Engineering Mathematics 1

[CHAPTER 4: COMPLEX NUMBER]

LET’S PRACTICE 12

1. Solve the following expression in an exponential form.
25(cos180  i sin180)  8(cos12  i sin12)
20(cos 50  i sin 50)

Ans: 10e 2.48i

2. Given that Z1  8(cos 34  i sin 34) and Z2  6040 . Solve Z2 in trigonometry form
Z1

Ans: 7.5(cos 6  i sin 6)

98

[CHAPTER 4: COMPLEX NUMBER]

3.Given z1  3  2i and z2  3(cos30  i sin 30)
a. Calculate z1  z2 and express the answer in the form of a  bi and exponential form

Ans:

Cartesian form: 4.8  9.7i

Exponential form:10.8e1.11i

b.. Calculate z1 and express the answer in the form of a  bi and exponential form
z2

Ans:

Cartesian form:1.2  0.077i
Exponential form:1.2e 0.064i

99

[CHAPTER 4: COMPLEX NUMBER]

4. Given z1  8145 , z2  2460
a. Calculate z1  z2 and express the answer in the form of a  bi

Ans:  174  81.14i

b. State 2z2 in trigonometric form. Illustrate the answer on Argand diagram.
z1

 Ans: 6 cos 850  i sin 850

100

[CHAPTER 5: MATRIX]

MATRIX

OBJECTIVES:
At the end of this topic, students should be able to understand about:

i. matrices definition
ii. dimension or order of matrices
iii. types of matrices
iv. operation of matrices (addition, subtraction, multiplication and division)
v. simultaneous equation using matrix (inverse matrix method and Cramer’s rule)

5.1 INTRODUCTION OF MATRIX

Introduction including of:
 Meaning of matrices
 Dimension or order of matrices
 Matrices Notation
 Types of matrices

 A matrix is an ordered rectangular array of numbers or functions.
 The numbers or functions are called the elements or the entries of the matrices.
 The matrices are denoting by capital letters.

A matrices is a rectangular arrangement of numbers in rows
and columns.

Figure 1

Referring on Figure 1 above;

 Its dimension are 2  3

 2 rows and 3 columns

 The entries of the matrices are 2,5,10,4,19,4

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[CHAPTER 5: MATRIX]

MATRICES NOTATION

 In order to identify an entry in a matrix, we simply write a subscript of the respective entry’s
row followed by the column.

 In general, the matrices can be denoted by as in Figure 2.

Figure 2

- Each aij is called an element of the matrices (or an entry of the matrices).

- This denotes the element in row i and column j.
- The entries of the matrices are organized in horizontal rows and vertical

columns.

DIMENSION OR ORDER OF A MATRICES

 The number of ROWS in matrices is represented with ‘m, and the number of COLUMNS is
represented with ‘n’.

 Hence the matrices can be called (m x n) order matrices.
 The numbers m and n are called dimensions/ order/ size of the matrices.
 Example:

MATRICES SIZE/ ORDER

A  1 1 0 4 1 4

- A row & 4 columns

2 1 3 2
B   3 1
4 - 3 rows & 2 columns
 0

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[CHAPTER 5: MATRIX]

EXAMPLE 1

34  5 31
A  22 15 
55 

52 13  9

i. What are the dimensions of the matrices below?

ii. Identify entry A23

SOLUTION:

i. Dimension of a matrices; 3 3
ii. 55

LET’S PRACTICE 1

 2  5 14 23  9
1. A   4 19 
4 34 9 

 41 5 30 1  6

i. What are dimension of A?

ii. Identify entry A34

iii. Identify entry A12

Ans: 3 5 , 1,  5

12 7 21 31 11
 
2. V   45 2 14 27 19 
15
 3 36 71 26
 
 4 13 55 34 15 

i. What are the dimensions of V above?

ii. Identify the entry V14
iii. What is the matrices notations to donate the entry 15

