ENGINEERING MATHEMATICS 1 by PolisasLib3

ENGINEERING MATHEMATICS 1

WAN AZLIZA BINTI WAN ZAKARIA RAHIMAH BT MOHD ZAIN @ AB. RAZAK NIK NOOR SALISAH BINTI NIK ISMAIL

MATHEMATIC, SCIENCE & COMPUTER

ENGINEERING MATHEMATICS 1

Wan Azliza Binti Wan Zakaria Rahimah Binti Mohd Zain @ Ab. Razak Nik Noor Salisah Binti Nik Ismail

Politeknik Sultan Haji Ahmad Shah

Published by POL/TEKNIK SULTAN HAJI AHMAD SHAH SEMAMBU 25350 KUANTAN

Materials published in this book under the copyright of Politeknik Sultan Haji Ahmad Shah. All rights reserved. No part of this publication may be reproduced or distributed in any form or by means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers.

Preface

Alhamdulillah, with the blessing of Allah S.W.T., the ebook of Engineering Mathematics 1 can be accomplished written as planned. This ebook has been prepared by consulting the Polytechnic’s syllabus in order to facilitate the lecture session and as one of the meaningful references to the students. Most of the contents of the ebook have been written by referring to various text books and course’s website. Nevertheless, because of the limited knowledge from us, we feel that some improvement and additional contents can be made in order to enhance the ebook’s quality in future. Thus, any comments for improvement are highly appreciated. Engineering Mathematics 1 consists of topics such as Basic Algebra, Partial Fraction, Trigonometry, Complex Number, Matrices and Scalar and Vector. Various type of exercise are provided for students to improve their skill in solving problems related to topics they have learned. We also would like to thank to all the friend lecturers, the Head of Mathematics Unit and also the Head of Mathematics, Science and Computer Department for their support and encourage us in accomplishing this ebook. We really hope that this ebook will be the important reference hence can help the students to enhance their understanding towards the excellent achievement especially in engineering discipline. Lastly, we also hope that this ebook will benefit to all lecturers and students.

Rahimah Bt Mohd Zain @ Ab. Razak Wan Azliza Binti Wan Zakaria Nik Noor Salisah Binti Nik Ismail

CONTENTS TOPIC

ITEMS

PAGES

BASIC ALGEBRA

1.1 Introduction of Algebra 1.2 Algebraic Expression 1.3 Quadratic equation 1.3.1 Solving an equation by using factorization method 1.3.2 Solving an equation by using quadratic formula 1.3.3 Solving an equation by using completing the squares

1 7 9 9 11 13

PARTIAL FRACTION

2.1 Introduction of Partial Fraction 2.2 Proper fraction 2.2.1 Proper fraction with linear factor 2.2.2 Proper fraction with repeated linear factor 2.2.3 Proper fraction with quadratic factor 2.3 Improper fraction

15 21 21 25 28 32

TRIGONOMETRY

3.1 Introduction of trigonometry 3.1.1 Angles and their measure 3.1.2 Trigonometric raatios 3.2 Graph of sine, cosine and tangent 3.3 Positive and negative value of trigonometric 3.4 Trigonometric equations and identities 3.5 Sine and cosine rules 3.5.1 Sine rule 3.5.2 Cosine rule 3.6 Area of triangle

36 36 38 40 42 46 51 51 54 57

COMPLEX NUMBER

4.1 Introduction of complex number 4.2 The operation of complex number 4.2.1 Addition 4.2.2 Subtraction 4.2.3 Multiplication 4.2.4 Division 4.3 Conjugate of a complex numbers 4.4 Graphical representation of a complex number through Argand diagram 4.4.1 Argand's diagram to represent a complex number 4.4.2 Modulus and argument 4.5 Complex number in other form 4.5.1 Complex number in polar form 4.5.2 Complex number in trigonometric form 4.5.3 Complex number in exponential form 4.5.4 Complex number in cartesian form 4.6 Multilpication and division of complex number in polar form

59 62 62 64 66 68 70 72 72 73 75 76 77 78 79 80

MATRIX 5.1 Introduction of matrix 5.2 Operation of matrices 5.2.1 Addition 5.2.2 Subtracting 5.2.3 Multiplication 5.3 Determinant 5.4 Inverse matrix using minor, adjoin and cofactor 5.5 Simultaneous linear equation using matrix 5.5.1 Simultaneous equation using inverse matrix 5.5.2 Simultaneous equation using Cramer's rule VECTOR AND SCALAR 6.1 Introduction of vector 6.2 The operation of vector 6.2.1 Vector addition 6.2.2 Addition and subtraction of vector using parallelogram method 6.2.3 Addition and subtraction of vector using triangle rule Apply scalar (dot) product of two vectors 6.3 6.4 Apply vector (cross) product of two vectors 6.4.1 Application of the vector (cross) product References

84 88 88 88 90 94 97 100 100 104

107 114 114 115 116 120 124 126 129

[CHAPTER 1: BASIC ALGEBRA]

BASIC ALGEBRA OBJECTIVES: At the end of this topic, students should be able to: i. ii. iii.

understand basic algebra understand the algebraic expression solve quadratic equation using: a. factorization method b. quadratic formula  b  b 2  4ac 2a c. completing the square method x

b  b    x  2    2   c  0

1.1 ▪

INTRODUCTION OF ALGEBRA

Algebra is the use of letters and symbols to represent values and their relations, especially for solving equations. The combination of each letters and symbols are called “Algebraic Expressions’’.

BASIC ALGEBRA

  

Basic algebra is the field of mathematics that it one step more abstract than arithmetic. Remember that arithmetic is the manipulation of numbers through basic math functions. Algebra introduces a variable, which stands for an unknown number or can be substituted for an entire group of numbers.

Algebraic expression 



In mathematics, an algebraic expression is an expression built up from integer constants, variables, and the algebraic operations (addition, subtraction, multiplication, division and exponentiation by an exponent that is a rational number). Examples of algebraic expressions:

7 x, x  4,5  x, x 2  9,

x 3 1

[CHAPTER 1: BASIC ALGEBRA] Parts of an Algebraic Expression

Terms

  ii.

iv.

7 n is the product of 7 and n which are the factors of 7 n .



The number placed before the variable or the numerical factor of the term is called Coefficient of that variable.



7 is the numerical factor of 7 n so 7 is coefficient here.

Variable 

Any letter like x , y etc. are called Variables. The variable in the above figure is n .

Operations 

vi.

Every term is formed by the product of the factors.

Coefficient



Like 7 n and 2 are the two terms in the above figure.

Factors

  iii.

Every expression is separated by an operation which is called Terms.

Addition, subtraction etc. are the operations which separate each term.

Constant

 

The number without any variable is constant.

2 is constant here.

