General Physics (Phys 1011): Vectors Basics

Debre Berhan University
College of Natural and Computational Sciences
Department of physics
General Physics (Phys 1011)
Chapter 1: Vectors
28 February, 2023
DBU, Department of physics General Physics (Phys 1011) 28 February, 2023 1 / 62

Vectors
☞ Physical quantities can be divided into two distinct groups with respect
to measurement: scalar and vector.
☞ A scalar is a quantity that is completely specified by a number and
unit.
• Scalars obey the rules of ordinary algebra
• Examples: mass, time, volume, speed, · · ·
☞ A vector is a quantity that is specified by both a magnitude and di-
rection in space.
• Vectors obey the laws of vector algebra.
• Examples: displacement, velocity, acceleration, · · ·

Vector Representation
☞ Algebraic representation:
• Vectors are represented algebraically by a letter (or symbol) with
an arrow over its head, ⃗
A
• the magnitude of a vector is a positive scalar and is written
either A or |⃗
A|
☞ Geometrical representation:
• In diagrams, a vector is represented by an arrow drawn in the
direction of the vector.
• The length of the arrow is a measure of the magnitude of the
vector.

Vectors
Trigonometry review
The three most basic trigonometric functions of a right triangle are the
sine, cosine, and tangent, defined as follows:
sin θ =
side opposite to θ
hypotenuse
=
y
r
cos θ =
side adjacent to θ
hypotenuse
=
x
r
tan θ =
side opposite to θ
side adjacent to θ
=
y
x
The Pythagorean theorem is :
r2
= x2
+ y2

Components of Vector
Consider a vector ⃗
A lying in the xy plane and making an angle θ
⃗
A can be expressed as the sum of
two component vectors: ⃗
Ax & ⃗
Ay,
⃗
A = ⃗
Ax + ⃗
Ay
where
☞ ⃗
Ax is the projection of ⃗
A along
x-axis and is called the
x-component of ⃗
A, and
☞ ⃗
Ay is the projection of ⃗
A along
y-axis and is called the
y-component of ⃗
A
☞ |⃗
Ax + ⃗
Ay| ̸= |⃗
Ax| + |⃗
Ay|
From trignometric functions:
Ax = A cos θ and Ay = A sin θ
the magnitude of ⃗
A:
A =
q
A2
x + A2
y

Example
Calculate the x- and y- component of the vector with magnitude 24.0 m
and direction 56.0◦.
Solution
Ax = A cos θ = (24.0m) × cos(56.0◦
) = +13.4m
Ay = A sin θ = (24.0m) × sin(56.0◦
) = +19.9m

Components of Vector
The direction angle or simply di-
rection is the angle the vector
forms with the +x-axis. Hence, the
direction of ⃗
A is
θ = tan−1

Ay
Ax

Conventionally, θA is positive for
counterclockwise rotation and neg-
ative for clockwise.
The signs of Ax and Ay depend on the direction angle θA: for vectors in
1st/4th quadrant, Ax 0 and θ = θA; for vectors in the 4th quadrant,
θ 0, since θA is measured clockwise from the +x-axis; for vectors 2nd
quadrant, θ 0, for vectors in 2nd/3rd quadrant, Ax 0, θA = θ + 180◦

Example
☞ Find the magnitude and
direction of the vector ⃗
A
represented by the following
pairs of components:
a. (Ax, Ay) = (3.0, 4.0)m
b. (Ax, Ay) = (−3.0, 4.0)m
c. (Ax, Ay) = (−3.0, −4.0)m
d. (Ax, Ay) = (3.0, −4.0)m
Solution
The magnitude of ⃗
A:
A =
q
A2
x + A2
y
Hence,
A =
q
(3.0m)2 + (4.0m)2
=
q
(−3.0m)2 + (4.0m)2
=
q
(−3.0m)2 + (−4.0m)2
=
q
(3.0m)2 + (−4.0m)2
= 5.0m

Example
a. θ = θA = tan−1

4.0m
3.0m

= +53.1◦
b. θ = tan−1

4.0m
−3.0m

= −53.1◦
θA = θ + 180◦
= −53.1◦
+ 180◦
= +126.9◦
c. θ = tan−1

−4.0m
−3.0m

= +53.1◦.
θA = θ + 180◦
= +53.1◦
+ 180◦
= +233.1◦
d. θ = θA = tan−1

−4.0m
3.0m

= −53.1◦.

