BEKP 2453
Electromagnetics Theory
Chapter 1 Vector Analysis
Part 1: Vector Algebra
NUR HAKIMAH AB AZIZ
1
Learning Outcomes
NUR HAKIMAH AB AZIZ
2
SCALARS AND VECTORS
• A quantity can either be a scalar or a vector.
• Scalar - a quantity that only has magnitude.
– e.g. time, mass, distance, temperature, electric potential.
Ԧ 𝐵, 𝑈or
– Represented by a letter with an arrow on top of it e.g. 𝐴,
by a letter in boldface type e.g. A, B, U.
• Vector – a quantity that has both magnitude and direction.
– e.g. velocity, force, displacement, electric field intensity.
– Represented simply by a letter A, B, U etc
NUR HAKIMAH AB AZIZ
3
Position and Distance Vectors
Cartesian Coordinate
𝑃1 = (1,2,3)
𝑃1 = (𝑥1 , 𝑦1 , 𝑧1 ) = (1, 2, 3)
𝑃2 =(𝑥2 , 𝑦2 , 𝑧2 ) = (3, 5, 6)
𝑃2 = (3,5,6)
Position Vector
Starts from the origin
ෝ + 𝑧1 𝐳ො = 𝐱ො + 2ෝ
𝑂𝑃1 = 𝑅1 = 𝐑 𝟏 = 𝑥1 𝐱ො + 𝑦1 𝒚
𝒚 + 3ො𝐳 = 1, 2, 3
𝐑 𝟏 = 𝒂𝒙 + 2𝒂𝒚 + 3𝒂𝒛
ෝ + 𝑧2 𝐳ො = 3ො𝐱 + 5ෝ
𝑂𝑃2 = 𝑅2 = 𝐑 𝟐 = 𝑥2 𝐱ො + 𝑦2 𝒚
𝒚 + 6ො𝐳 = 3, 5, 6
𝐑 𝟐 = 3𝒂𝒙 + 5𝒂𝒚 + 6𝒂𝒛
Distance Vector
From one point to another
𝑃1 𝑃2 = 𝑅12 = 𝑅2 − 𝑅1 = 3, 5, 6 − 1, 2, 3 = 2, 3, 3
= 2ො𝐱 + 3ෝ
𝒚 + 3ො𝐳
NUR HAKIMAH AB AZIZ
4
Magnitude of A Vector
𝐴Ԧ = 𝐴𝑥 𝑥ො + 𝐴𝑦 𝑦ො + 𝐴𝑧 𝑧Ƹ = 4, 5, 3
𝐴 =
=
𝐴𝑟 2 + 𝐴𝑧 2
2
2
2
𝐴𝑥 + 𝐴𝑦 + 𝐴𝑧 =
42 + 52 + 32 = 50
NUR HAKIMAH AB AZIZ
𝑄1 (4,5,3)
5
Unit Vector
𝑥ො = 1𝑥ො , 𝑦ො = 1𝑦,
ො
𝑧Ƹ = 1𝑧Ƹ
𝑥ො = 𝑦ො = 𝑧Ƹ = 1
Unit vector is a vector with magnitude 1
𝐴Ԧ = 4, 5, 3
Unit vector of 𝐴Ԧ
𝐴Ԧ = 𝐴𝑥 𝑥ො + 𝐴𝑦 𝑦ො + 𝐴𝑧 𝑧Ƹ =
𝑎ො =
𝐴Ԧ
𝐴𝑥 2 + 𝐴𝑦 2 + 𝐴𝑧 2
4
50
5
,
50
,
3
(4,5,3)
50
Magnitude of 𝑎ො
𝑎ො =
4
50
2
+
5
50
2
+
3
2
50
Magnitude of a unit vector is always 1
=
50
=1
50
NUR HAKIMAH AB AZIZ
6
Try this..