Ans: 4  5 , 31, V32 and V45

103

TYPES OF MATRICES [CHAPTER 5: MATRIX]
i. Rectangular matrices vi. Scalar matrices
ii. Square matrices vii. Transpose matrices
iii. Row matrices viii. Zero matrix/ Null matrices
iv. Column matrices ix. Identity matrix/ Unit matrices
v. Diagonal matrices

RECTANGULAR MATRICES

 A matrices in which number of rows is not 1 2 3
equal to number of columns. 3 1
A  4 2 2
3
1 2 3

ROW MATRICES COLUMN MATRICES SQUARE MATRICES

 A matrices with single
 A matrices with single column and any  A matrices with equal number
row and any number of number of rows. of rows and columns.
columns.
1  m  n
A  1 2 3 4 2
5 A  3 1 2 1
DIAGONAL MATRICES A  2 1 2
4
1 2 1
SCALAR MATRICES
ZERO MATRICES

 Is a square matrices in  A diagonal matrices in  A matrices in which every
which all the elements which all the diagonal element is zero.
except those on the elements are equal.
leading diagonal are
zero. 4 0 0 0 0 0
A  0 4 0 A  0 0 0
2 0 0
A  0 5 0 0 0 4 0 0 0

0 0 7

IDENTITY MATRICES /UNIT MATRICES TRANPOSE MATRICES

 When the diagonal elements  A matrices which is formed by turning all the
are one and nondiagonal rows of a given matrices into columns and
elements are zero. vice versa.

 A unit matrices is always a  The transpose of matrices A is written as AT
square matrices.
1 2 3T  1 4
1 0 0 4 5 6 2 5
A  0 1 0 3 6

0 0 1

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[CHAPTER 5: MATRIX]

5.2 OPERATION OF MATRICES

 Addition DON’T A
 Subtraction MATRICES

 Multiplication DEVIDE
- By scalar
- By matrices

5.2.1 ADDITION

 The two matrices must be the same size.
 The rows must match in size
 The columns must match in size

 Add the numbers in the matching positions.

EXAMPLE 2

These are the calculations:

3 8  4 0  7 8 347 80  8
4 6 1  9 5  3 41 5
6   9  3

5.2.2 SUBTRACTING

 The two matrices must be the same size.
 The rows must match in size
 The columns must match in size

 Subtract the numbers in the matching positions.

EXAMPLE 3

These are the calculations:

3 8  4 0   1 8 3  4  1 80  8
4 6 1  9  15 4 1 3
 3 6   9  15

105

[CHAPTER 5: MATRIX]

LET’S PRACTICE 2

1. Find 2  3  1  5
 4 2   3  2

Ans: 1  8
 1 0 

2. If A  2  3 and B   1  5 . Find A B.
 4    2
2   3

Ans: 3 2
 7 4

3. If A 3 5 4 and B  1 4 2 . Find A  B
 1 4 6  5 2 3

Ans: 2 1 6
 6 2 9

4. If A 3 5 4 and B  1 4 2 . Find A  B
 1 4 6  5 2 3

Ans: 4 9 2
4 6 3

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[CHAPTER 5: MATRIX]

5.2.3 MULTIPLICATION

MULTIPLICATION BY SCALAR
 To multiply a matrices by a single number is easy.

EXAMPLE 4

These are the calculations:

2  4 0  8 0 24 8 20  0 MU
1  9 2 18 21  2 LTIP
2 9  18 LYIN
GA
MA

TRICES BY ANOTHER MATRIX
 To multiply a matrices by another matrices, we need to do the “dot product”.