[CHAPTER 1: BASIC ALGEBRA] Examples of Algebraic Expressions

▪

An algebraic expression consists of numbers, variables, and operations. Here are a few examples:

ALGEBRAIC EXPRESSION

MEANING 8 times n

3x  1

3 times x plus 1

9 y 1

9 times y minus 1

n 1 4

N divided by 4 plus 1

EXAMPLE 1 1. Simplify 5 x  2  2 x SOLUTION:  5x  2 x  2

EXPLANATION: a. Move the like terms together. b. Add or subtract their coefficients.

 3x  2 2. Simplify 35  x   x 1  2 x   7 SOLUTION:  35  x   x1  2 x   7  15  3 x  x  2 x 2  7  15  7  3x  x  2 x 2  22  4 x  2 x 2

EXPLANATION: a. Clear the parentheses. b. Combine like terms by adding coefficients. c. Combine the constants. d. Rearrange sequence of power.

 2 x 2  4 x  22

3. Simplify  2a 3b  3ab 2 c SOLUTION:

 2a b  3ab c 3

 2  3  a 3  a  b  b 2  c  6  a 4  b 3  c

EXPLANATION: a. Multiply the numbers (coefficients). b. Multiply the variables - exponents can be combines if the base is the same.

 6 a 4 b 3 c

[CHAPTER 1: BASIC ALGEBRA] 3 4. Simplify  8a bc  2ab

SOLUTION:

8a 3bc − = 2ab

EXPLANATION: a. Write the division of the algebraic terms

as a fraction. b. Simplify the coefficient. c. Cancel variables of the same type in the numerator and denominator.

= - 4a2c

LET’S PRACTICE 1

Express each of the following expressions into a single algebraic fraction. a) p 3  q   q  p  3

Ans: 3 p  q  b) 3mn  1  1  m n

Ans: 3n  m  c) 3  p p 4

2 Ans: 12  p

[CHAPTER 1: BASIC ALGEBRA] d) 4 p 2 q  2 p   2 pq  1 pq

Ans: 2 p 2 q e) 4 x 2  3 x  9 x  6 

f) 5m n  6mn  4m n  3  6 2

Ans: x  23 x  18 

Ans: 33m 2 n 3  2mn  1 g)

x2  x  6 5  x3 3x  6

Ans:

5 3

[CHAPTER 1: BASIC ALGEBRA] h) 6n  2a   5n  3a 

Ans: n  27 a i)

7 p4q3 z 28 p 2 q 5

Ans: j)

2 1  m3 3m

Ans: k)

p2z 4q 2

1 m3

3x  3 2  x  3x  2 x  1 2

Ans:

x 1 x  1x  2

[CHAPTER 1: BASIC ALGEBRA]

1.2

ALGEBRAIC EXPRESSION

SOLVING LINEAR EQUATION

Linear Equation: A mathematical expression that has an equal sign and linear expressions. Variable: A number that you don't know, often represented by " x " or " y " but any letter will do! Variable(s) in linear expressions i.

Cannot have exponents (or powers).

For example, x squared or x 2 ii. Cannot multiply or divide each other. 



For example: " x " times " y " or xy ; " x " divided by " y " or

x y

iii. Cannot be found under a root sign or square root sign (sqrt). 

For example:

x or the "square root x "; sqrt (x)

Linear Expression: A mathematical statement that performs functions of addition, subtraction, multiplication, and division. EXAMPLE 2 1. Solve the following equation: x  12  25 SOLUTION: x  12  25

x  25  12 x  13

EXPLANATION: a. Isolate " x " to one side of the equation. b. Subtract 12 from each side to get constants on the right  12 . c. Check the solution. d. The result x equal to 13 .

2. Solve the following equation: y  6  42 SOLUTION:

y  6  42 y  42  6 y  48

EXPLANATION: a. Isolate "y" to one side of the equation. b. Add 6 from each side to get constants on the right (+6). c. Check the solution. d. The result y equal to 48. 7

[CHAPTER 1: BASIC ALGEBRA] 3. Solve the following equation: 3 z  6 SOLUTION: 3z  6

3z 6  3 3 z2

EXPLANATION: 1. Divide both sides by 3 to isolate the z. 2. Check the solution 3. The result z equal to 2

4. Solve the following equation: 42 x  9   4 x  4  6 x SOLUTION:

42 x  9   4 x  4  6 x

8 x  36  4 x  4  6 x 8 x  4 x  6 x  4  36 10 x  40 10 x 40  10 10 x4

EXPLANATION: 1. Expand the brackets and simplify. 2. Isolate " x" to one side of the equation. 3. Add 36 from each side to get constants on the right  36 . 4. Divide both sides by 10 to isolate the x . 5. Check the solution 6. The result x equal to 4

LET’S PRACTICE 2

2 x  4  30 5

Ans: 65 b) 8 

1 q  q3 4

Ans: 4 c) 2 z  3  1  2 z   25

Ans: 5 8

[CHAPTER 1: BASIC ALGEBRA] d) 4 y  8   y  3  15

Ans:

4 3

Ans: 1

6 7

e) 6  2 p  3  5 p  2 

1.3 1.3.1

QUADRATIC EQUATION

SOLVING AN EQUATION BY USING FACTORIZATION METHOD

EXAMPLE 3 2 Solve the following equations by factoring method x  8x  12  0 .

SOLUTION

EXPLANATION

x 2  8x  12  0

a. This equation is already in general form of ax 2  bx  c  0 .

Solution by cross multiply:

b. To get x , x must be multiply with x . c. Factor completely,  x  6 x  2  0 . d. Apply the Zero Product Rule x  6  0 or x  2  0. e. Solve each factor that was set equal to zero by getting the x on one side and the answer on the other. x  6 or x   2

x  6x  2  0 x6 0 x  6

x2 0 x  2

[CHAPTER 1: BASIC ALGEBRA] LET’S PRACTICE 3 Solve these equations by using factorization a) 5x 2  3x  2  0

2 5

Ans: x   , x  1 b) 4r 2  3r  10

5 4

Ans: r   , r  2 c) p 2  7  8 p

Ans: p  7, p  1 d) 5t 2  14t  3  0

Ans: t 

1 , t  3 5

e) y 2  6 y  8  0

Ans: y  4, y  2

[CHAPTER 1: BASIC ALGEBRA] 1.3.2



SOLVING AN EQUATION BY USING QUADRATIC FORMULA

When “Completing the Square” procedure is applied to a quadratic equation in general form, ax 2  bx  c  0 , then we receive the Quadratic formula.