Some Properties of Vectors
A vector changes if its magnitude and/or direction change
• The negative of the vector ⃗
B is the vector
that when added to ⃗
B gives zero. That is,
⃗
B + (−⃗
B) = 0
The vecotrs ⃗
B and −⃗
B have the same magni-
tude but opposite directions.
• Two vectors are equal if they have the
same magnitude and the same direction,
no matter where they are located.

Some Properties of Vectors
Multiplying a Vector by a Scalar
☞ when a vector ⃗
A is multiplied by a scalar quantity α, the direction of
the new vector ⃗
B depends on the sign of the scalar
⃗
B = α⃗
A
If α 0, then ⃗
B is parallel to ⃗
A and if α 0, then ⃗
B is directed to
the opposite of ⃗
A.
☞ Moreover, multiplication by a scalar is distributive:
α1
⃗
A + α2
⃗
A = (α1 + α2)⃗
A
☞ scalar multiplication by a sum of vectors is distributive:
α(⃗
A + ⃗
B) = α⃗
A + α⃗
B

Vectors
Geometric Vector Addition: Head-to-tail method
☞ A single vector that is obtained by adding two or more vectors is called
resultant vector, ⃗
R.
☞ Two vectors can be added geometrically using the tail-to-head method
and the parallelogram rule.
1. Head-to-tail method
Graphically vectors can be added by joining their head to tail and in
any order their resultant vector is the vector drawn from the tail of the
first vector to the head of the last vector.

Vectors
Suppose a particle undergoes a displacement
⃗
A followed by a second displacement ⃗
B. The
resultant vector, ⃗
R is:
⃗
R = ⃗
A + ⃗
B
If we make the displacements ⃗
A and ⃗
B in re-
verse order, thus,
⃗
R = ⃗
B + ⃗
A
Therefore, vector addition obeys the com-
mutative law, i.e.,
⃗
A + ⃗
B = ⃗
B + ⃗
A

Vectors
The method can be extended to
add morethan two vectors:
Suppose, we have four vecotrs as
shown in the figure.
The resultant vector sum is the vector
drawn from the tail of the first vector
to the tip of the last.
⃗
R = ⃗
A + ⃗
B + ⃗
C + ⃗
D
their sum is independent of the way
in which the individual vectors are
grouped together.
⃗
A + (⃗
B + ⃗
C) = (⃗
A + ⃗
B) + ⃗
C
This property is called the associa-
tive law of addition

Vectors
2. Parallelogram rule
It states that the resultant ⃗
R of two vectors ⃗
A and ⃗
B is the diagonal
of the parallelogram for which the two vectors ⃗
A and ⃗
B becomes
adjacent sides
Cosine law: |⃗
R| =
q
|⃗
A|2 + |⃗
B|2 − 2|⃗
A||⃗
B| cos(θ)
Sine law:
sin θ
|⃗
R|
=
sin α
|⃗
A|
=
sin β
|⃗
B|

Vectors
Geometric Vector Subtraction
☞ We define the difference ⃗
A − ⃗
B of two vectors ⃗
A and ⃗
B to be the
vector sum of ⃗
A and −⃗
B
⃗
A − ⃗
B = ⃗
A + (−⃗
B)

Examples
Example-1: parallelogram method
A car travels 20.0 km due north and then 35.0 km in a direction 60.0◦
west of north.
a. Find the magnitude and direction of the car’s
resultant displacement.
b. Suppose the trip were taken with the two
vectors in reverse order: 35.0 km at 60.0◦
west of north first and then 20.0 km due
north. How would the magnitude and the
direction of the resultant vector change?

Examples
Example-1: parallelogram method
Solution
Given that: |⃗
A| = 20.0 km, |⃗
B| = 35.0 km, θ = 180◦ − 60◦ = 120◦
Cosine law: |⃗
R| =
q
|⃗
A|2 + |⃗
B|2 − 2|⃗
A||⃗
B| cos(θ) = 48.2 km
Sine law:
sin β
|⃗
B|
=
sin θ
|⃗
R|
sin β =
|⃗
B|
|⃗
R|
sin θ =
35.0 km
48.2 km
sin 120◦
= 0.629
β = 38.9◦
The resultant displacement of the car is 48.2 km in a direction 38.9◦ west
of north.