𝐁𝐂 = 𝐶 − 𝐵
𝐴(2, −3, 1)
= 4, −7, −5
𝐵(3, 2, 0)
𝐶(7, −5, −5)
𝐵𝐶 =
42 + 72 + 52
= 90
4, −7, −5
෢=
𝑏𝑐
90
𝐂𝐀 = 𝐴 − 𝐶
Find the unit vector of AB, BC and CA
𝐀𝐁 = 𝐵 − 𝐴
= 1, 5, −1
𝐴𝐵 =
12 + 52 + 12
= 27
1, 5, −1
෢
𝑎𝑏 =
27
= −5, 2, 6
𝐶𝐴 =
52 + 22 + 62
= 65
−5, 2, 6
𝑐ෞ
𝑎=
65
NUR HAKIMAH AB AZIZ
7
Vector Addition
Vector Subtraction
𝐷 = 𝐴Ԧ − 𝐵
𝐷 = 𝐴𝑥 − 𝐵𝑥 𝑥ො + 𝐴𝑦 − 𝐵𝑦 𝑦ො + 𝐴𝑧 − 𝐵𝑧 𝑧Ƹ
= 𝐴𝑥 − 𝐵𝑥 , 𝐴𝑦 − 𝐵𝑦 , 𝐴𝑧 − 𝐵𝑧
𝐴Ԧ = 𝐴𝑥 𝑥ො + 𝐴𝑦 𝑦ො + 𝐴𝑧 𝑧Ƹ
𝐵 = 𝐵𝑥 𝑥ො + 𝐵𝑦 𝑦ො + 𝐵𝑧 𝑧Ƹ
Commutative Property
𝐶Ԧ = 𝐴Ԧ + 𝐵 = 𝐵 + 𝐴Ԧ
𝐶Ԧ = 𝐴𝑥 + 𝐵𝑥 𝑥ො + 𝐴𝑦 + 𝐵𝑦 𝑦ො + 𝐴𝑧 + 𝐵𝑧 𝑧Ƹ
= 𝐴𝑥 + 𝐵𝑥 , 𝐴𝑦 + 𝐵𝑦 , 𝐴𝑧 + 𝐵𝑧
NUR HAKIMAH AB AZIZ
8
Vector Addition: Head-To-Tail Rule
B
A
+
ෝ + 𝐴𝑧 𝐳ො
𝐴Ԧ = 𝐴𝑥 𝐱ො + 𝐴𝑦 𝒚
ෝ + 𝐵𝑧 𝐳ො
𝐵 = 𝐵𝑥 𝐱ො + 𝐵𝑦 𝒚
B
=
A
A+B
=
A+B
A
B
𝐴Ԧ + 𝐵 = 𝐵 + 𝐴Ԧ = (𝐴𝑥 + 𝐵𝑥 )ො𝐱 + (𝐴𝑦 + 𝐵𝑦 )ෝ
𝒚 + (𝐴𝑧 + 𝐵𝑧 )ො𝐳
9
Vector Addition: Parallelogram Rule
B
A
+
ෝ + 𝐴𝑧 𝐳ො
𝐴Ԧ = 𝐴𝑥 𝐱ො + 𝐴𝑦 𝒚
=
ෝ + 𝐵𝑧 𝐳ො
𝐵 = 𝐵𝑥 𝐱ො + 𝐵𝑦 𝒚
A
B
A+B
𝐴Ԧ + 𝐵 = 𝐵 + 𝐴Ԧ = (𝐴𝑥 + 𝐵𝑥 )ො𝐱 + (𝐴𝑦 + 𝐵𝑦 )ෝ
𝒚 + (𝐴𝑧 + 𝐵𝑧 )ො𝐳
10
Vector Subtraction
B
A
-
ෝ + 𝐴𝑧 𝐳ො
𝐴Ԧ = 𝐴𝑥 𝐱ො + 𝐴𝑦 𝒚
ෝ + 𝐵𝑧 𝐳ො
𝐵 = 𝐵𝑥 𝐱ො + 𝐵𝑦 𝒚
+
-B
Head-To-Tail Rule
Parallelogram Rule
-B
=
A
A-B
=
-B
A-B
𝐴Ԧ − 𝐵 = 𝐴Ԧ + (−𝐵) = (𝐴𝑥 − 𝐵𝑥 )ො𝐱 + (𝐴𝑦 − 𝐵𝑦 )ෝ
𝒚 + (𝐴𝑧 − 𝐵𝑧 )ො𝐳
A
11
Try this..
𝐴
𝐀𝐁
a) 𝐂𝐀 + 𝐀𝐁 = 𝐂𝐁
𝐴(2, −3, 1)
= −𝐁𝐂
𝐵(3, 2, 0)
= −4, 7, 5
𝐵
𝐂𝐀
𝐂𝐁
𝐶
𝐶(7, −5, −5)
𝐴
b) 𝐀𝐁 − 𝐂𝐁 = 𝐀𝐁 + 𝐁𝐂
Find :
= 𝐀𝐂
a) CA + AB
= −𝐂𝐀
b) AB - CB
= 5, −2, −6
𝐀𝐁
𝐵
𝐀𝐂
𝐁𝐂
𝐶
NUR HAKIMAH AB AZIZ
12
Vector Multiplication – Scalar/Dot Product
𝐀 ∙ 𝐁 = 𝐴 𝐵 cos 𝜃𝐴𝐵
Where 𝜃𝐴𝐵 is the smaller angle between A and B,
drawn tail to tail.