The expression for the solutions of a Quadratic Equation through coefficients of this equation.

ax 2  bx  c  0 x

 b  b 2  4ac 2a

Solution set for a Quadratic Equation may contain i. Two distinct real numbers ii. One real number (repeated root) iii. No real solution

EXAMPLE 4 Solve the following equations using quadratic formula  3x 2  6 x  5  0 . SOLUTION

EXPLANATION

 3x 2  6 x  5  0

x x

 6  

62  4 35 2 3

 6   36  60 6

 6  96 or 6 x  0.633 x

 6  96 6 x  2.633

a. This equation is already in general form of ax 2  bx  c  0 . b. Identify a , b and c , then plug them into the quadratic formula. In this case, a   3 , b  6 , and c  5 . c. Use the order of operations to simplify the quadratic formula. d. Then, solve the equation either positive or negative. e. There will usually be two answers.

x

[CHAPTER 1: BASIC ALGEBRA] LET’S PRACTICE 4 Solve the equations below using quadratic formula to find the roots value for each of the following equation. a)

3t  5  t 2

Ans: t  1.19, t  4.19 b) 2a 2  8a  4  0

Ans: a  0.586, a  3.414 c) 2 p 2  3 p  1

Ans: p  0.28, p  1.78 d) 5t 2  14t  4  0

Ans: t  3.061, t  0.261

e) y 2  6 y  28  0

Ans: y  9.08, y  3.08 f) 2 x 2  5x  3  0

Ans: x 

1 , x  3 2 12

[CHAPTER 1: BASIC ALGEBRA] 1.3.3

▪

SOLVING AN EQUATION BY COMPLETING THE SQUARES

If u is an algebraic expression and d is a positive real number, then the equation

u 2  d has exactly two solutions: and

u d

u d

u   d  Formula

b b  x     c  0 2  2 

where a  1

Completing the square procedure ▪

Change the quadratic equation in the ax 2  bx  c  0 to an equivalent equation in the form a  x  d 2  k which then can be solved using the Square Root Method.

EXAMPLE 5

Solve the following equations using completing square 5x 2  4 x  2  0 . SOLUTION

EXPLANATION

5x 2  4 x  2  0

a. Divide all terms by 5 . b. Identify a, b, and c and plug them into the quadratic formula. In this case a  1 , b  0.8 , and c   0 .4 . c. Move the number term to the right side of the equation. d. Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation. e. Take the square root on both sides of the equation.

x 2  0.8 x  0.4  0 2

 0 .8    0 .8   x     0.4  0 2   2   2

0.8   x   0.16  0.4  0 2   2

0.8   x   0.56 2   x  0.4   0.56 x   0.56  0.4 x1  0.56  0.4 x1  1.148

x2   0.56  0.4

f. Add 0.4 on right side.

x2  0.348

[CHAPTER 1: BASIC ALGEBRA] LET’S PRACTICE 5

By using completing the square, find the roots value for each of the following question. a) 3 x 2  10 x  2  0

Ans: x  0.19, x  3.52 b) 20r 2  40r  8  0

Ans: r  1.77, r  0.23 c) 10 p 2  6  30 p

Ans: p  0.188, p  3.188 d) 5t 2  16t  3

Ans: t  0.178, t  3.378 e) y 2  6 y  7  0

Ans: y  4.414, y  1.586

[CHAPTER 2:PARTIAL FRACTION]

PARTIAL FRACTION OBJECTIVES: At the end of this topic, students should be able to understand about: i. ii. iii. iv. v.

Proper and Improper Fraction Proper Fraction with Linear Factor Proper Fraction with Repeated Linear Fraction Proper Fraction with Quadratic Factor Improper Fraction

2.1

INTRODUCTION OF PARTIAL FRACTION

DEFINE ALGEBRAIC FRACTIONS

a , the dividend a is called the numerator and the divisor b is b



In the algebraic fraction

 

called the denominator. The numerator and denominator are called the terms of the algebraic fraction. The simplest way to define a numerator and a denominator is the following:

 

Numerator: the top number of a fraction Denominator: the bottom number of a fraction

What Are Algebraic Fractions? 

Algebraic fractions are fractions using a variable in the numerator or denominator, such as

4 because division by 0 is impossible, variables in the denominator have certain x 

restrictions. The denominator can never equal 0.

[CHAPTER 2:PARTIAL FRACTION] PARTIAL FRACTION

EXPLANATION

2 x 4 x6 3 yz 5 x2 y

x cannot equal 0( x ≠ 0) x cannot equal 6( x ≠ 6) y-z cannot equal 0( y-z ≠ 0) so y cannot equal z(a ≠ b) Neither x or y cannot equal 0. ( x ≠ 0, ( y ≠ 0))

PROPER AND IMPROPER FRACTIONS



Fractions that are greater than 0 but less than 1 are called proper fractions.



In proper fractions, the numerator is less than the denominator.



When a fraction has a numerator that is greater than or equal to the denominator, the fraction is an improper fraction.



An improper fraction is always 1 or greater than 1.



Finally, a mixed number is a combination of a whole number and a proper fraction.

IDENTIFYING PROPER AND IMPROPER FRACTIONS 

In a proper fraction, the numerator is always less than the denominator.



Examples of proper fractions include



In an improper fraction, the numerator is always greater than or equal to the denominator.



Examples of improper fractions include

1 7 2 , , 2 13 905

5 10 30 , , 3 7 15

[CHAPTER 2:PARTIAL FRACTION] LET’S PRACTICE 1 Identify the fraction below: Equations

Fractions(Proper /Improper)

30 x 1 x x 1 5 2 x 1 3 9x  x  6 x2  5 2 x4  4 x  7 x5  8 x4 x 2  3x  1 6x 2 ( x  1)( x 2  3)

CHANGING IMPROPER FRACTIONS TO MIXED NUMBERS

 

An improper fraction can also be written as a mixed number. Mixed numbers contain both a whole number and a proper fraction.



Examples of mixed numbers include 6 ,1

1 4 6 ,3 5 10 18

Writing Improper Fractions as Mixed Numbers Step 1: Divide the denominator into the numerator. Step 2: The quotient is the whole number part of the mixed number. Step 3: The remainder is the numerator of the fractional part of the mixed number. Step 4: The divisor is the denominator of the fractional part of the mixed number.

[CHAPTER 2:PARTIAL FRACTION] EXAMPLE 1

Write the improper fraction SOLUTION: a. b. c. d.

47 as a mixed number. 7

47  7  6 remainder 5 .Divide the denominator into the numerator. The quotient, 6 , becomes the whole number. The remainder, 5 , becomes the numerator. The denominator, which is also used as the divisor, remains as 7 .

Answer:

47 5 =6 7 7

Writing Mixed Numbers as Improper Fractions Step 1: Multiply the denominator of the fraction by the whole number. Step 2: Add this product to the numerator of the fraction. Step 3: The sum is the numerator of the improper fraction. Step 4: The denominator of the improper fraction is the same as the denominator of the fractional part of the mixed number.

[CHAPTER 2:PARTIAL FRACTION] EXAMPLE 2

3 4

Write 4 as an improper fraction SOLUTION: a. Multiply the denominator of the fraction by the whole number. b. Add this result to the numerator of the fraction. c. This answer becomes the numerator of the improper fraction. Notice that the denominator of the improper fraction is the same as the denominator that was in the fractional part of the mixed number.

