg · m/s2
. What are the SI units of the constant G? [Ser4 1-17]
If we denote the dimensions of F by [F] (and the same for the other quantities) then
then dimensions of the quantities in Newton’s Law are:
[M] = M (mass) [m] = M [r] = L [F] :
ML
T2
What we don’t know (yet) is [G], the dimensions of G. Putting the known dimensions into
Newton’s Law, we must have:
ML
T2
= [G]
M · M
L2
since the dimensions must be the same on both sides. Doing some algebra with the dimen-
sions, this gives:
[G] =

ML
T2

L2
M2
=
L3
MT2
so the dimensions of G are L3
/(MT2
). In the SI system, G has units of
m3
kg · s3
16. In quantum mechanics, the fundamental constant called Planck’s constant,
h, has dimensions of [ML2
T−1
]. Construct a quantity with the dimensions of
length from h, a mass m, and c, the speed of light. [FGT 1-54]
The problem suggests that there is some product of powers of h, m and c which has
dimensions of length. If these powers are r, s and t, respectively, then we are looking for
values of r, s and t such that
hr
ms
ct
has dimensions of length.
What are the dimensions of this product, as written? We were given the dimensions of
h, namely [ML2
T−1
]; the dimensions of m are M, and the dimensions of c are L
T
= LT−1
(it
is a speed). So the dimensions of hr
ms
ct
are:
[ML2
T−1
]r
[M]s
[LT−1
]t
= Mr+s
L2r+t
T−r−t
where we have used the laws of combining exponents which we all remember from algebra.

Now, since this is supposed to have dimensions of length, the power of L must be 1 but
the other powers are zero. This gives the equations:
r + s = 0
2r + t = 1
−r − t = 0
which is a set of three equations for three unknowns. Easy to solve!
The last of them gives r = −t. Substituting this into the second equation gives
2r + t = 2(−t) + t = −t = 1 =⇒ t = −1
Then r = +1 and the first equation gives us s = −1. With these values, we can confidently
say that
hr
ms
ct
= h1
m−1
c−1
=
h
mc
has units of length.
1.2.4 Vectors; Vector Addition
17. (a) What is the sum in unit–vector notation of the two vectors a = 4.0i + 3.0j
and b = −13.0i + 7.0j? (b) What are the magnitude and direction of a + b? [HRW5
3-20]
(a) Summing the corresponding components of vectors a and b we find:
a + b = (4.0 − 13.0)i + (3.0 + 7.0)j
= −9.0i + 10.0j
This is the sum of the two vectors is unit–vector form.
(b) Using our results from (a), the magnitude of a + b is
|a + b| =
q
(−9.0)2 + (10.0)2 = 13.4
and if c = a + b points in a direction θ as measured from the positive x axis, then the
tangent of θ is found from
tan θ =

cy
cx

= −1.11
If we naively take the arctangent using a calculator, we are told:
θ = tan−1
(−1.11) = −48.0◦
which is not correct because (as shown in Fig. 1.5), with cx negative, and cy positive, the

c
x
y
Figure 1.5: Vector c, found in Example 17. With cx = −9.0 and cy = +10.0, the direction of c is in the
second quadrant.
x
y
b
a
4.0 m
5.0 m
N
E
W
S
35o
Figure 1.6: Vectors a and b as given in Example 18.
correct angle must be in the second quadrant. The calculator was fooled because angles
which differ by multiples of 180◦
have the same tangent. The direction we really want is
θ = −48.0◦
+ 180.0◦
= 132.0◦
18. Vector a has magnitude 5.0 m and is directed east. Vector b has magnitude
4.0 m and is directed 35◦
west of north. What are (a) the magnitude and (b) the
direction of a + b? What are (c) the magnitude and (d) the direction of b − a?
Draw a vector diagram for each combination. [HRW6 3-15]
(a) The vectors are shown in Fig. 1.6. (On the axes are shown the common directions N, S,
E, W and also the x and y axes; “North” is the positive y direction, “East” is the positive
x direction, etc.) Expressing the vectors in i, j notation, we have:
a = (5.00 m)i
and
b = −(4.00 m) sin 35◦
+ (4.00 m) cos 35c
irc
= (−2.29 m)i + (3.28 m)j
So if vector c is the sum of vectors a and b then:
cx = ax + bx = (5.00 m) + (−2.29 m) = 2.71 m
cy = ay + by = (0.00 m) + (3.28 m) = 3.28 m