Unit Vector
☞ A unit vector is a vector that has a magnitude of 1, with no units.
☞ its purpose is to point a given vector in specified direction.
☞ It is usually denoted with a hat (b)
☞ We use î, ĵ and k̂ to represent unit vectors pointing in the +x, +y
and +z directions, respectively.
☞ The unit vectors î, ĵ, and k̂ form a set of mutually perpendicular
vectors.
☞ |î| = |ĵ| = |k̂| = 1

Unit Vector, Cont’d
Vectors in 2D
☞ Consider a 2D vector ⃗
A,
⃗
Ax = Axî
⃗
Ay = Ayĵ
☞ Interms of its components,
⃗
A = Axî + Ayĵ
☞ magnitude of ⃗
A,
A =
q
A2
x + A2
y
☞ The direction of ⃗
A from the
+x-axis
tan θ = Ay/Ax

Vectors in 3D
To specify the location of a point in space, we need three coordinates (x,
y, z), where x and y specify locations in a plane, and z gives a vertical
position above or below the plane.

Vectors in 3D
Consider the vector ⃗
A : ⃗
Ax = Axî, ⃗
Ay = Ayĵ and ⃗
Az = Azk̂
☞ ⃗
A = Axî + Ayĵ + Azk̂
☞ |⃗
A| =
q
A2
x + A2
y + A2
z
☞ direction of ⃗
A
cos θx =
Ax
|⃗
A|
cos θy =
Ay
|⃗
A|
cos θz =
Az
|⃗
A|
☞ The unit vector in the same direction as ⃗
A is: Â =
⃗
A
|⃗
A|

Vector Addition: Analytical/Component Method
Resultant of Vectors
☞ A single vector that is obtained by adding two or more vectors is
called resultant vector, ⃗
R. Equivalently, we can represent:
⃗
R = Rxî + Ryĵ + Rzk̂
☞ Suppose we have two vectors
⃗
⃗
B = Bxî + Byĵ + Bzk̂
to find the resultant of the two vectors, we simply add them
component by component,
⃗
R = ⃗
A + ⃗
B

Therefore,
⃗
R = (Axî + Ayĵ + Azk̂) + (Bxî + Byĵ + Bzk̂)
= (Ax + Bx)î + (Ay + By)ĵ + (Az + Bz)k̂
Comparison with the resultant vector representation gives
Rx = Ax + Bx
Ry = Ay + By
Rz = Az + Bz

☞ Analytical methods can be used to find components of a resultant of
many vectors.
⃗
R = ⃗
R1 + ⃗
R2 + ⃗
R3 + · · · + ⃗
RN
=
N
X
ℓ=1
⃗
Rℓ
=
N
X
ℓ=1

Rℓxî + Rℓyĵ + Rℓzk̂

=
N
X
ℓ=1
Rℓx
!
î +
N
X
ℓ=1
Rℓy
!
ĵ +
N
X
ℓ=1
Rℓz
!
k̂

Therefore, scalar components of the resultant vector are
Rx =
N
X
ℓ=1
Rℓx = R1x + R2x + · · · + RNx
Ry =
N
X
ℓ=1
Rℓy = R1y + R2y + · · · + RNy
Rz =
N
X
ℓ=1
Rℓz = R1z + R2z + · · · + RNz
Having found the scalar components, we can write the resultant in vector
component form:
⃗
R = Rxî + Ryĵ + Rzk̂

Example-1
☞ A shopper leaves home and drives to a store located 7.00 km away in
a direction 30.0° north of east. Leaving the store, the shopper drives
5.00 km in a direction 50.0° west of north to a restaurant.
a. Find the distance and direction from the
shopper’s home to the restaurant.
b. What are the answers if both distances in
the problem statement are doubled but the
directions are not changed?
c. Find the distance and direction from the
shopper’s home to the restaurant if the store
is located 7.00 km due east from home and
the restaurant remains 5.00 km from the
store in a direction 50.0° west of north.

Example-1: Solution
Given: |⃗
A| = 7.00km, |⃗
B| = 5.00km, θA = 30.0◦ and θB = 50.0◦
The components of ⃗
A and ⃗
B:
Ax = |⃗
A| cos θA = 7.00 km × cos(30.0◦
) = 6.06 km
Ay = |⃗
A| sin θA = 7.00 km × sin(30.0◦
) = 3.50 km
Bx = |⃗
B| cos θB = −5.00 km × cos(40.0◦
) = −3.83 km
By = |⃗
B| sin θB = 5.00 km × sin(40.0◦
) = 3.21 km
The vector sum, ⃗
R = ⃗
A + ⃗
B. Interms of x- and y-component,
Rx = Ax + Bx = 2.23km
Ry = Ay + By = 6.71km
Therefore, R =
q
R2
x + R2
y = 7.07km and θ = tan−1

Ry
Rx

= 71.6◦

Example-2
Five coplanar forces shown in the figure act on an object. Find their
resultant by using the component method.