EXAMPLE 5

The “dot product” is multiply matching members, then sum up:
 First row, first column:

1,2,3 7,9,11  58

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[CHAPTER 5: MATRIX]

 First row, second columns:

1,2,3  8,10,12  64

 Second rows, first column:

4,5,6  7,9,11  139

 Second rows, second columns:

4,5,6  8,10,12  154

 THE ANSWER:

1 2 3 7 8  58 64 
4 5 6  10 139 154
  9 12 

11

REMEMBER

 In arithmetic:

3  5  5  3 [Commutative Law]

 But this is NOT generally true for matrices (matrix multiplication is not commutative).
AB  BA

108

[CHAPTER 5: MATRIX]

EXAMPLE 6

Given A  1 2 and B  2 0 . Prove that AB  BA .
3 4 1 2

A  B  1 2  2 0
3 4 1 2

A  B  4 4
10 8
The answer is
DIFFERENT

B  A  2 0  1 2 [PROVED]
1 2 3 4

B  A  2 4
7 8

LET’S PRACTICE 3

1. Given C  1 2 0 . Find 2C
0 1  3

Ans: 2 4 0
0 2  6

2. If A  2 0 1 Find 5A  (3A)
1 3 2
.

Ans: 4 0 2
 2 6 4

3. Given A 3 1 and B  2 3 . Find AB
4 2 1 5

Ans: 7 14
10 22

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[CHAPTER 5: MATRIX]

X  3 1 Y   2 4
2   3 1 . Find XY .
4. If 5  and

Ans:  3 11
 19 13

3 1 2 2 0
 2 4 0  4 . Find AB
5. If A and B   1 2

 3

Ans: 1 0
 8 16

 2 0 1  5 1  2
  1 
6. If A  3 5 2  and B  0 4  . Find AB

 4 1 4   2  3 3 

 8 5  7
 
Ans:  14 3 20 

 13  16 24 

 3 2 5 2  1 0
  3 2 . Find AB
7. If A  0 1 6  and B  5

 4 2  1 1 4 2

 5 7 6
 19 10
Ans:  3

 3 10 2 

110

[CHAPTER 5: MATRIX]

5.3 DETERMINANT

DETERMINANT OF A MATRICES
 Determinant of a matrices is a special number that can be calculated from a square
matrix.
 The symbol for determinant is two vertical lines either side.
 The determinant of a matrices may be negative or positive.
 Example:

A - means the determinant of the matrices A

CALCULATING THE DETERMINANT

 First of all the matrices must be square.
 have the same number of rows as columns.

i. FOR A 2 2 MATRICES

 For a 2 2 matrices (2 rows and 2 columns)

 Formula:

A  a b The determinant is;
c  A  ad  bc
d 

“BUTTERFLY” rule for matrices 2 2

111

[CHAPTER 5: MATRIX]

EXAMPLE 7

Given B  4 6 . Find the determinant of matrices B
3 8

SOLUTION

A  48  6 3  32 18

A  14

ii. FOR A 3 3MATRICES

 For a 3 3 matrices (3 rows and 3 columns)

 Formula

a b c The determinant is;
A  d 
e f  A  aei  fh  bdi  fg  cdh  eg

g h i 

EXAMPLE 8

6 1 1
Given C  4  2 5

2 8 7 .
Find the determinant of matrices C.

SOLUTION:

C  6 14  40  128  10  132  4

C  306

112

[CHAPTER 5: MATRIX]

LET’S PRACTICE 4

1. Find the determinant of matrices A.

A  2 5
1  3

2. Find the determinant of matrices B. Ans: 11
Ans: 10
B  3 2
1 4 Ans:  44

3. Find the determinant of matrices M. Ans:  4
Ans:  266
 3 0 1
  113
M   2 5 4 

 3 1 3 

4. Find the determinant of matrices N.

1 1 1 
N  2 5 
7 

2 1 1

5. Find the determinant of matrices Z.

 7 5 4
 2 6
Z   4

 2  3 4

[CHAPTER 5: MATRIX]

5.4 INVERSE MATRICES USING MINOR, COFACTOR AND ADJOIN

 Determinant of a matrices
 Minor of a matrices
 Cofactor of a matrices
 Adjoin of a matrices
 Inverse matrices

 Ignore the values on the current row and column
 Calculate the determinant of the remaining values