x
y
b
a
N
E
W
S
x
y
b
-a
N
E
W
S
(a) (b)
c
d
Figure 1.7: (a) Vector diagram showing the addition a + b. (b) Vector diagram showing b − a.
The magnitude of c is
c =
q
c2
x + c2
y =
q
(2.71 m)2 + (3.28 m)2 = 4.25 m
(b) If the direction of c, as measured counterclockwise from the +x axis is θ then
tan θ =
cy
cx
=
3.28 m
2.71 m
= 1.211
then the tan−1
operation on a calculator gives
θ = tan−1
(1.211) = 50.4◦
and since vector c must lie in the first quadrant this angle is correct. We note that this angle
is
90.0◦
− 50.4◦
= 39.6◦
just shy of the +y axis (the “North” direction). So we can also express the direction by
saying it is “39.6◦
East of North”.
A vector diagram showing a, b and c is given in Fig. 1.7(a).
(c) If the vector d is given by d = b − a then the components of d are given by
dx = bx − ax = (−2.29 m) − (5.00 m) = −7.29 m
cy = ay + by = (3.28 m) − (0.00 m) + (3.28 m) = 3.28 m
The magnitude of c is
d =
q
d2
x + d2
y =
q
(−7.29 m)2 + (3.28 m)2 = 8.00 m
(d) If the direction of d, as measured counterclockwise from the +x axis is θ then
tan θ =
dy
dx
=
3.28 m
−7.29 m
= −0.450

y
x
30o
105o
a
b
Figure 1.8: Vectors for Example 19.
Naively pushing buttons on the calculator gives
θ = tan−1
(−0.450) = −24.2◦
which can’t be right because from the signs of its components we know that d must lie in the
second quadrant. We need to add 180◦
to get the correct answer for the tan−1
operation:
θ = −24.2◦
+ 180.0◦
= 156◦
But we note that this angle is
180◦
− 156◦
= 24◦
shy of the −y axis, so the direction can also be expressed as “24◦
North of West”.
A vector diagram showing a, b and d is given in Fig. 1.7(b).
19. The two vectors a and b in Fig. 1.8 have equal magnitudes of 10.0 m. Find
(a) the x component and (b) the y component of their vector sum r, (c) the
magnitude of r and (d) the angle r makes with the positive direction of the x
axis. [HRW6 3-21]
(a) First, find the x and y components of the vectors a and b. The vector a makes an angle
of 30◦
with the +x axis, so its components are
ax = a cos 30◦
= (10.0 m) cos 30◦
= 8.66 m
ay = a sin 30◦
= (10.0 m) sin 30◦
= 5.00 m
The vector b makes an angle of 135◦
with the +x axis (30◦
plus 105◦
more) so its components
are
bx = b cos 135◦
= (10.0 m) cos 135◦
= −7.07 m
by = b sin 135◦
= (10.0 m) sin 135◦
= 7.07 m
Then if r = a + b, the x and y components of the vector r are:
rx = ax + bx = 8.66 m − 7.07 m = 1.59 m
ry = ay + by = 5.00 m + 7.07 m = 12.07 m

A
12.0 m
C
15.0 m
40.0o
20.0o
y
x
Figure 1.9: Vectors A and C as described in Example 20.
So the x component of the sum is rx = 1.59 m, and. . .
(b) . . . the y component of the sum is ry = 12.07 m.
(c) The magnitude of the vector r is
r =
q
r2
x + r2
y =
q
(1.59 m)2 + (12.07 m)2 = 12.18 m
(d) To get the direction of the vector r expressed as an angle θ measured from the +x axis,
we note:
tan θ =
ry
rx
= 7.59
and then take the inverse tangent of 7.59:
θ = tan−1
(7.59) = 82.5◦
Since the components of r are both positive, the vector does lie in the first quadrant so that
the inverse tangent operation has (this time) given the correct answer. So the direction of r
is given by θ = 82.5◦
.
20. In the sum A + B = C, vector A has a magnitude of 12.0 m and is angled 40.0◦
counterclockwise from the +x direction, and vector C has magnitude of 15.0 m
and is angled 20.0◦
counterclockwise from the −x direction. What are (a) the
magnitude and (b) the angle (relative to +x) of B? [HRW6 3-22]
(a) Vectors A and C are diagrammed in Fig. 1.9. From these we can get the components
of A and C (watch the signs on vector C from the odd way that its angle is given!):
Ax = (12.0 m) cos(40.0◦
) = 9.19 m Ay = (12.0 m) sin(40.0◦
) = 7.71 m
Cx = −(15.0 m) cos(20.0◦
) = −14.1 m Cy = −(15.0 m) sin(20.0◦
) = −5.13 m
(Note, the vectors in this problem have units to go along with their magnitudes, namely m
(meters).) Then from the relation A + B = C it follows that B = C − A, and from this we
find the components of B:
Bx = Cx − Ax = −14.1 m − 9.19 m = −23.3 m