Example-2: Solution
(a) First we find the x- and y-components of each force.
Force x-component y-component
19.0 N 19.0 N 0 N
15.0 N 7.50 N 13.0 N
16.0 N -11.3 N 11.3 N
11.0 N -9.53 N -5.50 N
22.0 N 0 N -22.0 N
(b) The resultant ⃗
R has components Rx =
P
Fx and Ry =
P
Fy
Rx = 19.0N + 7.50N + (−11.3N) + (−9.53N) + 0N = +5.7N
Ry = 0N + 13.0N + 11.3N + (−5.50N) + (−22.0N) = −3.2N

Example-2: Solution
(c) The magnitude of the resultant is
|⃗
R| =
q
R2
x + R2
y =
q
(+5.7N)2 + (−3.2N)2 = 6.5N
(d) Finally, we sketch the resultant as
shown in the fig. and find its
angle
tan ϕ =
−3.2N
5.7N
= −0.56
from which ϕ = 29◦. Then,
θ = 360◦ − 29◦ = 331◦
Therefore, the resultant is 6.5 N at 331◦ from +x-axis

Example-3
If the velocity vector of the military convoy is
⃗
v = (4.0î + 3.0ĵ − 0.1k̂)km/h
what is the unit vector of its direction of motion?
Solution
The magnitude of the vector ⃗
v is
|⃗
v| =
q
v2
x + v2
y + v2
z =
q
(4.0)2 + (3.0)2 + (0.1)2km/h = 5.0km/h
the unit vector v̂
v̂ =
⃗
v
|⃗
v|
=
4.0î + 3.0ĵ − 0.1k̂
5.0
= (80.0î + 60.0ĵ + 2.0k̂) × 10−2

Example-4
Given the two displacements
⃗
A = (6.0î + 3.0ĵ − 1.0k̂)m
⃗
B = (4.0î − 5.0ĵ + 8.0k̂)m
find the magnitude of the displacement 2⃗
A − ⃗
B

Example-4: Solution
We multiply the vector ⃗
A by 2 (a scalar) and subtract the vector ⃗
B from
the result, so as to obtain the vector ⃗
C = 2⃗
A − ⃗
B
⃗
C = 2(6.0î + 3.0ĵ − 1.0k̂)m − (4.0î − 5.0ĵ + 8.0k̂)m
= [2(12.0 − 4.0)î + (6.0 + 5.0)ĵ + (−2.0 − 8.0)k̂]m
= (8.0î + 11.0ĵ − 10.0k̂)m
the magnitude of ⃗
C
|⃗
C| =
q
C2
x + C2
y + C2
z =
q
(8.0m)2 + (11.0m)2 + (−10.0m)2 = 16.9m

Exercise-1
☞ Arrange the following vectors in order of their magnitude, with the
vector of largest magnitude first.
⃗
A = (3.0î + 5.0ĵ − 2.0k̂)m
⃗
B = (−3.0î + 5.0ĵ − 2.0k̂)m
⃗
C = (3.0î − 5.0ĵ − 2.0k̂)m
⃗
D = (−3.0î + 5.0ĵ + 2.0k̂)m

Exercise-2
☞ A person going for a walk follows the path shown in Figure. The total
trip consists of four straight-line paths. At the end of the walk, what
is the person’s resultant displacement measured from the starting
point?

Products of Vectors
☞ Vectors are not ordinary numbers, so ordinary multiplication is not
directly applicable to vectors.
☞ A vector can be multiplied by another vector but may not be divided
by another vector.
☞ There are two kinds of products of vectors:
a) scalar product - yields a result that is a scalar quantity
b) vector product - yields another vector.
Suppose we have two vectors ⃗
A and
⃗
B with θ being the angle between
the vectors as shown in the
following figure.