Prove that:
𝐱ො ∙ 𝐱ො = 𝐱ො 𝐱ො cos 0°
= 1 1 (1)
=1
𝐱ො ∙ 𝐲ො = 𝐱ො 𝐲ො cos 90°
= 1 1 (0)
=0
NUR HAKIMAH AB AZIZ
13
Vector Multiplication – Scalar/Dot Product
Given:
𝐴Ԧ = 𝐴𝑥 , 𝐴𝑦 , 𝐴𝑧
= 0, −2, 5
𝐵 = 𝐵𝑥 , 𝐵𝑦 , 𝐵𝑧
= 3, −3, −2
𝐀∙𝐁=𝐁∙𝐀
Commutative Property
𝐀∙ 𝐁+𝐂 =𝐀∙𝐁+𝐀∙𝐂
Distributive Property
𝐀 ∙ 𝐀 = 𝐴2
ො + 𝐴𝑧 𝐵𝑧 (ො𝒛 ∙ 𝒛ො )
𝐀 ∙ 𝐁 = (𝐴𝑥 𝐵𝑥 )(ො𝐱 ∙ 𝐱ො ) + 𝐴𝑦 𝐵𝑦 )(𝐲ො ∙ 𝐲)
𝐀 ∙ 𝐁 = 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 +𝐴𝑧 𝐵𝑧 = 𝐴 𝐵 cos 𝜃𝐴𝐵
𝜃𝐴𝐵 = cos −1
𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 + 𝐴𝑧 𝐵𝑧
𝐀∙𝐁
= cos −1
𝐴 𝐵
𝐴 𝐵
Vectors A and B are said to be orthogonal or
perpendicular if
𝐀∙𝐁=0
since cos 90 = 0
𝐀 ∙ 𝐁 = 0 3 + −2 −3 + (5)(−2) = −4
𝐴 =
22 + 52 = 29
32 + 32 + 22 = 22
−4
= cos −1
= 99.11°
29 22
𝐵 =
𝜽𝑨𝑩
NUR HAKIMAH AB AZIZ
14
Vector Multiplication – Vector/Cross Product
ෝ 𝐀×𝐁 =𝐧
ෝ 𝐴 𝐵 sin 𝜃𝐴𝐵
𝐀×𝐁=𝐧
Prove that:
ෝ = 𝒛ො
𝐱ො × 𝒚
ෝ
𝐲ො × 𝒛ො = 𝒙
xyzxyz cyclic order
ෝ = −ො𝒛
𝐲ො × 𝒙
ෝ = −ෝ
𝐳ො × 𝒚
𝒙
ෝ 𝐱ො × 𝐱ො = 𝐧
ෝ 𝐱ො 𝐱ො sin 0°
𝐧
ෝ 1 1 (0)
=𝐧
=0
ෝ=𝒚
ෝ
𝐳ො × 𝒙
𝐱ො × 𝒛ො = −ෝ
𝒚
𝜃𝐴𝐵 measured from the tail of 𝐴Ԧ
to the tail of 𝐵
𝜃𝐴𝐵 = 𝜃𝐵 − 𝜃𝐴
ෝ is a unit vector normal to the
𝐧
plane containing 𝐴Ԧ and 𝐵
The magnitude of the cross product,
equals the area of the parallelogram
defined by the two vectors
ෝ 𝐱ො × 𝒚
ෝ = 𝐳ො 𝐱ො 𝐲ො sin 90°
𝐧
= 𝐳ො 1 1 (1)
= 𝐳ො
ෝ 𝐲ො × 𝒙
ෝ = 𝐳ො 𝐱ො 𝐲ො sin −90°
𝐧
= 𝐳ො 1 1 (-1)
= −ො𝐳
NUR HAKIMAH AB AZIZ
15
Vector Multiplication – Vector/Cross Product
Given
𝐴Ԧ = 𝐴𝑥 , 𝐴𝑦 , 𝐴𝑧
= 3, −3, −2
= 0, −2, 5
𝐱ො
𝐀 × 𝐁 = 𝐴𝑥
𝐵𝑥
𝐱ො 𝐲ො
𝒛ො 𝐱ො
𝐀 × 𝐁 = 0 −2 5 อ 0
3 −3 −2 3
= −2 −2 − 5 −3 , 5 3
𝐵 = 𝐵𝑥 , 𝐵𝑦 , 𝐵𝑧
𝐲ො
𝐴𝑦
𝐵𝑦
𝒛ො 𝐱ො
𝐴𝑧 ቮ 𝐴𝑥
𝐵𝑧 𝐵𝑥
𝐲ො
𝐴𝑦
𝐵𝑦
𝐲ො
−2
−3
− 0 −2 , 0 −3 − 3(−2)
= 19, 15, 6
𝐀×𝐁 =
𝜽𝑨𝑩 =
ෝ 𝐀 × 𝐁 = 𝐴𝑦 𝐵𝑧 − 𝐴𝑧 𝐵𝑦 𝐱ො + 𝐴𝑧 𝐵𝑥 − 𝐴𝑥 𝐵𝑧 𝐲ො + 𝐴𝑥 𝐵𝑦 − 𝐴𝑦 𝐵𝑥 𝐳ො
𝐧
192 + 152 + 62 = 622
sin−1
622
29 22
= 80.89° @ 99.11°
ෝ 𝐀×𝐁 =𝐧
ෝ 𝐴 𝐵 sin 𝜃𝐴𝐵
𝐧
𝜃𝐴𝐵 = sin−1
𝐀×𝐁
𝐴 𝐵
The magnitude of the cross product,
equals the area of the parallelogram
defined by the two vectors
𝐀 × 𝐁 = −𝐁 × 𝐀
Anticommutative Property
𝐀 × 𝐁 + 𝐂 = 𝐀 × 𝐁 + 𝐀 × 𝐂 Distributive Property
𝐀×𝐀=0