3 4 4  4  16 16  3  19 4

19 4 3 4

Answer: 4 =

19 4

PARTIAL FRACTIONS  

To express a single rational fraction into the sum of two or more single rational fractions is called Partial fraction resolution. For example,

2x  x 2  1 1 1 1    2 x x 1 x 1 x( x  1)

2x  x2 1 1 1 1 is the resultant fraction and  are its partial fractions.  2 x( x  1) x x 1 x 1

[CHAPTER 2:PARTIAL FRACTION] PROCEDURES TO FIND PARTIAL FRACTION PARTIAL FRACTION DECOMPOSITION The method is called “Partial Fraction Decomposition” and goes like this:Step 1: Factor the bottom

6x  6 6x  6  ( x  2 x  3) ( x  3)( x  1) 6x  6 A B   2 ( x  2 x  3) ( x  3) ( x  1) 6 x  6  A( x  1)  B( x  3)..........equation1 2

Step 2: Write one partial fraction for each of those factors Step 3: Multiply through by the bottom so we no longer Step 4: Now find the Constants, A & B

Let ( x  3)  0 Thus, x  3 Substituting into equation 1

6( 3)  6  A( 3  1)  B (3  3) 12  A(4)  B (0)

A3 Let ( x  1)  0 Thus, x  1

Substituting into equation 1

6(1)  6  A(1  1)  B (1  3)

12  A(0)  B (4) B3 Step 4: Rewrite the value of A and B into the partial fraction

6x  6 3 3   ( x  2 x  3) ( x  3) ( x  1) 2

DEFINE TYPES OF PARTIAL FRACTION

[CHAPTER 2:PARTIAL FRACTION]

2.2 2.2.1

PROPER FRACTION

PROPER FRACTION WITH LINEAR FACTOR

FORMULA

EXAMPLE 3

Determine the partial fraction decomposition of each of the following: a.

8x  4

x  1x  2 

SOLUTION:

8x  4 A B   x  1x  2  x  1 x  2 

8 x  4  A( x  2)  B( x  1) …………… equation 1 if ( x  2)  0 x  2

Substitute x  2 into equation 1

8 2   4  A 2  2  B 2  1  12  A0  B 3 B4

if ( x  1)  0 x 1 Substitute x  1 into equation 1 81  4  A1  2  B1  1

12  A3  B0 A4

Therefore the partial fraction:-

8x  4 4 4   ( x  1)( x  2) ( x  1) ( x  2)

[CHAPTER 2:PARTIAL FRACTION]

x 2  11 ( x  1)( x  2)( x  3)

SOLUTION

x 2  11 A B C    ( x  1)( x  2)( x  3) ( x  1) ( x  2) ( x  3)

x 2  11  A( x  2)( x  3)  B( x  1)( x  3)  A( x  2)( x  1) ……..equation 1 if ( x  1)  0 x 1 Substitute x  1 into equation 1

(1) 2  11  A(1  2)(1  3)  B(1  1)(1  3)  C (1  1)(1  2) 12  A(6)  B(0)  C (0) A

12  2 6

if  x  2  0 x  2 Substitute x  2 into equation 1 (2) 2  11  A(2  2)(2  3)  B(2  1)(2  3)  C (2  1)(2  2) 15  A(0)  B(15)  C (0) 15 B 1 15 if x  3  0 x3 Substitute x  1 into equation 1 (3) 2  11  A(3  2)(3  3)  B(3  1)(3  3)  C (3  1)(3  2) 20  A(0)  B(0)  C (10) 20 C 2 10 Therefore the partial fraction:-

x 2  11 2 1 2    ( x  1)( x  2)( x  3) ( x  1) ( x  2) ( x  3)

[CHAPTER 2:PARTIAL FRACTION] c.

3x  1 ( x  5 x  6) 2

SOLUTION:

3x  1 A B * Must be factorized   ( x  5 x  6) ( x  2) ( x  3) 2

3x  1  A( x  3)  B( x  2) ……..equation 1 According to denominator, x  3  0

x3

Substitute x  3 into equation 1

3(3)  1  A(3  3)  B(3  2) 10  A(0)  B(1) 1 B 10

According to denominator, x  2  0

x2

Substitute x  2 into equation 1

3(2)  1  A(2  3)  B(2  2) 7  A(1)  B(0) 1 A 7 3x  1 1 1 Therefore the partial fraction:  2 7( x  2) 10( x  3) ( x  5 x  6)

LET’S PRACTICE 2 Determine the partial fraction decomposition of each of the following. a.

x4 ( x  1)( x  8)

Ans:

5 4  9( x  1) 9( x  8) 23

[CHAPTER 2:PARTIAL FRACTION] b.

6x  4 ( x  9 x  8) 2

Ans: c.

44 2  7( x  8) 7( x  1)

9  9x (2 x  7 x  4) 2

Ans:

1 5  (2 x  1) ( x  4)

x2  5 x( x  3)( x  6)

Ans: 

5 14 41   18x 27( x  3) 54( x  6) 24

[CHAPTER 2:PARTIAL FRACTION]

2x2  x  2 ( x  3)(2 x  1)( x  1)

Ans:

2.2.2

1 1 1   2( x  3) 32 x  1 6( x  1)

PROPER FRACTION WITH REPEATED LINEAR FACTOR

FORMULA

EXAMPLE 4 Determine the partial fraction decomposition below.

3x 2  2 ( x  2)( x  1) 2 SOLUTION:

3x 2  2 A B C    2 ( x  2) ( x  1) ( x  1) 2 ( x  2)( x  1)

3x 2  2  A( x  1) 2  B( x  1)( x  2)  C ( x  2) ……..equation 1 According to denominator, x  2  0 Substitute x  2 into equation 1

x2

3(2) 2  2  A(2  1) 2  B(2  1)(2  2)  C (2  2) 14  A(1)  B(0)  C (0) A  14 25

[CHAPTER 2:PARTIAL FRACTION] According to denominator, x  1  0

x 1

Substitute x  1 into equation 1

3(1) 2  2  A(1  1) 2  B(1  1)(1  2)  C (1  2) 5  A(0)  B(0)  C (1) C  5 To find the value of B we can use the method of Equating Coefficients. We take equation 1 and multiply-out the right-hand side, and then collect up like terms.

3x 2  2  A( x  1) 2  B( x  1)( x  2)  C ( x  2) ……..equation 1 3x 2  2  A( x  1)( x  1)  B( x  1)( x  2)  C ( x  2) Multiplying out:

3x 2  2  ( A  B) x 2  (2 A  3B  C ) x  ( A  2B  2C ) Left

Right

Compare the same coefficients right and left side: Use Equating coefficients Coefficients

Right

x x

( A  B)

Left 3

(2 A  3B  C )

Constant,k

( A  2 B  2C )

Substitute value of A,……… A  14

A B  3 14  B  3 B  11 Then; A  14, B  11, C  5

Therefore the Partial Fraction:-

3x 2  2 14 11 5    2 ( x  2) ( x  1) ( x  1) 2 ( x  2)( x  1)

[CHAPTER 2:PARTIAL FRACTION] LET’S PRACTICE 3 Determine the partial fraction decomposition of each of the following. a.

x2 x( x  4) 2

Ans: b.

3  x2 x 2 ( x  6)

Ans:  c.