EXAMPLE9

3 0 2 
Given A  2 0  2

0 1 1  . Find the inverse matrices of A
SOLUTION:
STEP 1: DETERMINANT

A  30  2 0  22  0  6  4

A  10

STEP 2: MINOR
M indicates the minor of matrices A

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[CHAPTER 5: MATRIX]

m11  0  2  0   21  2 m12  2  2  21   20  2 m13  2 0  21  0  2
1 1  0 1  0 1

m21  0 2  0  2  2 m22  3 2  31  20  3 m23  3 0  31  0  3
1 1 0 1 0 1

m31  0 2  0  0  0 m32  3 2  3 2  22  10 m33  3 0  30  0  0
0  2 2  2 2 0

 2 2 2
M   2 3 3

 0  10 0
STEP 3: COFACTOR

 2 2 2   2  2 2 
Cofactor   2 3 3    2 3  3

 0 10 0   0 10 0 
STEP 4: ADJOIN

 “Transpose” all elements in previous matrices.
 The elements of row will be elements of columns.

 2 2 0
AdjA   2 3 10

 2  3 0 

STEP 5: INVERSE MATRICES

 Formula;

A 1  1  AdjA
A

A 1  1 2 2 0
10  2 3 10
 2 3 0 

A1  111555 15 0
310 
 310
1
0

115

[CHAPTER 5: MATRIX]

LET’S PRACTICE 5

1. Find the inverse matrices

1 0  3
C  2  2 
1 

0 1 3 

Using minor, cofactor and adjoin.

 5 3  6
Ans:  6 3  7

 2 1  2
116

[CHAPTER 5: MATRIX]

2. Find the inverse matrices

 2 1 0 
 3 1
M   1

 3 0 1 

Using minor, cofactor and adjoin.

3 1 1
 
Ans:  4 4 4 
 1 1 1 

 2 2 2 
 9 3 7 

 4 4 4

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[CHAPTER 5: MATRIX]

3. Find the inverse matrices

 3 1  6
 
B   2 0 4 

1 2  3

Using minor, cofactor and adjoin.

 4 92 2
 32 
Ans:  1 52
 0
 2  1

118

[CHAPTER 5: MATRIX]

4. Find the inverse matrix

 3 7  5
 
F   4 1 12 

 2 9 1

Using minor, cofactor and adjoin.

 107  38 79 
 
Ans:  327 327 327 
 28 13 16 
 327 327 327 
 38 41  25 

 327 327 327 

119

[CHAPTER 5: MATRIX]

5.5 SIMULTANEOUS LINEAR EQUATION USING MATRICES

 Inverse Matrices Method
 Cramer’s Rule

REMEMBER:-
i. The system must have the same number of equations as variables, that is, the
coefficient matrices of the system must be square.
ii. The determinant of the coefficient matrices must be non-zero. The reason, of course, is
that the inverse of a matrices exist precisely when its determinant is non-zero.

5.5.1 SIMULTANEOUS EQUATION USING INVERSE MATRICES METHOD

EXAMPLE 10

 x 3y  z 1
Given 2x  5y  3

3x  y  2z  2 Solve the simultaneous equation by using inverse matrices method.

SOLUTION:

STEP 1: Rewrite the system using matrices multiplication

1 3 1 x  1 
  y  
 2 5 0    3 

 3 1  2 z   2

STEP 2: Writing the coefficient matrices as A

x  1  1 3 1 
A y  
  3  Where A   2 5 0 
 
 z   2  3 1  2

120

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STEP 3: Formula

A1  1 AdjA
A

x  1 
 y  
  A 1  3 

 z   2

STEP 4: Determinant of A

A  15  2  0 1 32  2  0  312 1  5 3

A 9

STEP 5: Minor of A

m11  5 2  01  10 m12  2 2  03  4 m13  21  53  13
m21  3 2 11  7 m22  1 2  13  1 m23  11  33  10
m31  30  15  5 m32  10  12  2 m33  15  32  11