NUR HAKIMAH AB AZIZ
16
Parallel Vectors
Vector A and B are parallel and in the same direction, if
• 𝐀 = 𝑘𝐁
(The simplest approach)
• 𝐀 ∙ 𝐁 = 𝐀 𝐁 cos 0 = 𝐀 𝐁 ; (𝜃𝐴𝐵 = 0°, cos 𝜃𝐴𝐵 = 1)
ෝ 𝐴 𝐵 sin 0 = 0
• 𝐀×𝐁 =𝐧
; (𝜃𝐴𝐵 = 0°, sin 𝜃𝐴𝐵 = 0)
*If k = -ve number, it means the vectors are parallel but in the opposite direction where 𝜃𝐴𝐵 = 180°
NUR HAKIMAH AB AZIZ
17
Mid-Term Exam Sem 2-2020/2021
i) Perimeter = 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐴 = −6, −2, −3 + 0, 5, −6 + −24, −8, −12 + −30, −5, −21
= 62 + 22 + 32 + 02 + 52 + 62 + 242 + 82 + 122 + 302 + 52 + 212 = 7 + 61 + 784 + 1366
= 79.77
ii) 𝐴𝐵 = 𝑘 𝐶𝐷
−6, −2, −3 = −24𝑘1 , −8𝑘2 , −12𝑘3
𝐵𝐶 = 𝑘 𝐷𝐴
0, 5, −6 = −30𝑘1 , −5𝑘2 , −21𝑘3
𝑘1 = 𝑘2 = 𝑘3 = 0.25
𝑘1 = 0, 𝑘2 = −1, 𝑘3 = 0.29
𝐴𝐵 ∥ 𝐶𝐷
𝑘1 ≠ 𝑘2 ≠ 𝑘3
𝐵𝐶 ∦ 𝐷𝐴
NUR HAKIMAH AB AZIZ
18
Mid-Term Exam Sem 1-2022/2023
(b)(ii)
𝐐𝐏 = 6, 3, 6
(b) Three point-charges are located at P(10, 5, 11), Q(4, 2, 5)
and R(2, 1, z), where y depends on the phase angle between
𝐑𝐐 = 2, 1, 5 − 𝑧
If QP is perpendicular to RQ,
𝐐𝐏 ∙ 𝐑𝐐 = 0
vectors QP and RQ. Solve for y if:
6, 3, 6 ∙ 2, 1, 5 − 𝑧 = 0
6 2 +3 1 +6 5−𝑧 =0
i. QP and RQ are parallel.
45 − 6𝑧 = 0
ii. QP and RQ are perpendicular.
𝑧=
iii. The angle between QP and QR is 60.
45
= 7.5
6
(b)(iii)
If the angle between QP and QR is 60
(b)(i)
6, 3, 6 ∙ −2, −1, 𝑧 − 5 = 𝐐𝐏 𝐐𝐑 cos 60°
𝐐𝐏 = 10, 5, 11 − 4, 2, 5 = 6, 3, 6
6𝑧 − 45 = 𝐐𝐏 𝐐𝐑 cos 60°
𝐑𝐐 = 4, 2, 5 − (2, 1, 𝑧) = 2, 1, 5 − 𝑧
𝐐𝐏 =
If QP is parallel with RQ,
62 + 32 + 62 = 9
𝐐𝐑 =
𝐐𝐏 = 𝑘𝐑𝐐
6𝑧 − 45 = 9(0.5) 𝑧 2 − 10𝑧 + 30
6, 3, 6 = 𝑘 2, 1, 5 − 𝑧
6𝑧 − 45
6
𝑘 = = 3 thus,
2
6=3 5−𝑧
𝑧=
15 − 6
=3
3
5 + (𝑧 − 5)2
2
= 4.52 (𝑧 2 − 10𝑧 + 30)
36𝑧 2 − 540𝑧 + 2025 = 20.25𝑧 2 − 202.5𝑧 + 607.5
15.75𝑧 2 − 337.5𝑧 + 1417.5 = 0
𝑧 = 5.735 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 120° @𝟏𝟓. 𝟔𝟗𝟒
NUR HAKIMAH AB AZIZ
19
Vector Component
Given vector A and B where the component of A along vector B (AB ) is given as
መ cos 𝜃𝐴𝐵 = 𝐀 ∙ 𝐛
መ
𝐀𝐁 = 𝐀 cos 𝜃𝐴𝐵 = 𝐀 𝐛
መ = 𝐀∙𝐛
መ 𝐛
መ = 𝐀∙
𝐀𝐁 = 𝐀𝐁 𝐛
𝐁
𝐁