1 4  ( x  4) ( x  4) 2

1 1 13  2 12 x 2 x 12( x  6)

2x  1 (1  x)(5 x  1) 2

Ans:

3 15 7   16(1  x) 16(5 x  1) 4(5 x  1) 2 27

[CHAPTER 2:PARTIAL FRACTION] 2.2.3

PROPER FRACTION WITH QUADRATIC FACTOR

FORMULA

Note: (cx 2  dx  c) cannot be factorized UNFACTORIZED

FACTORIZED

( x  1)

( x 2  1)

( x 2  4)

( x 2  4)

( x 2  16)

( x 2  16)

( x 2  64)

( x 2  64)

( x 3  1)  ( x  1)( x 2  x  1)

EXAMPLE 5

Determine the partial fraction decomposition below.

6x2  8 x ( x 2  2) SOLUTION:

6x2  8 A Bx  C   2 2 x( x  2) x ( x  2)

6 x 2  8  A( x 2  2)  ( Bx  C )( x) ……………..equation 1 According to denominator, x  0 Substitute x  0 into ……………………………equation 1

6(0)  8  A(0  2)  ( B(0)  C )(0)  8  A(2)  B(0)  C (0) A  4

To find the value of B and C we can use the method of Equating Coefficients. We take equation 1 and multiply-out the right-hand side, and then collect up like terms.

[CHAPTER 2:PARTIAL FRACTION]

6 x 2  8  A( x 2  2)  ( Bx  C )( x) ………….equation 1 6 x 2  8  Ax 2  2 A  Bx 2  Cx Multiplying out:

6 x 2  8  ( A  B) x 2  Cx  2 A Left

Right

Compare the same coefficients right and left side: Use Equating coefficients Coefficients

Right

x x

Constant,k

( A  B)

Left 6

C 2A

0 -8

Substitude value of A,……… A  4

A B  6 4B 6 B  10 C 0 Then; A  4, B  10, C  0 Therefore the Partial Fraction:-

6x2  8 4 10 x   2 2 x ( x  2) x ( x  2)

[CHAPTER 2:PARTIAL FRACTION] LET’S PRACTICE 4 Determine the partial fraction decomposition of each of the following. a.

2x  6 ( x  4)( x 2  5)

Ans: 

2 2x  2  3( x  4) 3( x 2  5)

Ans:

2x  5 2  2 3 x  2 31  x 

3 x ( x  2)(1  x) 2





2x  3 ( x 2  1) x

Ans:

3x  2 3  ( x 2  1) x 30

[CHAPTER 2:PARTIAL FRACTION] d.

9 x( x  6) 2

Ans: e.

3x  8 ( x  4 x  1)( x  2) 2

Ans: f.

3 3x  2 x 2( x 2  6)

x  51 2  2 13( x  4 x  1) 13( x  2)

2 x 2  4 x  10 ( x 2  1) x 2

Ans:

 4 x  12  4  10 ( x 2  1)

[CHAPTER 2:PARTIAL FRACTION]

2.3

IMPROPER FRACTION

FORMULA

Note * Degree of numerator is the same or higher than the denominator n  k

EXAMPLE 6 Determine the partial fraction decomposition below.

4x 3  5 x 3 1 SOLUTION:

4 Use Long Division technique ( x  1) (4 x  5) 3

( 4 x 3  4)  _______ 9

4x 3  6 9 9  4 3 So, 3 …………… 3 must be state in form partial fraction x 1 x 1 x 1 Partial fraction for

9 is…. x 1 3

9 A Bx  C   2 x  1 ( x  1) ( x  x  1) 3

9  A( x 2  x  1)  ( Bx  C )( x  1) ……..equation 1 According to denominator, x  1  0

x 1

[CHAPTER 2:PARTIAL FRACTION] Substitute x  1 into equation 1

9  A((1) 2  1  1)  ( B(1)  C )(1  1) 9  A(3)  B(0)  C (0) A3 To find the value of B and C we can use the method of Equating Coefficients. We take equation 1 and multiply-out the right-hand side, and then collect up like terms.

9  A( x 2  x  1)  ( Bx  C )( x  1) ……..equation 1 9  Ax 2  Ax  A  Bx 2  Bx  Cx  C Multiplying out:

9  ( A  B) x 2  ( A  B  C ) x  ( A  C ) Left

Right

Compare the same coefficients right and left side: Use Equating coefficients Coefficients

Right

x2 x

A B

Left 0

A BC AC

0 9

Constant,k

A B  0 B  A B  3

ABC  0

3  (3)  C  0 C  6

Then; A  3; B   3; C  6

The Partial Fraction:-

9 3 3x  6   2 x  1 ( x  1) ( x  x  1) 3

Therefore the Partial Fraction:-

4x3  5 3 3x  6  4  2 3 ( x  1) ( x  x  1) x 1

[CHAPTER 2:PARTIAL FRACTION] LET’S PRACTICE 5

Express of the following in partial fractions. a.

4 x 3  10 x  4 (2 x  1) x

Ans: 2 x  1  b.

3 4  (2 x  1) x

2x 2 ( x  3)(2 x  1)

Ans: 1 

18 1  5( x  3) 5(2 x  1) 34

[CHAPTER 2:PARTIAL FRACTION]

4 x 3  3x  2 ( x  2)(2 x  1)

Ans: d.

2x  3 

24 2  5( x  2) 5(2 x  1)

x3 ( x  2)( x  3)

Ans: x  1 

8 27  5( x  2) 5( x  3)

[CHAPTER 3: TRIGONOMETRY]

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

Introduction of trigonometric

ii.

Define sine, cosine, tangent, cosecant, secant and cotangent.

iii.

Graph of sine, cosine and tangent.

iv.

Positive and negative value of trigonometric.

Apply sine and cosine rules.

3.1

INTRODUCTION OF TRIGONOMETRY

3.1.1 ANGLES AND THEIR MEASURE ANGLES IN UNIT DEGREES AND RADIANS



Angles are measured in either degrees or radians (rad). The size of a radian is determined by the requirement that there are 2 radians in a circle. Thus 2 radians equals 360 degrees.



This mean that 1 radian 

 

180



, and 1 degree 



180

radians.

EXAMPLE 1 

1. Express in radian measure:  (a) 54



(b) 135

SOLUTION:  (a) 54  54

 180

 (b) 135  135

 0.9425 rad



3   rad 180 4

[CHAPTER 3: TRIGONOMETRY]

2. Express each angle in degree measure: (a)

 3

(b)

rad

5 rad 9

SOLUTION: (a) (b)

 3

rad 

 3



180



 60

5 5 180 rad    100 9 9 

LET’S PRACTICE 1

1. Express in radian measure: a) 330 b) 160 c) 42

Ans: 5.76,  2.79 & 0.733

2. Express each angle in degree measure: 4 a) rad 3 21 b)  5

Ans: 76.39° , 756°

RIGHT ANGLE TRIANGLE 

In mathematics, the pythagorean theorem, also known as Pythagoras’ theorem, is a fundamental relation in Euclidean geometry among the three sides of a right triangle.