10  4 13
 1 10
Minor    7

  5  2 11

STEP 6: Cofactor of A

10  4 13   10 4 13
 10   
Cofactor    7 1    7 1 10 

  5  2 11     5 2 11

STEP 7: Adjoin of A

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10 7  5 
 
AdjA   4 1 2 

13 10 11

STEP 8: Inverse matrix of A

A1  1 AdjA
A

1  10 7 5
9  1 
A1   4 10 2 

  13  11

STEP 9: Solve the equation

 x 1  10 7 5  1 
 y 9  1  
 z    4 10 2  3 

  13 11 2

 x 1  10  21  10
 y 9  
 z    4 34 

 13  30  22

 x  1  21
 y 9  3
 z   39 


STEP 10: Find the x, y and z

x 7133 
 y 
  

 13 
 z   3 

122

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LET’S PRACTICE 6

Solve the simultaneous equation below by using inverse matrices method.

x  y  z  3
1. 2x  3y  4z  23

 3x  y  2z  15

Ans: 2,1,4

123

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x  2y  z  7
2. 2x  3y  4z  3

xyz 0

Ans: 1,3,2

124

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4x  2 y  2z  10
3. 2x  8y  4z  32

30x  12 y  4z  24

Ans:  2,6,3

125

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3x  2 y  z  24
4. 2x  2 y  2z  12

x  5 y  2z  31

Ans: 3,4,7

126

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5x  2y  4x  0
5. 2x  3y  5z  8

3x  4 y  3z  11

Ans:  2,1,3

127

[CHAPTER 5: MATRIX]
5.5.2 SIMULTANEOUS EQUATION USING CRAMER’s RULE
EXAMPLE 11

x  2 y  3z  5
Given 3x  y  3z  4

 3x  4 y  7z  7

Solve the simultaneous equation by using Cramer’s Rule.

SOLUTION:
STEP 1: Rewrite the system using matrix multiplication

 1 2 3 x  5
 3    
 3 1   y    4 

 3 4 7  z   7

STEP 2: Find the determinant

A  117   34 237  3 3 334 1 3

A  19  24  45
A  40

STEP 3: Construct another 3 matrices

 5 2 3   1 5 3   1 2  5
 1  3  4  3  
Ax   4 Ay   3 Az   3 1 4 

 7 4 7   3  7 7   3 4  7

STEP 4: Find the determinant of each matrices

 5 2 3 
 1  3  517   34  247   3 7 344 1 7 95 14  69  40
Ax   4

 7 4 7 

 1 5 3 
 4  3  147  3 7  537  3 3 33 7 4 3  7  60  27  40
Ay   3

 3  7 7 

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 1 2  5
 
Az   3 1 4   11 7 44  23 7  4 3  534 1 3  23 18  75  80

 3 4  7

STEP 5: Solve the equation

x Ax   40  1
A 40

y AY  40 1
A 40

z Az   80  2
A 40

129

[CHAPTER 5: MATRIX]
LET’S PRACTICE 7
Solve the simultaneous equation below by using Cramer’s Rule

 2x  2 y  4z  22
1.  3x  5y  2z  35

 6x  3y  3z  21

Ans: 2,9,2

130

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4x  y  2z  1
2.  2x  y  6z  3

4x  6 y  3z  7

Ans: 1,1,1

131

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 5x  9 y  3z  40
3. 9x  4 y  2z  4
 5x  7 y  4z  15

Ans:  2,3,1

132

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4x  2 y  2z  4
4.  7x  y  2z  22
2x  6 y  z  26

Ans: 2,4,2

133

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4x  7 y  3z  37
5.  2x  3y  6z  61

2x  2 y  4z  44

Ans:  5,5,6

134

[CHAPTER 5: VECTOR AND SCALAR]

VECTOR AND SCALAR

OBJECTIVES:
At the end of this topic, students should be able to:

i. define vector
ii. understand the operation of vector
iii. apply scalar (dot) product of two vectors
iv. apply vector (cross) product of two vectors
v. understand area of parallelogram

6.1 INTRODUCTION OF VECTOR

1. Basic vector definition.
 Vector notation
 Vector representation
 Equality of vectors
 Negative vector

 Physical quantities can be classified under two main headings – vectors and scalars.