Products of Vectors: Dot Product
☞ As the name suggests, the dot product between the two vectors is
denoted by ⃗
A · ⃗
B
☞ ⃗
A · ⃗
B can be interpreted as
either the product of |⃗
B| with
the orthogonal projection ⃗
A⊥
of vector ⃗
A onto the direction
of vector ⃗
B or vice versa.
⃗
A · ⃗
B = |⃗
B|(|⃗
A| cos θ) = BA⊥
⃗
B · ⃗
A = |⃗
A|(|⃗
B| cos θ) = AB⊥
Hence, as a general rule:
⃗
A · ⃗
B = |⃗
A||⃗
B| cos θ

☞ The dot product between two vectors returns a scalar values. Hence,
dot product is also known as scalar product
☞ Since |⃗
A| and |⃗
B| are positive quantities, the sign of the dot product
depends on θ, i.e.,
(i) 0◦ ≤ θ 90◦, ⃗
A · ⃗
B 0
(ii) 90◦ θ ≤ 180◦, ⃗
A · ⃗
B 0
(iii) θ = 90◦, ⃗
A · ⃗
B = 0 (the vectors are orthogonal)
The dot product is maximum when θ = 0◦ and minimum when
θ = 180◦
☞ The dot product obeys the laws of algebra:
(i) ⃗
A · ⃗
B = ⃗
B · ⃗
A · · · commutative
(ii) ⃗
A · (⃗
B + ⃗
C) = ⃗
A · ⃗
B + ⃗
A · ⃗
C · · · distributive

☞ î, ĵ and k̂ are mutually perpendicular to each other. Hence,
î · ĵ = î · k̂ = k̂ · ĵ = 0
î · î = ĵ · ĵ = k̂ · k̂ = 1
☞ in the rectangular coordinate system in a plane, the scalar
x-component of a vector is its dot product with î, and the scalar
y-component of a vector is its dot product with ĵ :
⃗
A · î = |⃗
A||î| cos θA = A cos θA = Ax
⃗
A · ĵ = |⃗
A||ĵ| cos(90◦
− θA) = A sin θA = Ay
☞ the scalar product of two vectors is the sum of the products of their
respective components.
⃗
A · ⃗
B = AxBx + AyBy + AzBz (proof!)

Example-1
Suppose the magnitudes of the vectors are given as |⃗
A| = 4.00 and
|⃗
B| = 5.00. Find the scalar product between the vectors.

Example-1
Solution:
method-1 ⃗
A · ⃗
B = |⃗
A||⃗
B| cos θ
= 4.00 × 5.00 × cos 77.0◦
= 4.50

Example-1
Solution:
method-1 ⃗
A · ⃗
B = |⃗
A||⃗
B| cos θ
= 4.00 × 5.00 × cos 77.0◦
= 4.50
method-2 Ax = |⃗
A| cos θ = (4.00) cos(53.0◦
) = 2.407
Ay = |⃗
A| sin θ = (4.00) sin(53.0◦
) = 3.195
Bx = |⃗
B| cos θ = (5.00) cos(130.0◦
) = −3.214
By = |⃗
B| sin θ = (5.00) sin(130.0◦
) = 3.830
Therefore, ⃗
A · ⃗
B = AxBx + AyBy
= (2.407)(−3.214) + (3.195)(3.830)
= 4.50

Example-2
The vectors ⃗
A and ⃗
B are given by
⃗
A = 2î + 3ĵ + k̂
⃗
B = −4î + 2ĵ − k̂
a. Determine the scalar product between ⃗
A and ⃗
B
b. Find the angle θ between ⃗
A and ⃗
B

Example-2: Solution
a. The scalar product:
⃗
A · ⃗
B = (2î + 3ĵ + k̂) · (−4î + 2ĵ − k̂)
= (2î) · (−4î) + (2î) · (2ĵ) + (2î) · (−k̂)
= −8 + 6 − 1
= −3

Example-2: Solution
a. The scalar product:
⃗
A · ⃗
B = (2î + 3ĵ + k̂) · (−4î + 2ĵ − k̂)
= (2î) · (−4î) + (2î) · (2ĵ) + (2î) · (−k̂)
= −8 + 6 − 1
= −3
b. The angle θ can be found from: ⃗
A · ⃗
B = |⃗
A||⃗
B| cos θ
|⃗
A| =
q
A2
x + A2
y + A2
z =
√
14
|⃗
B| =
q
B2
x + B2
y + B2
z =
√
21
cos θ =
⃗
A · ⃗
B
|⃗
A||⃗
B|
=
−3
√
14
√
21
θ = cos−1