By = Cy − Ay = −5.13 m − 7.71 m = −12.8 m
Then we find the magnitude of vector B:
B =
q
B2
x + B2
y =
q
(−23.3)2 + (−12.8)2 m = 26.6 m
(b) We find the direction of B from:
tan θ =

By
Bx

= 0.551
If we naively press the “atan” button on our calculators to get θ, we are told:
θ = tan−1
(0.551) = 28.9◦
(?)
which cannot be correct because from the components of B (both negative) we know that
vector B lies in the third quadrant. So we need to ad 180◦
to the naive result to get the
correct answer:
θ = 28.9◦
+ 180.0◦
= 208.9◦
.
This is the angle of B, measured counterclockwise from the +x axis.
21. If a − b = 2c, a + b = 4c and c = 3i + 4j, then what are a and b? [HRW5 3-24]
We notice that if we add the first two relations together, the vector b will cancel:
(a − b) + (a + b) = (2c) + (4c)
which gives:
2a = 6c =⇒ a = 3c
and we can use the last of the given equations to substitute for c; we get
a = 3c = 3(3i + 4j) = 9i + 12j
Then we can rearrange the first of the equations to solve for b:
b = a − 2c = (9i + 12j) − 2(3i + 4j)
= (9 − 6)i + (12 − 8)j
= 3i + 4j
So we have found:
a = 9i + 12j and b = 3i + 4j
22. If A = (6.0i − 8.0j) units, B = (−8.0i + 3.0j) units, and C = (26.0i + 19.0j) units,
determine a and b so that aA + bB + C = 0. [Ser4 3-46]

A
B
C
y
x
450
450
Figure 1.10: Vectors for Example 23
The condition on the vectors given in the problem:
aA + bB + C = 0
is a condition on the individual components of the vectors. It implies:
aAx + bBx + Cx = 0 and aAy + bBy + Cy = 0 .
So that we have the equations:
6.0a − 8.0b + 26.0 = 0
−8.0a + 3.0b + 19.0 = = 0
We have two equations for two unknowns so we can find a and b. The are lots of ways
to do this; one could multiply the first equation by 4 and the second equation by 3 to get:
24.0a − 32.0b + 104.0 = 0
−24.0a + 9.0b + 57.0 = = 0
Adding these gives
−23.0b + 161 = 0 =⇒ b =
−161.0
−23.0
= 7.0
and then the first of the original equations gives us a:
6.0a = 8.0b − 26.0 = 8.0(7.0) − 26.0 = 30.0 =⇒ a =
30.0
6.0
= 5.0
and our solution is
a = 7.0 b = 5.0
23. Three vectors are oriented as shown in Fig. 1.10, where |A| = 20.0 units,
|B| = 40.0 units, and |C| = 30.0 units. Find (a) the x and y components of the

resultant vector and (b) the magnitude and direction of the resultant vector.
[Ser4 3-47]
(a) Let’s first put these vectors into “unit–vector notation”:
A = 20.0j
B = (40.0 cos 45◦
)i + (40.0 sin 45◦
)j = 28.3i + 28.3j
C = (30.0 cos(−45◦
))i + (30.0 sin(−45◦
))j = 21.2i − 21.2j
Adding the components together, the resultant (total) vector is:
Resultant = A + B + C
= (28.3 + 21.2)i + (20.0 + 28.3 − 21.2)j
= 49.5i + 27.1j
So the x component of the resultant vector is 49.5 and the y component of the resultant is
27.1.
(b) If we call the resultant vector R, then the magnitude of R is given by
R =
q
R2
x + R2
y =
q
(49.5)2 + (27.1)2 = 56.4
To find its direction (given by θ, measured counterclockwise from the x axis), we find:
tan θ =
Ry
Rx
=
27.1
49.5
= 0.547
and then taking the inverse tangent gives a possible answer for θ:
θ = tan−1
(0.547) = 28.7◦
.
Is this the right answer for θ? Since both components of R are positive, it must lie in the
first quadrant and so θ must be between 0◦
and 90◦
. So the direction of R is given by 28.7◦
.
24. A vector B, when added to the vector C = 3.0i+4.0j, yields a resultant vector
that is in the positive y direction and has a magnitude equal to that of C. What
is the magnitude of B? [HRW5 3-26]
If the vector B is denoted by B = Bxi + Byj then the resultant of B and C is
B + C = (Bx + 3.0)i + (By + 4.0)j .
We are told that the resultant points in the positive y direction, so its x component must
be zero. Then:
Bx + 3.0 = 0 =⇒ Bx = −3.0 .