It states that the area of the square whose side is the hyotenuse is equal to the sum of the areas of the squares on the other two sides.



Formula;

By the Pythgorean theorem that the sum of the squares of each of the smallest sides equals the square of the hypotenuse.



Trigonometric functions are how the relationships amongst the length of the sides of a right triangle vary as the other angles are changed.

c  a2  b2 Figure 3.1: Pythagoras’ Theorem

[CHAPTER 3: TRIGONOMETRY]

3.1.2 TRIGONOMETRIC RATIOS



The ratios of the sides of a right triangle are called trigonometric ratios.



Three common trigonometric ratios are the sin (sin), cosine (cos) and tangent (tan).



The three others trigonometric ratios are the cosecant (csc), secant (sec) and cotangent (cot).

Figure 3.2: Trigonometric ratios in right triangles

Figure 3.3: Basic and Pythagorean Identities

SOH-CAH-TOA : an easy way to remember trigonometric ratios 

The word SOHCAHTOA helps us remember the definitions of sine, cosine and tangent. Table 3.1: Trigonometric ratios

ACRONYM PART

VERBAL DESCRIPTION

SOH

Sine is Opposite over Hypotenuse

CAH

Cosine is Adjacent over Hypotenuse

TOA

Tangent is Opposite over Adjacent

MATHEMATICAL DEFINITION

opposite hypotenuse adjacent cos A  hypotenuse opposite tan A  adjacent sin A 

[CHAPTER 3: TRIGONOMETRY] EXAMPLE 2

Figure 3.4

Find sin A and tan A SOLUTION: opposite hypotenuse 3 sin A  5

opposite adjacent 3 tan A  4

sin A 

tan A 

LET’S PRACTICE 2

Find: a) cos F b) sin F c) tan F Ans:

5 12 12 , & 13 13 5

a) sin H b) cos H c) tan H

d) csc G e) sec G f) cot G

Ans:

8 15 8 17 17 8 , , , , & 17 17 15 15 8 15 39

[CHAPTER 3: TRIGONOMETRY]

Find: a) sin B b) cos B c) tan B

d) cosecB e) sec B f) cot B

Ans:

5 12 5 13 13 12 , , , , & 13 13 12 5 12 5

Given BAC  60 0 . Find the length of BC.

3.2

Ans: 22.52

GRAPH OF SINE, COSINE AND TANGENT

GRAPH OF SINE y  sin x

𝑥°

x rad



−90°  

0° 0

1

90° 

180°



270° 3

360° 2

1

Figure 3.5: Graph of sine

[CHAPTER 3: TRIGONOMETRY]

GRAPH OF COSINE

y  cos x 𝑥°

x rad



−90°  

0° 0

90° 

180°



360° 2

1

270° 3

Figure 3.6: Graph of cosine

GRAPH OF TANGENT y  tan x

𝑥°

x rad



−90°  

 

−45°  

0° 0

1

45° 

90° 



180°



270° 3 2



Figure 3.7: Graph of tangent



An asymptote is an imaginary line that a curve get closer and closer to but never touches.

[CHAPTER 3: TRIGONOMETRY]

3.3

POSITIVE AND NEGATIVE VALUE OF TRIGONOMETRIC

FOUR QUADRANTS

Figure 3.8a: Four quadrants

RELATED ANGLE 

Reference angles allow us to evaluate more complex angles and makes easier when evaluating angles.



A reference angle

x and

x ' , is the positive acute angle made by the terminal side of

the angle x and x-axis. 

The figure shows differing angles that lies in quadrants I, II, III and IV.



Remember that in quadrant I, an angle and its related angle are the same measure.

𝜃 = 180° − 𝛼

𝜃=𝛼

𝜃 = 180° + 𝛼

𝜃 = 360° − 𝛼

Figure 3.8b: Four quadrants

where; = Actual angle 𝛼 = Related angle 42

[CHAPTER 3: TRIGONOMETRY]

EXAMPLE 3

Find the value of  for sin   0.5 where 0° ≤ 𝜃 ≤ 360° SOLUTION:

sin   0.5

Sine positive (Quadrant 1 and 2)

Quadrant 1:   30 0

  sin 1 0.5

Quadrant 2:

  30 0

  180 0  30 0  150 0

Therefore;   30 0 & 150

LET’S PRACTICE 3

Find the value of  for the followings where 0° ≤ 𝜃 ≤ 360° a. cos  0.8264

sin   0.8660

cos ec  5.461

sec   1.5642

Ans: 34.27° & 325.73°

Ans: 60° & 120°

Ans: 10.55° & 169.45°

Ans: 50.26° & 309.74° 43

[CHAPTER 3: TRIGONOMETRY]

EXAMPLE 4

Find the value of  for 𝑐𝑜𝑠𝜃 = −0.3255, where 0° ≤ 𝜃 ≤ 360° SOLUTION: 𝑐𝑜𝑠𝜃 = −0.3255 𝜃 = 𝑐𝑜𝑠 𝜃 = 71°

Cos is negative value (Quadrant 2 and 3)

0.3255

Quadrant 2:

𝜃 = 180° − 71° = 109° Quadrant 3:

𝜃 = 180° + 71° = 251° Therefore; 𝜃 = 109° & 251°

LET’S PRACTICE 4 Find the value of  for the followings where 0° ≤ 𝜃 ≤ 360° ; a. cos  0.9061

tan  1.364

sin   0.5246

cot  2.53

Ans: 154.97° & 205.03°

Ans: 126.25° & 306.25°

Ans: 211.64° & 328.26°

Ans: 158.43° & 338.43° 44

[CHAPTER 3: TRIGONOMETRY]

EXAMPLE 5 Find the value of  for sin = 0.5870, where 0° ≤ 𝜃 ≤ 360° SOLUTION:

sin

 2

 0.5870

𝜃 = sin 2

0.5870

Sin is positive value (Quadrant 1 and 2)

QUADRANT 1:

𝜃 = 35.94° 2 𝜃 = 2(35.94° ) = 71.89°

𝜃 = 35.94° 2

QUADRANT 2:

𝜃 = 180° − 35.94° = 144.06° 2 𝜃 = 2(144.06° ) = 288.12° Therefore; 𝜽 = 𝟕𝟏. 𝟖𝟗° & 𝟐𝟖𝟖. 𝟏𝟐°

LET’S PRACTICE 5 Find the values of  in the range 0° ≤ 𝜃 ≤ 360° for each of the following, 1 a. cos    0 .7384 2

cos 2  0.4756

Ans: 275.2

Ans: 30.8 , 149.2  , 210.8 & 329.2  45

[CHAPTER 3: TRIGONOMETRY]

3.4

TRIGONOMETRIC EQUATIONS AND IDENTITIES

1. Basic Identities Sin  Tan Cos 

2. Trigonometric Identities

cos2   sin2   1 1  tan2   sec2  cot2   1  cosec 2

3. Angle Sum and Difference Identities

sin     sin  cos   cos sin 

cos     cos cos   sin  sin  tan    

tan   tan  1  tan tan 

4. Double Angle Identities

sin 2 A  2 sin A cos A cos 2 A  cos 2 A  sin 2 A cos 2 A  1  2 sin 2 A cos 2 A  2 cos 2 A  1 2 tan A tan 2 A  1  tan 2 A

[CHAPTER 3: TRIGONOMETRY]

EXAMPLE 6 1.