−3
√
14
√
21

= 100◦

Exercise
Given two vectors
⃗
A = 4.0î + 7.0ĵ
⃗
B = 5.0î + 2.0ĵ
Calculate the following
(a) the magnitude of each vector
(b) the magnitude and direction of the vector difference |⃗
A − ⃗
B|
(c) the scalar product of the two vectors ⃗
A and ⃗
B
(d) the angle between these two vectors

Products of Vectors: Cross Product
☞ Suppose the tails of two vectors ⃗
A and ⃗
B are connected with each
other tail-to-tail. Thus, the cross product of the two vectors is
denoted by ⃗
A × ⃗
B
☞ Since cross product is itself a vector, it is referred to as the vector
product.
☞ |⃗
A × ⃗
B| is defined as the product of |⃗
A| with the magnitude of the
perpendicular component of ⃗
B to that of ⃗
A or vise versa:
☞ Therefore, |⃗
A × ⃗
B| = |⃗
A||⃗
B| sin θ
|⃗
A × ⃗
B| = |⃗
A|(|⃗
B| sin θ) |⃗
A × ⃗
B| = (|⃗
A| sin θ)|⃗
B|

☞ The direction of ⃗
A × ⃗
B is perpendicular to the plane that contains
vectors ⃗
A and ⃗
B as determined by the right-hand rule.
☞ Right-hand rule: Curl the fingers of your right hand around the
perpendicular line so that the finger point in the direction of rotation;
your thumb will then point in the direction of the cross product.
Example: the direction of ⃗
A × ⃗
B and ⃗
B × ⃗
A is determined as follows.
⃗
A × ⃗
B = −⃗
B × ⃗
A

The right-hand rule is sometimes called as a corkscrew right-hand rule.
A corkscrew is placed in a direction perpendicular to the plane that
contains vectors ⃗
A and ⃗
B , and its handle is turned in the direction from
the first to the second vector in the product. The direction of the cross
product is given by the progression of the corkscrew.
☞ Upward movement means the
cross-product vector points up.
☞ Downward movement means
the cross-product vector points
downward.

Some Properties
☞ The vector product is not commutative.
⃗
A × ⃗
B = −⃗
B × ⃗
A
☞ the cross product has distributive property:
⃗
A × (⃗
B + ⃗
C) = ⃗
A × ⃗
B + ⃗
A × ⃗
C
☞ the cross product of any vector with itself is zero
î × î = ĵ × ĵ = k̂ × k̂ = 0
i.e., each product is a zero vector - that is all components equal to
zero and an undefined direction

Some Properties
☞ The cross product of two different unit vectors is always a third unit
vector.
Example: Consider the pairs of î and ĵ
• The magnitude
|î × ĵ| = |ĵ||ĵ| sin 90◦
= (1)(1)(1) = 1
• The direction of the vector product î × ĵ must be orthogonal to
the xy-plane, which means it must be along the z-axis. The only
unit vectors along the z-axis are −k̂ or +k̂. By the corkscrew
rule, the direction of vector î × ĵ is identical to +k̂
î × ĵ = +k̂ = −ĵ × î
ĵ × k̂ = +î = −k̂ × ĵ
k̂ × î = +ĵ = −î × k̂

Some Properties
Notice that, the three unit vectors î, ĵ, and k̂ appear in the cyclic order as
shown in the figure below
When unit vectors in the cross product appear in a different order, the
result is a unit vector that is antiparallel to the remaining unit vector

Example-1
Suppose we want to find the cross product ⃗
A × ⃗
B for vectors
⃗
⃗
B = Bxî + Byĵ + Bzk̂
Hence,
⃗
A × ⃗
B = (Axî + Ayĵ + Azk̂) × (Bxî + Byĵ + Bzk̂)
Using the distributive and anticommutative properties of vector products
and cyclic order of unit vectors, we can find (show!)
⃗
A × ⃗
B = (AyBz − AzBy)î + (AzBx − AxBz)ĵ + (AxBy − AyBx)k̂

Example-1
Thus the components of ⃗
C = ⃗
A × ⃗
B are
Cx = AyBz − AzBy
Cy = AzBx − AxBz
Cz = AxBy − AyBx
The vector product can also be expressed in determinant form as
⃗
A × ⃗
B =
+î −ĵ +k̂
Ax Ay Az
Bx By Bz
=
Ay Az
By Bz
î +
Ax Az
Bx Bz
(−ĵ) +
Ax Ay
Bx By
k̂
= (AyBz − AzBy)î + (AzBx − AxBz)ĵ + (AxBy − AyBx)k̂