VECTOR
 Vector is a physical quantity which is specified by magnitude or length and a direction in
space.
 For example, displacement and velocity are both specified by a magnitude and a direction
and are therefore examples of a vector quantities.

SCALAR
 Scalar is a physical quantity which is specified by just its magnitude.
 For example, distance and speed are both fully specified by a magnitude and are therefore
examples of scalar quantities.

VECTOR NOTATION

 Vectors are written as Y, y, ỹ or Y   .
 The magnitude of a vector Y is written as Y .

135

[CHAPTER 5: VECTOR AND SCALAR]

VECTOR REPRESENTATION
 A vector is represented by a straight line with an arrowhead.

 In the Figure 1, the line OA represents a Figure 1

vector OA

 You can write:

OA   24

which means that to go from O to A,
move 4 units in the positive x direction
and 2 units in the positive y direction.
This is called the column vector or matrix
form.

In Cartesian form, the vector OA can be
represented as

OA  4i  2 j

Then,

OA   4   4i  2 j
2

VECTOR MAGNITUDE

 The magnitude or modulus of the vector OA is represented by the length OA and its
donated by OA .

EXAMPLE 1

 By referring Figure 2;

Figure 2

136

[CHAPTER 5: VECTOR AND SCALAR]

The magnitude of a vector; Angle;

OA  x 2  y 2   tan 1  y 
OA  42  2 2  x 
OA  20units
  tan 1  2 
 4 

  26.570

LET’S PRACTICE 1

1. Find the magnitude of each of these vectors.

a. 4i  3 j b. 2i  2 j  k

c. 79 d.  57 

 3 

Ans: 5 , 3, 11.40 & 9.11

EQUALITY OF VECTORS

 Two vectors are said to be equal if they have the same
magnitude and direction.

NEGATIVE VECTOR

 A vector having the same magnitude but opposite
direction.

137

[CHAPTER 5: VECTOR AND SCALAR]

UNIT VECTOR

 A unit vector is a vector of length 1.

unit vector, uˆ  u
u

EXAMPLE 2

1. Find the unit vector in the direction of v = 5i – 2j +4k.

SOLUTION v  52   22  42  45
magnitude of v,

if vˆ is a unit vector in the direction of v:

vˆ  5i  2 j  4k 4k
45 45

vˆ  5 i  2 j 
45 45

LET’S PRACTICE 2
1. Find a unit vector in the direction of the vector 8i  6 j .

Ans: 4i  3 j
5

2. Find a unit vector in the direction of v  3i  2 j  5k .

Ans: 3i  2 j  5k
38
138

[CHAPTER 5: VECTOR AND SCALAR]

3. Find a unit vector in the direction of the vector  7 
9

Ans:  7i  9 j
130

4. Find a unit vector in the direction of the vector  123

  4

Ans:  3i  12 j  4k
13

POSITION VECTORS

 By referring to the Figure 3, the Figure 3
position vector of a point P with
respect to a fixed origin O is the vector
OP.

 This is not a free vector, since O is a
fixed point.

 It can be write as;

OP  P

 Then,

PQ  PO  OQ

PQ  P  Q

PQ  Q  P

139

[CHAPTER 5: VECTOR AND SCALAR]

EXAMPLE 3

7  5
   
1. Given that a   3  and b  2  , find ab and ab . Hence, find unit vector in

 2  3 

the direction of a  b .

Solution:

 7   5 2
    5
a  b   3    2  

 2  3  1

a  b  22  52 12

a  b  30

a    a  
b a  b
b
Unit vector,

a b  2i 5j k  2 i 5 j 1k
30 30 30
30

2. Given that OX  6i  3 j  k and OY  2i  4 j  5k . Find the unit vector in the

direction of XY .