NUR HAKIMAH AB AZIZ
𝐁
(𝐀 ∙ 𝐁)𝐁
=
𝐁
𝐁𝟐
20
Quiz 1 Sem 3-2021/2022
Figure Q1 shows 4 points P, Q, R and S in Cartesian coordinate where
abc is the last three digits of your Matrix Number. Determine:
a) Vector PS
b) The magnitude of PS
𝐐𝐏 = 7, 2, 3 − 10, 0, −2 = −3, 2, 5
Let’s say, abc=001
𝐐𝐑 = 0, 0, 1 − 10, 0, −2 = −10, 0, 3
𝐐𝐑 2 = 102 + 0 + 32 = 109
𝐐𝐏 ∙ 𝐐𝐑 = −3 −10 + 2 0 + 5 3 = 45
𝐐𝐏 ∙ 𝐐𝐑 𝐐𝐑 45 −10, 0, 3
𝐐𝐒 =
=
= −4.13, 0,1.24
𝐐𝐑 2
109
𝐏𝐒 = 𝐐𝐒 − 𝐐𝐏 = −4.13, 0,1.24 − −3, 2, 5
= −1.13, −2, −3.76
𝐏𝐒 =
NUR HAKIMAH AB AZIZ
1.132 + 22 + 3.762 = 4.4
21
Triple Product
Scalar Triple Product
𝐀 ∙ 𝐁 × 𝐂 = 𝐁 ∙ 𝐂 × 𝐀 = 𝐂 ∙ (𝐀 × 𝐁)
ABCABC.. Cyclic order is preserved
Vector Triple Product
𝐀 × 𝐁 × 𝐂 = 𝐁 𝐀 ∙ 𝐂 − 𝐂(𝐀 ∙ 𝐁)
known as the “bac-cab” rule
𝐀 × 𝐁 × 𝐂 ≠ (𝐀 × 𝐁) × 𝐂
NUR HAKIMAH AB AZIZ
22
Mid-Term Exam Sem 1-2022/2023
(a)
(a) Prove the “BAC-CAB” rule of the vector triple product
using the three vector fields given as follows:
𝒂𝒓 𝒂𝜽
𝐊× 𝐂×𝐓 = 𝐊× 0
3
−1 0
𝒂𝝓 𝒂𝒓
−2 0
4 −1
𝒂𝜽
3
0
= 𝐊 × 12 + 0,2 − 0,0 + 3
෡ + (3) 𝛟
෡
෡ + −5 𝛉
𝐊= 4 𝐑
= 𝐊 × 12, 2, 3
෡ + (2) 𝛟
෡
𝐂= 3 𝛉
𝒂𝒓 𝒂𝜽 𝒂𝝓 𝒂𝒓 𝒂𝜽
= −4 5
2 −4 5
12 2
3 12 2
= 15 − 4, 24 + 12, −8 − 60
෡
෡ + (−4) 𝛟
𝐓= 1 𝐑
= 11, 36, −68
𝐂 𝐊∙𝐓 −𝐓 𝐊∙𝐂
= 𝐂 −4(−1) + 5(0) + 2(4) − 𝐓 −4(0) + 5(3) + 2(−2)
= 12𝐂 − 11𝐓
= 12 0, 3, −2 − 11 −1, 0, 4
= 0, 36, −24 − −11, 0, 44
= 11, 36, 68
NUR HAKIMAH AB AZIZ
23
Exercise 1
Given
𝐀 = 3𝐚𝐱 − 5𝐚𝐲 − 7𝐚𝐳
𝐁 = −2𝐚𝐱 − 4𝐚𝐲 − 8𝐚𝐳
𝐂 = 𝐚𝐱 + 2𝐚𝐲 + 4𝐚𝐳
𝐃 = 2𝐚𝐱 + 𝐚𝐲 + 6𝐚𝐳
Find:
P(0, -1, 1)
Q(5, 5, 5)
R(x, 8, 7)
a) The magnitude of 𝐀 − 2𝐚𝐱 − 𝟐𝐁
(10.724)
b) |𝐀 − 2𝐚𝐱 − 2B||2C|A
<294.86,-491.4,-688>
c) The component of B along 𝐚𝐱 , 𝐚𝐲 , and 𝐚𝐳
(-2 𝐚𝐱 ; −4𝐚𝐲 ; −8𝐚𝐳 )
d) A unit vector in the direction of 𝐀 − 2𝐚𝐱 − 2𝐁
<0.466,0.28,0.84>
e) 𝐀 ∙ 𝐂
(-35)
f) The area of parallelogram formed by 𝐀 and 𝐂
(22.76)
g) 𝜃𝐴𝐶 using both dot product and cross product
(147)
h) Whether vector 𝐂 parallel to A and B
(no,yes opposite direction)
i) The component of B along A
<2.53,-4.22,-5.9>
j) The end point of vector A if the vector starts at P.