Example-2
Given
⃗
A = −2î − 3ĵ
⃗
B = −î − 2ĵ + 3k̂
⃗
C = −î + 2ĵ + 3k̂
Calculate the following
(a) ⃗
A × ⃗
B
(b) the angle between ⃗
A and ⃗
B
(c) the unit vector orthogonal to the plane containing ⃗
A and ⃗
C
(d) the area of the parallelogram formed by ⃗
B and ⃗
C
(e) 3⃗
C · (2⃗
A × ⃗
B)

Example-2: Solution
(a)
⃗
A × ⃗
B =
+î −ĵ +k̂
−2 −3 0
−1 −2 3
=
−3 0
−2 3
î +
−2 0
−1 3
(−ĵ) +
−2 −3
−1 −2
k̂
= [(−3)3 − 0(−2)]î − [(−2)3 − 0(−1)]ĵ + [(−2)(−2) − (−3)(−1
= −9î + 6ĵ + k̂

Example-2: Solution
(b) We use |⃗
A × ⃗
B| = |⃗
A||⃗
B| sin θ
|⃗
A × ⃗
B| =
q
(−9)2 + 62 + 12 =
√
118
|⃗
A| =
q
(−2)2 + (−3)2 =
√
13
|⃗
B| =
q
(−1)2 + (−2)2 + 32 =
√
14
Hence,
θ = sin−1 |⃗
A × ⃗
B|
|⃗
A||⃗
B|
!
= sin−1
s
118
13 × 14
!
= 53.6◦

Example-2: Solution
(c) We first calculate the vector orthogonal to ⃗
A and ⃗
C
⃗
U = ⃗
A × ⃗
C =
+î −ĵ +k̂
−2 −3 0
−1 2 3
=
−3 0
2 3
î +
−2 0
−1 3
(−ĵ) +
−2 −3
−1 2
k̂
= −9î + 6ĵ − 7k̂
Then, the unit vector orthogonal to ⃗
A and ⃗
C
Û =
⃗
U
|⃗
U|
=
−9î + 9ĵ − 7k̂
p
(−9)2 + 62 + (−7)2
=
−9î + 9ĵ − 7k̂
√
166

Example-2: Solution
(d) the area of the parallelogram formed by ⃗
B and ⃗
C
⃗
B × ⃗
C =
+î −ĵ +k̂
−1 −2 3
−1 2 3
=
−2 3
2 3
î +
−1 3
−1 3
(−ĵ) +
−1 −2
1 2
k̂
= [(−2)3 − 3(2)]î − [(−1)3 − 3(−1)]ĵ + [(−1)2 − (−2)1]k̂
= −12î − ĵ
The area is given by
|⃗
B × ⃗
C| =
q
(−12)2 + 12 =
√
145

Example-2: Solution
(e) from the properties of scalar multiplication we can combine the
factors in the desired vector product to give:
3⃗
C · (2⃗
A × ⃗
B) = 6⃗
C · (⃗
A × ⃗
B)
From the solution in part (a),
⃗
A × ⃗
B = −9î + 6ĵ + k̂
Then:
⃗
C · (⃗
A × ⃗
B) = (î + 2ĵ + 3k̂) · (−9î + 6ĵ + k̂) = 6
Hence,
6⃗
C · (⃗
A × ⃗
B) = 6(6) = 36

Exercises
(1) Given two vectors ⃗
A and ⃗
B such that
Ax = −2.0, Ay = 3.0, Ay = 4.0
Bx = 3.0, By = 1.0, Bx = −3.0
(a) Write the vecotrs ⃗
A and ⃗
B in unit vector notation.
(b) Find the magnitude of each vector
(c) Express a vector ⃗
C, where ⃗
C = 3.0⃗
A − 4.0⃗
B
(d) Find the magnitude and direction of ⃗
A − ⃗
B and ⃗
B − ⃗
A
(e) Calculate the scalar product ⃗
A · ⃗
B
(f) Find the magnitude and direction of ⃗
A × ⃗
B and ⃗
B × ⃗
A
(g) Find the angle between the directions ⃗
A and ⃗
B
(h) Obtain a unit vector perpendicular to the two vectors

General Physics (Phys 1011): Vectors Basics

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