Solution:

XY  OY  OX

 2   6   4
   3  
XY   4     7 

 5  1   6

XY   42  7 2   62  101

Unit vector, XY  XY
XY

XY   4i  7 j  6k
101

XY   4 i  7 j  6 k
101 101 101

140

[CHAPTER 5: VECTOR AND SCALAR]

LET’S PRACTICE 3

1
1. If given B  3 , find the unit vector of B .

2

Ans: 1 i 3 j 2k
14 14 14
2. If given vector A  2i  3 j  6k and B  i  j  2k . Find
a. 2A  B
b. 2A  B

c. Unit vector of 2A  B

Ans: 5i  5 j  14k , 246 & 5 i  5 j  14 k
246 246 246

141

[CHAPTER 5: VECTOR AND SCALAR]

3. Find the unit vector of BA if given OA i  5 j 12k and OB  3i  5 j  k .

Ans:  2 i  11 k
125 125

142

[CHAPTER 5: VECTOR AND SCALAR]

6.2 THE OPERATION OF VECTOR

2. Operation
 Addition and subtraction of a vector
 Demonstrate addition and subtraction of vector using parallelogram method
 Demonstrate addition and subtraction of vector using triangle construction
method

6.2.1 VECTOR ADDITION

Figure 4

By referring the Figure 4 above;

OR  OP  PR OR  OQ  QR

i) OR  A  B  C ii) OR  A  B  C

From i) and ii)

OR  A  B  C  A  B  C

143

[CHAPTER 5: VECTOR AND SCALAR]
6.2.2 ADDITION AND SUBTRACTION OF VECTOR USING PARALLELOGRAM METHOD

 PARALLELOGRAM is a quadrilateral with both pairs of opposite sites parallel.
 Quadrilateral are four side polygons.
 Congruent refer to a shape (in mathematics) that has the same shape and size as another.

Properties of Parallelogram

Figure 5

PROPERTIES

i. If a quadrilateral is a parallelogram, then its opposite sides are congruent.

PQ  RS and SP  QR

ii. If a quadrilateral is a parallelogram, then its opposite angles are congruent.

P  R and Q  S

iii. If a quadrilateral is a parallelogram, then its consecutive angles are supplementary.

P  Q  180 0 R  S  180 0

Q  R  180 0 S  P  180 0

iv. If a quadrilateral is a parallelogram, then its diagonals bisect each other

QM  SM and PM  RM

144

[CHAPTER 5: VECTOR AND SCALAR]

EXAMPLE 4

Figure 6
ABCD is a parallelogram, find the sum of vectors below in unit of vector guide.

i. DA AC

ii. AD  AB

iii. AB  CB

Solution:

i. DA AC  DC

ii. AD AB  AC

iii. AB  CB  DB

6.2.3 ADDITION AND SUBTRACTION OF VECTOR USING TRIANGLE RULE

 Another way to define addition of two vectors is by a head-to-tail construction that
creates two sides of a triangle.

 The third side of the triangle determines the sum of the two vectors.

Figure 7

The vector u  v is defined to be the vector OP

145

[CHAPTER 5: VECTOR AND SCALAR]

EXAMPLE 5

Figure 8

ABCD is a triangle where BC  CD  DE . Given AB  6a  4b and BC  a  b . Express the

followings in term of a and b.

i. ED

ii. AC

iii. DA

Solution:

i. ED  CB  BC  a  b  b  a

AC  AB  BC

ii. AC  6a  4b a  b

AC  7a  3b

DA  DB  BA

iii. DA  2a  b  6a  4b

DA  2a  2b  6a  4b
DA  8a  2b

LET’S PRACTICE 4

146

[CHAPTER 5: VECTOR AND SCALAR]
1. Given that FGHJ is a parallelogram, find MH and FH. Given FM  5

Figure 9

Ans: 5,10

2.

Figure 10
ABCD is a parallelogram, find the sum of vectors below in unit of vector guide.
a. AB  BD

Ans: AD

b. CO OD

Ans: CD

c. CA BC

Ans: BA

d. OB DO

LET’S PRACTICE 5 Ans: DB

147