(3,-6,-6)
Then, find the perpendicular distance from point Q to A
(7.4)
k) Find x of R if
• PQ||QR
(7.5)
• QP is perpendicular to QR
(-0.2)
• Angle between QP and QR is 150
(5.29@12.65)
l)Prove that 𝐀 ∙ 𝐁 × 𝐃 = 𝐁 ∙ 𝐃 × 𝐀 = 𝐃 𝐀 × 𝐁 ≠ 𝐁 ∙ 𝐀 × 𝐃
-70=-70=-7070
m) Prove that 𝐀 × 𝐁 × 𝐃 = 𝐁 𝐀 ∙ 𝐃 − 𝐃 𝐀 ∙ 𝐁 ≠ 𝐀 × 𝐁 × 𝐃
<-58,94,-92>=<-58,94,-92>  <250,-116,-64>
NUR HAKIMAH AB AZIZ
24
Example 1
Given the vectors M = -10ax + 4ay - 8az and N = 8ax + 7ay - 2az, find:
a) a unit vector in the direction of -M + 2N
b) the magnitude of 5ax + N - 3M
c) |M||2N|(M + N)
a) Let’s say 𝐴Ԧ = −𝑀 + 2𝑁
𝐴Ԧ = 10, −4, 8 + 16, 14, −4 = 26, 10, 4
Unit vector of 𝐴Ԧ therefore,
𝐴Ԧ
26, 10, 4
26, 10, 4
𝑎ො =
=
=
= 0.924, 0.355, 0.142
2
2
2
Ԧ
792
26 + 10 + 4
𝐴
b) Let’s say 𝐵 = 5𝑥ො + 𝑁 − 3𝑀
𝐵 = 5, 0, 0 + 8, 7, −2 − −30, 12, −24 = 43, −8, 22
𝐵 = 432 + 82 + 222 = 48.96
c) 𝑀 2𝑁 𝑀 + 𝑁 = 102 + 42 + 82 162 + 142 + 42 −10, 4, −8 + 8, 7, −2
= −580.5, 3192.7, −2902.4
NUR HAKIMAH AB AZIZ
= 180 468 −2, 11, −10
25
Example 2
The three vertices of a triangle are located at A(-1, 2, 5), B(-4, -2, -3), and C(1, 3, -2).
a) Find the length of the perimeter of the triangle
b) Find a unit vector that is directed from the midpoint of
side AB to the midpoint of side BC
c) Show that this unit vector is parallel to AC.
A(-1, 2, 5)
MPAB
𝑉
C(1, 3, -2)
B(-4, -2, -3)
a) Perimeter = 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐴 = −3, −4, −8 + 5, 5, 1 + −2, −1, 7
MPBC
= 32 + 42 + 82 + 52 + 52 + 12 + 22 + 12 + 72 = 89 + 51 + 54
= 23.92
b) 𝑀𝑃𝐴𝐵 =
𝐴+𝐵
2
−1,2,5 + −4,−2,−3
2
=
= (−2.5, 0, 1)
𝑀𝑃𝐵𝐶 =
𝐵+𝐶
2
=
−4, −2, −3 +(1, 3, −2)
2
= −1.5, 0.5, −2.5
𝑉 = −1.5, 0.5, −2.5 − −2.5, 0, 1 = 1, 0.5, −3.5
𝑣ො =
𝑉
𝑉
=
1, 0.5, −3.5
12
+
0.52
+
3.52
= 0.272, 0.136, −0.953
c) From a) , 𝐴𝐶 = 2, 1, −7 = 𝑘𝑣ො = 0.272𝑘1 , 0.136𝑘2 , −0.953𝑘3
0.272𝑘1 = 2
𝑘1 = 7.35
0.136𝑘2 = 1
𝑘2 = 7.35
− 0.953𝑘3 = −7
𝑘 = 7.35
3 HAKIMAH AB AZIZ
NUR
𝑘1 = 𝑘2 = 𝑘3
∴ 𝐴𝐶 ∥ 𝑣ො
26
Example 3
A(6, −2, −4)
The vector from the origin to the point A is given as 6, −2, −4 , and
the unit vector directed from the origin towards point B is
2, −2, 1 /3. If points A and B are ten units apart, find the
coordinates of point B.