Now, the magnitude of C is
C =
q
C2
x + C2
y =
q
(3.0)2 + (4.0)2 = 5.0
so that if the magnitude of B + C is also 5.0 then we get
|B + C| =
q
(0)2 + (By + 4.0)2 = 5.0 =⇒ (By + 4.0)2
= 25.0 .
The last equation gives (By + 4.0) = ±5.0 and apparently there are two possible answers
By = +1.0 and By = −9.0
but the second case gives a resultant vector B + C which points in the negative y direction
so we omit it. Then with By = 1.0 we find the magnitude of B:
B =
q
(Bx)2 + (By)2 =
q
(−3.0)2 + (1.0)2 = 3.2
The magnitude of vector B is 3.2.
1.2.5 Multiplying Vectors
25. Vector A extends from the origin to a point having polar coordinates (7, 70◦
)
and vector B extends from the origin to a point having polar coordinates (4, 130◦
).
Find A · B. [Ser4 7-13]
We can use Eq. 1.7 to find A · B. We have the magnitudes of the two vectors (namely
A = 7 and B = 4) and the angle φ between the two is
φ = 130◦
− 70◦
= 60◦
.
Then we get:
A · B = AB cos φ = (7)(4) cos 60◦
= 14
26. Find the angle between A = −5i − 3j + 2k and B = −2j − 2k. [Ser4 7-20]
Eq. 1.7 allows us to find the cosine of the angle between two vectors as long as we know
their magnitudes and their dot product. The magnitudes of the vectors A and B are:
A =
q
A2
x + A2
y + A2
z =
q
(−5)2 + (−3)2 + (2)2 = 6.164
B =
q
B2
x + B2
y + B2
z =
q
(0)2 + (−2)2 + (−2)2 = 2.828
and their dot product is:
A · B = AxBx + AyBy + AzBz = (−5)(0) + (−3)(−2) + (2)(−2) = 2

Then from Eq. 1.7, if φ is the angle between A and B, we have
cos φ =
A · B
AB
=
2
(6.164)(2.828)
= 0.114
which then gives
φ = 83.4◦
.
27. Two vectors a and b have the components, in arbitrary units, ax = 3.2,
ay = 1.6, bx = 0.50, by = 4.5. (a) Find the angle between the directions of a and
b. (b) Find the components of a vector c that is perpendicular to a, is in the xy
plane and has a magnitude of 5.0 units. [HRW5 3-51]
(a) The scalar product has something to do with the angle between two vectors... if the
angle between a and b is φ then from Eq. 1.7 we have:
cos φ =
a · b
ab
.
We can compute the right–hand–side of this equation since we know the components of a
and b. First, find a · b. Using Eq. 1.8 we find:
a · b = axbx + ayby
= (3.2)(0.50) + (1.6)(4.5)
= 8.8
Now find the magnitudes of a and b:
a =
q
a2
x + a2
y =
q
(3.2)2 + (1.6)2 = 3.6
b =
q
b2
x + b2
y =
q
(0.50)2 + (4.5)2 = 4.5
This gives us:
cos φ =
a · b
ab
=
8.8
(3.6)(4.5)
= 0.54
From which we get φ by:
φ = cos−1
(0.54) = 57◦
(b) Let the components of the vector c be cx and cy (we are told that it lies in the xy plane).
If c is perpendicular to a then the dot product of the two vectors must give zero. This tells
us:
a · c = axcx + aycy = (3.2)cx + (1.6)cy = 0
This equation doesn’t allow us to solve for the components of c but it does give us:
cx = −
1.6
3.2
cy = −0.50cy

