Chapter 3 : Vectors
- Introduction
- Addition of Vectors
- Subtraction of Vectors
- Scalar Multiplication of Vectors
- Components of Vectors
- Magnitude of Vectors
- Product of 2 Vectors
- Application of Scalar/Dot Product & Cross Product
Introduction
Vectors
• Has magnitude
(represent by length of
arrow) .
• direction (direction of
the arrow either to the
right, left, etc).
• Eg: move the brick 5m to
the right
Scalars
• Has magnitude only.
• Eg: move the brick
5m.
Introduction
Vectors Representation
• Use an arrow connecting an initial point A to terminal point B.
• Denote AB
• Written as
• Magnitude of AB  AB
Introduction
Vectors Negative
• Vector in opposite direction, a , but has same magnitude as a .
Introduction
Equal Vectors
• If we have 2 vectors, with same magnitude & direction
.
Addition of Vectors
1. The Triangle Law
• Any 2 vectors can be added by joining the initial point of b to the
terminal point of a .
• Eg:
Addition of Vectors
2. The Parallelogram Law
• If 2 vector quantities are represented by 2 adjacent sides of a
parallelogram, then the diagonal of parallelogram will be equal to
the summation of these 2 vectors.
• Eg:
• The parallelogram law is affected by the triangle law.
Addition of Vectors
The sum of a number of vectors
Subtraction of Vectors
• Is a special case of addition.
• Eg:
Scalar Multiplication
• k a ; vector a multiply with scalar, k.
• .
Parallel
Vectors
Parallel
Vectors
Scalar Multiplication
Components of Vectors – Unit Vectors
Vectors in 2 Dimensional (R )
2
Vectors in 3 Dimensional (R )
3
Exercise :
Draw the vector
i. 2 i  6 j
ii. 4 i  5 j  2k
Components of Vectors
Magnitude of Vectors
1. For Any Vector
Example:
Exercise:
Magnitude of Vectors
2. From one point to another point of vector
Example:
 
- point / coordinate
 vector
Magnitude of Vectors
Solution:
i) P to Q = PQ  OQ  OP
=  9  1, 2  5, 4  7 
= 8, 3, 3
 PQ  82  (3) 2  (3) 2  82
ii) Q to R =QR  OR  OQ
=  3  9, 2  2, 6  4 
= 6, 0, 2
 QR  (6) 2  0  22  40
Do Exercise 3.3 in Textbook page 70.
Unit Vectors
Example:
Do Exercise 3.4 in Textbook page 70.
Direction Angles & Cosines
 ,  ,  : direction angles of vector OP
cos  , cos  , cos  : direction cosines of the vector OP
cos  
x
OP
, cos  
y
OP
, cos  
z
OP
Direction Angles & Cosines
Example:
Solution (i):
Direction cosines
Direction angles
90.77
Direction Angles & Cosines
Solution (ii)
Direction cosines
PQ  OQ  OP
=  3  5, 4  7, 1  2 
= 8, 3,3
PQ  (8) 2  (3) 2  32  82
 cos  
8
3
3
, cos  
, cos  
,
82
82
82
Direction angles
 8 
  cos 1 
  152.06
 82 
 3 
  cos 1 
  109.35
 82 
 3 
  cos 1 
  70.65
82


Do Exercise 3.5 in Textbook page 72.
Do Tutorial 3 in Textbook page 85 :
• No. 2 (i)
• No. 3 (i)
• No. 4
• No. 5 (iii)
• No. 6 (i)
Operations of Vectors by Components
Example:
Solution:
Do Exercise 3.6 in Textbook page 72.
Product of 2 Vectors
Dot Product / Scalar Product
Example:
Solution:
Do Exercise 3.7 in Textbook page 73.
Find Angle Between 2 Vectors
Example:
Solution:
Do Exercise 3.8 in Textbook page 74.
Product of 2 Product
Cross Product / Vector Product
Example:
Product of 2 Product
Cross Product / Vector Product
Solution:
i
j
k
i) u  v  4 7 1  i (35  1)  j (20  2)  k (4  14)
2 1 5
=36i  22 j  10k
i j k
ii) v  u  2 1 5  i (1  35)  j (2  20)  k (14  4)
4 7 1
=-36i  22 j  10k
Do Exercise 3.9 in Textbook page 74.
Find Angle Between 2 Vectors
Applications of Vectors
• Projections
• The Area of Triangle & Parallelogram
• The Volume of Parallelepiped & Tetrahedron
• Equations of Planes
3
• Parametric Equations of Line in R
• Distance from a Point to the Plane
i. Projections
Scalar projection of b onto a:
a.b a
compab 
 .b  scalar
a
a
Vector projection of b onto a:
 a.b  a
a
proja b  
  compab   vector
a
 a a
Example :
i.
Given a  2 i  3 j  k and b  2 i  j  3k . Find the scalar projection
and vector projection of b onto a
ii. Find compab and projab given that a  4 i  3 j  k and b  2 i  j  k
Solutions:
ii. The Area of Triangle and Parallelogram
Area of triangle POQ = 1 / 2 a b sin   1 / 2 a  b
Area of parallelogram OQRP  a b sin   a  b
Note that parallelogram can be divided into 2 triangles.
Example :
Solutions:
Solutions:
iii. The Volume of Parallelepiped and
Tetrahedron
A parallelepiped is a three-dimensional formed by six parallelogram.
• Define three vectors a  a1 , a2 , a3 , b  b1 ,b2 ,b3 , c  c1 , c2 , c3
• To represent the three edges that meet at one vertex.
• The volume of the parallelepiped is equal to the magnitude of their
scalar triple product
V  a  b  c 
• Volume of Parallelepiped
V  a  b  c 
 b  c  a
= c  a  b
• Volume of Tetrahedron
a1
1
V  a   b  c   b1
6
c1
a2
b2
c2
a3
b3
c3
Example :
Solution:
iv. Equations of Planes
Example:
Solutions:
Example :
Solutions:
v. Parametric Equations of a Line in R
3
Parametric equations of a line :
Cartesian equations :
Example :
Solutions:
vi. Distance from a Point to the Plane
Example:
Solutions:
ii.
Vector n1  10, 2, 2
Vector n 2  5,1, 2
Let 1st equation to find the point
Let x=z=0
10(0)  2 y  2(0)  5
5
y
2
5
P  (0, ,0)
2
5
0(5)  (1)  0( 2)  1
2
D
 0.2887
2
2
2
5  1  (2)