Solve 𝑠𝑖𝑛𝜃 + 2 = 3, where 0° ≤ 𝜃 ≤ 360° SOLUTION: sin   2  3 sin   1

Quadrant 1: 𝜃 = 90°

Sine positive (Quadrant 1 and 2)

  sin 1 1   90 

Quadrant 2: x = 180° − 90° = 90°

Reference Angle

Therefore; 𝒙 = 𝟗𝟎° 2.

Solve 5𝑠𝑖𝑛𝜃 − 2𝑐𝑜𝑠𝜃 = 0, where 0° ≤ 𝜃 ≤ 360° . SOLUTION: Quadrant 1:

5𝑠𝑖𝑛𝜃 − 2𝑐𝑜𝑠𝜃 = 0

𝜃 = 21.80°

5𝑠𝑖𝑛𝜃 2𝑐𝑜𝑠𝜃 − =0 𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃 5𝑡𝑎𝑛𝜃 − 2 = 0 Tangent is positive 2 (Quadrant 1 and 3) 𝑡𝑎𝑛𝜃 = 5 2 𝜃 = 𝑡𝑎𝑛 = 21.80° 5

Quadrant 3: 𝜃 = 180° + 21.80° = 201.80°

Therefore; 𝜽 = 𝟐𝟏. 𝟖𝟎° & 𝟐𝟎𝟏. 𝟖𝟎°

LET’S PRACTICE 6

Solve each of the following equations for 0° ≤ 𝜃 ≤ 360° . a. 4 sin   3 cos 

Ans: 36.87 & 216.87 47

[CHAPTER 3: TRIGONOMETRY]

b. sin   cos  0

Ans: 135 & 315

c. 4 tan   3 sec 

Ans: 48.59 & 131.4

d. 2 tan  cos ec  3

Ans: 48.19 & 311.81 2.

Solve each of the following equations for 0° ≤ 𝑥 ≤ 360° . a. 2 cos 2 x  cos x

b. 5 sin x cos x  sin x

Ans: 60 , 90 , 270 & 300

Ans: 78.46 & 281.54 c. sin x  sin x  2  0 2

Ans: 270 d. 2 tan 2 x  sec x tan x

Ans: 30 & 150 48

[CHAPTER 3: TRIGONOMETRY]

EXAMPLE 7 Solve the equation 2 cos 2   3 cos   1  0 for the values of  in the range 0° ≤ 𝜃 ≤ 360°. SOLUTION:

2 cos 2   3 cos  1  0 2 cos  1cos  1  0 2 cos   1  0 cos  

1 2 1 2

Cos positive (Quadrant 1 and 4)

  cos 1     60 

cos   1  0 cos   1

  cos 1 1   0

Quadrant 1: θ = 60° Quadrant 4:

Reference angle

Cos positive (Quadrant 1 and 4) Reference angle

θ = 360° − 60° = 300°

Quadrant 1: 𝜃 = 0° Quadrant 4: 𝜃 = 360° − 0° = 360°

Therefore; ° ° ° 𝜃 = 0 , 60 , 300 & 360 °

[CHAPTER 3: TRIGONOMETRY]

LET’S PRACTICE 7

Solve the equation 1  2 sin   4 cos 2  0 for values of  in the range of 0° ≤ 𝜃 ≤ 360° .

Ans: 48.59 , 131.41 , 210 & 330 Solve each of the following equations for values of  in the range 0° ≤ 𝜃 ≤ 360° . a. 8 sin 2   2 cos   5  0

Ans: 41.41 , 120 , 240 & 318.59 b. 7 sin 2   cos 2   5 sin 

c. 3 cos   2 cot 

Ans: 19.47 , 30 , 150 & 160.53

Ans: 41.81 & 138.19

[CHAPTER 3: TRIGONOMETRY]

3.5

SINE AND COSINE RULES

3.5.1 SINE RULE Sine Rule:

a b c   sin A sin B sin C sin A sin B sin C   a b c

EXAMPLE 8 a.

Find the length of AC.

SOLUTION:



 ABC  180   75   40   ABC  65





AC 8.2   sin 65 sin 40  8.2 sin 65  AC   11.56cm sin 40  b. Solve the triangle with the following given dimensions: a  43 cm , b  75 cm and B  30 0

SOLUTION: sin A sin B  a b sin A sin 30   43 75  43 sin 30  A  sin 1  75 

   16.66  

[CHAPTER 3: TRIGONOMETRY]

C  180 0  30 0  16 . 66 0  133 . 34 

c b  sin C sin B c 75   sin 133.34 sin 30  75 sin 133.34  c  109.09cm sin 30 

LET’S PRACTICE 8

Find the length of AC Ans: 4.52cm 2.

Find  ABC Ans: 31.69 52

[CHAPTER 3: TRIGONOMETRY]

Find  ABC and the length of AB

Ans: 11.82 & 10.02cm

Find the length of AC and BC. 5.

Ans: 5.19cm & 6.35cm

Solve the triangles with the given parts: a) 𝑎 = 45.7𝑐𝑚, 𝐴 = 68.20° , 𝐵 = 47° b) 𝑎 = 4.608𝑚, 𝑏 = 3.207𝑚, A = 18.23° c) 𝑏 = 742𝑚𝑚, B = 53° , C = 3.5°

a) C = 64.8° , 𝑏 = 36𝑐𝑚, 𝑐 = 44.54𝑐𝑚 Ans: b) B = 12.58° , C = 149.19° , c = 7.545𝑚 c) A = 123.5° , 𝑎 = 774.75𝑚𝑚, 𝑐 = 56.72𝑚𝑚

[CHAPTER 3: TRIGONOMETRY]

3.5.2 COSINE RULE Cosine Rule:

a 2  b 2  c 2  2bc cos A b 2  a 2  c 2  2ac cos B c 2  a 2  b 2  2ab cos C

EXAMPLE 9 1.