Since the vector c has magnitude 5.0, we know that
c =
q
c2
x + c2
y = 5.0
Using the previous equation to substitute for cx gives:
c =
q
c2
x + c2
y
=
q
(−0.50 cy)2 + c2
y
=
q
1.25 c2
y = 5.0
Squaring the last line gives
1.25c2
y = 25 =⇒ c2
y = 20. =⇒ cy = ±4.5
One must be careful... there are two possible solutions for cy here. If cy = 4.5 then we have
cx = −0.50 cy = (−0.50)(4.5) = −2.2
But if cy = −4.5 then we have
cx = −0.50 cy = (−0.50)(−4.5) = 2.2
So the two possibilities for the vector c are
cx = −2.2 cy = 4.5
and
cx = 2.2 cy = −4.5
28. Two vectors are given by A = −3i + 4j and B = 2i + 3j. Find (a) A × B and
(b) the angle between A and B. [Ser4 11-7]
(a) Setting up the determinant in Eq. 1.12 (or just using Eq. 1.11 for the cross product) we
find:
A × B =
i j k
−3 4 0
2 3 0
= (0 − 0)i + (0 − 0)j + ((−9) − (8))k = −17k
(b) To get the angle between A and B it is easiest to use the dot product and Eq. 1.7. The
magnitudes of A and B are:
A =
q
A2
x + A2
y =
q
(−3)2 + (4)2 = 5 B =
q
B2
x + B2
y =
q
(2)2 + (3)2 = 3.61

and the dot product of the two vectors is
A · B = AxBx + AyBy + AzBz = (−3)(2) + (4)(3) = 6
so then if φ is the angle between A and B we get:
cos φ =
A · B
AB
=
6
(5)(3.61)
= 0.333
which gives
φ = 70.6◦
.
29. Prove that two vectors must have equal magnitudes if their sum is perpen-
dicular to their difference. [HRW6 3-23]
Suppose the condition stated in this problem holds for the two vectors a and b. If the
sum a + b is perpendicular to the difference a − b then the dot product of these two vectors
is zero:
(a + b) · (a − b) = 0
Use the distributive property of the dot product to expand the left side of this equation. We
get:
a · a − a · b + b · a − b · b
But the dot product of a vector with itself gives the magnitude squared:
a · a = a2
x + a2
y + a2
z = a2
(likewise b · b = b2
) and the dot product is commutative: a · b = b · a. Using these facts,
we then have
a2
− a · b + a · b + b2
= 0 ,
which gives:
a2
− b2
= 0 =⇒ a2
= b2
Since the magnitude of a vector must be a positive number, this implies a = b and so vectors
a and b have the same magnitude.
30. For the following three vectors, what is 3C · (2A × B) ?
A = 2.00i + 3.00j − 4.00k
B = −3.00i + 4.00j + 2.00k C = 7.00i − 8.00j
[HRW6 3-36]

Actually, from the properties of scalar multiplication we can combine the factors in the
desired vector product to give:
3C · (2A × B) = 6C · (A × B) .
Evaluate A × B first:
A × B =
i j k
2.0 3.0 −4.0
−3.0 4.0 2.0
= (6.0 + 16.0)i + (12.0 − 4.0)j + (8.0 + 9.0)k
= 22.0i + 8.0j + 17.0k
Then:
C · (A × B) = (7.0)(22.0) − (8.0)(8.0) + (0.0)(17.0) = 90
So the answer we want is:
6C · (A × B) = (6)(90.0) = 540
31. A student claims to have found a vector A such that
(2i − 3j + 4k) × A = (4i + 3j − k) .
Do you believe this claim? Explain. [Ser4 11-8]
Frankly, I’ve been in this teaching business so long and I’ve grown so cynical that I don’t
believe anything any student claims anymore, and this case is no exception. But enough
about me; let’s see if we can provide a mathematical answer.
We might try to work out a solution for A, but let’s think about some of the basic
properties of the cross product. We know that the cross product of two vectors must be
perpendicular to each of the “multiplied” vectors. So if the student is telling the truth, it
must be true that (4i + 3j − k) is perpendicular to (2i − 3j + 4k). Is it?
We can test this by taking the dot product of the two vectors:
(4i + 3j − k) · (2i − 3j + 4k) = (4)(2) + (3)(−3) + (−1)(4) = −5 .
The dot product does not give zero as it must if the two vectors are perpendicular. So we
have a contradiction. There can’t be any vector A for which the relation is true.

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