𝐵 = 𝑘𝑏෠ =
B
𝑏෠ =
𝑘 2, −2, 1
3
𝐴𝐵 = 10 = 𝐵 − 𝐴Ԧ =
|AB|=10
2, −2, 1
3
𝑘 2, −2, 1
− 6, −2, −4
3
(2𝑘 − 18)2 + (−2𝑘 + 6)2 + (𝑘 + 12)2
= 10
3
(2𝑘 − 18)2 + (−2𝑘 + 6)2 + (𝑘 + 12)2 = 900
9𝑘 2 − 72𝑘 − 396 = 0
𝑘 = 11.75
𝐵 = (7.83, −7.83, 3.92)
NUR HAKIMAH AB AZIZ
27
Example 4
Points P and Q are located at (0, 2, 4) and (-3, 1, 5). Calculate
a) The position vector P
b) The distance vector from P to Q
c) The distance between P and Q
d) A vector parallel to PQ with magnitude of 10
a)𝑃 = 0, 2, 4
b) 𝑃𝑄 = −3, 1, 5 − 0, 2, 4 = −3, −1, 1
c) 𝑃𝑄 = 32 + 12 + 12 = 11 = 3.32
d) 𝐴Ԧ = 𝑘 −3, −1, 1
𝐴Ԧ = 𝑘 32 + 12 + 12 = 𝑘 11 = 10
𝑘 = 3.015
𝐴Ԧ = 3.015 −3, −1, 1 = −9.045, −3.015, 3.015
Ans:
b) rPQ = -3ax – ay + az
AB a
AZIZ- 3.015a + 3.015a )
c) 3.317 d) ANUR
= HAKIMAH
(-9.045
x
y
z
28
Example 5
ෝ + 𝒛ො and B = 2ෝ
Given vectors A = 3ො𝐱 + 4 𝒚
𝒚 - 5ො𝒛, find
xො
ii. 𝐀 × 𝐁 = 3
0
i. 𝐀 ∙ 𝐁
𝐀 ∙ 𝐁 = 3,4,1 ∙ 0,2, −5 = 0 + 8 − 5 = 3
𝑦ො
4
2
𝑧Ƹ
1
−5
𝑥ො
3
0
𝑦ො
4
2
= −20 − 2 xො + 0 + 15 yො + 6 − 0 𝑧Ƹ = −22,15,6
𝐀×𝐁 =
222 + 152 + 62 = 745
iii. 𝜃𝐴𝐵 using both dot product and cross product
𝐀 =
32 + 42 + 12 = 26
𝐁 =
02 + 22 + 52 = 29
𝜃𝐴𝐵 = COS −1
𝜃𝐴𝐵 = sin−1
𝐀∙𝐁
3
= COS −1
= 𝟖𝟑. 𝟕𝟑°
𝐀 𝐁
26 29
NUR HAKIMAH AB AZIZ
𝐀×𝐁
𝐀 𝐁
= sin−1
745
26 29
= 𝟖𝟑. 𝟕𝟑°
29
Example 6
In Cartesian coordinates, vector A points from the origin to
point P1 = (2, 3, 3), and vector B is directed from P1 to point
P2 = (1, -2, 2). Find:
ෝ,
(a) vector A, its magnitude A, and unit vector 𝒂
(b) the angle between A and the y axis,
(c) vector B,
(d) the angle 𝜃𝐴𝐵 between A and B, and
(e) perpendicular distance from the origin to vector B
𝐀 = 22 + 32 + 32 = 22
a) 𝐀 = 2,3,3
𝑎ො =
2,3,3
22
b) 𝜃𝐴𝑦 = COS −1
𝟑𝑦ො
𝐀
= COS −1
3
22
= 𝟓𝟎. 𝟐𝟒°
c) 𝐁 = 1, −2,2 − 2,3,3 = −1, −5, −1
𝐁 = 12 + 52 + 12 = 2𝟕
d) 𝜃𝐴𝐵 = COS−1
𝐀∙𝐁
𝐀 𝐁
= COS−1
−2−15−3
22 2𝟕
= 𝟏𝟒𝟓. 𝟏𝟓°
e) 𝐎𝐏𝟑 = 𝐀 sin 180 − 𝜃𝐴𝐵 = 2.68
NUR HAKIMAH AB AZIZ
30
Example 7
A vector field is specified as G = 24xyax + 12(x2 + 2)ay + 18z2az.
Given two points, P (1, 2, -1) and Q(-2, 1, 3), find:
a) 𝐆𝐏 = 24 1 2 , 12 12 + 2 , 18(−1)2 = 48,36,18
a) G at P
b) a unit vector in the direction of G at Q
b) 𝐆𝐐 = 24 −2 1 , 12 −22 + 2 , 18(3)2 = −48,72,162
c) a unit vector directed from Q toward P
d) the equation of the surface on which |G| = 60
−48,72,162
𝐠ෞ𝐐 =
482 + 722 + 1622
= −0.26,0.392,0.882
c) 𝐐𝐏 = 1,2, −1 − −2,1,3 = 3,1, −4
෢ =
𝐐𝐏
3,1,−4
32 +12 +42
d) 𝐆 = 𝟔𝟎 =
= 0.588,0.196, −0.784
(24𝑥𝑦)2 +(12 𝑥 2 + 2 )2 +(18𝑧 2 )2
(24𝑥𝑦)2 +(12 𝑥 2 + 2 )2 +(18𝑧 2 )2 = 3600
(24𝑥𝑦)2 +(12 𝑥 2 + 2 )2 +(18𝑧 2 )2 = 3600
576𝑥 2 𝑦 2 + 144 𝑥 2 + 2 2 + 324𝑧 4 = 3600
16𝑥 2 𝑦 2 + 4 𝑥 2 + 2 2 + 9𝑧 4 = 100
16𝑥 2 𝑦 2 + 4𝑥 4 + 16𝑥 2 + 16 + 9𝑧 4 = 100
4𝑥 4 + 16𝑥 2 (𝑦 2 + 1) + 9𝑧 4 = 84
NUR HAKIMAH AB AZIZ
31
Example 8
Given A = 𝐱ො - 𝐲ො + 2ො𝐳, B = 𝐲ො + 𝐳ො, and C = -2ො𝐱 + 3ො𝐳, find
(A × B) × C and compare it with A × (B × C)
xො 𝑦ො
𝐀 × 𝐁 = 1 −1
0 1
𝑧Ƹ
2
1
𝑥ො
1
0
xො
𝑦ො
(𝐀 × 𝐁) × 𝐂 = −3 −1
−2 0
xො 𝑦ො
𝐁×𝐂= 0 1
−2 0
𝑧Ƹ
1
3
𝑥ො
0
−2
xො 𝑦ො 𝑧Ƹ
𝐀 × (𝐁 × 𝐂) = 1 −1 2
3 −2 2
(𝐀 × 𝐁) × 𝐂 ≠ 𝐀 × (𝐁 × 𝐂)
𝑦ො
−1 = (−1 − 2)ොx + 0 − 1 yො + 1 − 0 𝑧Ƹ = −3, −1,1
1
𝑧Ƹ
1
3
𝑥ො
−3
−2
𝑦ො
−1 = (−3 − 0)ොx + −1 + 9 yො + 0 − 2 𝑧Ƹ = −3,7, −2
0
𝑦ො
1 = (3 − 0)ොx + −2 − 0 yො + 0 + 2 𝑧Ƹ = 3, −2,2
0
𝑥ො
1
3
𝑦ො
−1 = (−2 + 4)ොx + 6 − 2 yො + −2 + 3 𝑧Ƹ = 2,4,1
−2
NUR HAKIMAH AB AZIZ
32
Example 9
Given P = 2ො𝐱 - 𝐳ො, Q = 2ො𝐱 - 𝐲ො + 2ො𝐳, and R = 2ො𝐱 - 3𝐲ො + 𝐳ො,
a) Prove that 𝐐 ∙ 𝐑 × 𝐏 = 𝐑 ∙ 𝐏 × 𝐐 = 𝐏 ∙ 𝐐 × 𝐑 ≠ 𝐏 ∙ 𝐑 × 𝐐
b) Prove that 𝐏 × 𝐐 × 𝐑 = 𝐐 𝐏 ∙ 𝐑 − 𝐑 𝐏 ∙ 𝐐
c) The component of P along Q
xො 𝑦ො
𝑧Ƹ
𝑥ො 𝑦ො
xො 𝑦ො
𝑧Ƹ
𝑥ො 𝑦ො
a) 𝐑 × 𝐏 = 2 −3 1 2 −3 = 3,4,6 , 𝐏 × 𝐐 = 2 0 −1 2 0 = −1, −6, −2
2 0 −1 2 0
2 −1 2 2 −1
𝐐 ∙ 𝐑 × 𝐏 = 6 − 4 + 12 = 14
𝐑 ∙ 𝐏 × 𝐐 = −2 + 18 − 2 = 14
xො 𝑦ො 𝑧Ƹ 𝑥ො 𝑦ො
xො 𝑦ො 𝑧Ƹ 𝑥ො 𝑦ො
𝐐 × 𝐑 = 2 −1 2 2 −1 = 5,2, −4 , 𝐑 × 𝐐 = 2 −3 1 2 −3 = −5, −2,4
2 −3 1 2 −3
2 −1 2 2 −1
𝐏 ∙ 𝐐 × 𝐑 = 10 + 0 + 4 = 14
𝐏 ∙ 𝐑 × 𝐐 = −10 + 0 − 4 = −14
xො 𝑦ො 𝑧Ƹ
𝑥ො 𝑦ො
b) 𝐏 × 𝐐 × 𝐑 = 2 0 −1 2 0 = 2,3,4
5 2 −4 5 2
𝐐 𝐏 ∙ 𝐑 − 𝐑 𝐏 ∙ 𝐐 = 2, −1,2 4 + 0 − 1 − 2, −3,1 4 + 0 − 2 = 6, −3,6 − 4, −6,2 = 2,3,4
c) 𝐏𝐐 =
𝐏∙𝐐 𝐐
𝐐𝟐
=
4+0−2 2,−1,2
22 +12 +22
=
4,−2,4
9
NUR HAKIMAH AB AZIZ
33