Given 𝑎 = 16.4𝑐𝑚, 𝑏 = 11.8𝑐𝑚 and

SOLUTION:

∠C = 67° . Find the length of 𝑐. c 2  a 2  b2  2ab cos C c 2  16.42  11.82  216.411.8cos 670 c 2  256.97 c  256.97 c  16.03cm

2. Solve the triangle given that a  6.00cm , b  7.56cm and ∠C = 54° . SOLUTION: c 2  a 2  b2  2ab cos C c 2  62  7.562  267.56cos 540 c  6.31cm

For A , use the law of sines: sin A sin C  a c sin A sin 540  6 6.31 6 sin 540 sin A  6.31 A  50.290

𝐁 = 180° − (50.29° + 54° ) = 𝟕𝟓. 𝟕𝟏°

[CHAPTER 3: TRIGONOMETRY]

LET’S PRACTICE 9

1. Given 𝑎 = 7𝑐𝑚, 𝑏 = 5𝑐𝑚 and ∠C = 60° . Find the length of 𝑐 .

Ans:

6.24 cm

2. Given 𝑎 = 7𝑐𝑚, 𝑏 = 5𝑐𝑚 and 𝑐 = 3𝑐𝑚. Find B .

Ans: 38.21° 3. Given 𝑎 = 15𝑐𝑚, 𝑏 = 12𝑐𝑚 and 𝑐 = 14𝑐𝑚. Find A .

Ans: 69.99°

[CHAPTER 3: TRIGONOMETRY]

LET’S PRACTICE 10 1. Given ∠ABC = 64° . Find: a. A b. length of 𝑐

Ans: 74.05° , 6.4𝑐𝑚 2. Given the length of 𝐽𝐿 = 6.2𝑐𝑚, 𝐾𝐿 = 4.6𝑐𝑚 and ∠K = 52° . Find: a. the angle of ∠L b. length of 𝐽𝐾

Ans: 92.22° , 7.86𝑐𝑚 3. Given the length of 𝑂𝑃 = 9.1𝑐𝑚, 𝑂𝑄 = 5.9𝑐𝑚 and 𝑃𝑄 = 10.2𝑐𝑚. Find: a. P b.

Q

Ans: 35.01° , 62.25°

[CHAPTER 3: TRIGONOMETRY]

3.6

AREA OF TRIANGLE Area of triangle ABC:

1  ab sin C 2 1  bc sin A 2 1  ac sin B 2 EXAMPLE 10 Given 𝐴𝐶 = 15𝑐𝑚, 𝐵𝐶 = 9𝑐𝑚 and

SOLUTION:

∠ACB = 64° . Find the area of triangle.

Area of ABC; 1 159sin 640 2  60.67cm2 

2. Given 𝐴𝐶 = 6𝑐𝑚, 𝐵𝐶 = 5𝑐𝑚 and ∠ABC = 49° . Find the area of triangle 𝐴𝐵𝐶.

SOLUTION: 5 6 = sin 𝐴 sin 49° sin 𝐴 =

5 sin 49° = 0.6289 6

A = sin

0.6289 = 38.97°

C = 180° − 49° − 38.97° = 92.03° Area of ABC; =

1 (5)(6) sin 92.03° = 14.99𝑐𝑚 2

[CHAPTER 3: TRIGONOMETRY]

LET’S PRACTICE 11

1. Given 𝐷𝐹 = 8𝑐𝑚, 𝐸𝐹 = 12𝑐𝑚 and ∠DFE = 126° . Find the area of triangle 𝐷𝐸𝐹.

Ans:

38.83cm 2

Ans:

64.18cm2

Ans:

20.98cm2

2. Given 𝐴𝐶 = 11𝑐𝑚, 𝐵𝐶 = 14.5𝑐𝑚 and ∠B = 48° . Find the area of triangle.

3. Given 𝐸𝐷 = 7𝑐𝑚, 𝐷𝐹 = 9𝑐𝑚 and 𝐸𝐹 = 6𝑐𝑚. Find the area of triangle 𝐷𝐸𝐹.

[CHAPTER 4: COMPLEX NUMBER]

COMPLEX NUMBER OBJECTIVES: At the end of this topic, students should be able to understand about: i. ii. iii. iv. v.

Identify real part and imaginary part Recognize that i   1 Perform the operations of complex number Represent complex number using Argand Diagram Complex number in other form

4.1

INTRODUCTION OF COMPLEX NUMBER

Introduction including:    

Complex Number in general form Definition of  1 Real part Imaginary part

REAL NUMBER IMAGINARY

the numbers like 1, 10.158, -0.34, 2/5,

3 , or any number

when squared give a negative result

NUMBER



The "unit" imaginary number (like 1 for Real Numbers) is i, which is the square root of −1



Because when we square i we get −1; i2 = −1



And we keep that little "i" there to remind us we need to multiply by

1

[CHAPTER 4: COMPLEX NUMBER]

4.1.1 THE CONCEPT OF A COMPLEX NUMBER



A Complex Number is a combination of a Real Number and an Imaginary Number.



General form of complex number written as:

a + bi Examples of a (real number):-

10.15

2 5

-0.3462



Examples of b (Imaginary Numbers):3i

1.04i

3i 4

−2.8i

( 2 )i

1998i

Examples of complex number:

1+i

39 + 3i

−2 +  i

0.8 − 2.2i

Complex Number

Real Part

3 + 2i

−6i

−6

2 +

1 i 2

Imaginary Part 2 Purely Real Purely Imaginary

A Complex Number consists of real part and imaginary part. But either part can be 0.

[CHAPTER 4: COMPLEX NUMBER]

EXAMPLE 1

Simplify: 1.

4

7i  i 2

3. 4.

2 3

SOLUTION: 1.

 4  4(1)  4i 2  2i

7i  i 2  7i 3  7i (1)  7i

i 9  i 2  i 2  i 2  i 2  i   1   1   1   1  i

 2   3  2  3  1  2  3i

LET’S PRACTICE 1

Simplify: 1.

2 i  3i

−√−64

4 2

√−98

Ans:  6 , 4  2i , −8𝑖, 9.9𝑖

[CHAPTER 4: COMPLEX NUMBER]

4.2

THE OPERATIONS OF COMPLEX NUMBERS

4.2.1 ADDITION To add two or more complex numbers, we add each part separately.



Complex Number 2

Imaginary part

a  bi   c  di   a  c   b  d i Complex Number 1

Real part

EXAMPLE 2 a.

Perform the addition of 3  2i and 1  7i SOLUTION: ( 3  2i ) + (1  7i )

REAL PART

3  1  4

IMAGINARY PART

2  7i  9i

ANSWER

4  9i

Perform the addition of 3  5i and 4  3i SOLUTION: (3  5i ) + (4  3i) REAL PART

3  4  7

IMAGINARY PART

5  3i  2i

ANSWER

7  2i

[CHAPTER 4: COMPLEX NUMBER]

LET’S PRACTICE 2 Solve the following operation of complex number. a. 5  2i   1  8i 

b. 3  6i    2  9i 

c. 8i  1   5  6i 

d.  7  5i   5  4i 

e. 2  3i   7  6i 

f.  24  10i   12  14i 

g..  30  52i    17  16i 

h. 72  14i   21  43i 

Ans: 6  10i

Ans: 1  15i

Ans:  4  2i

Ans:  2  9i

Ans: 9  3i

Ans:  12  24i

Ans:  47  68i

Ans: 93  29i 63

[CHAPTER 4: COMPLEX NUMBER]

4.2.2 SUBSTRACTION 

To subtract two or more complex numbers, we do same steps as for addition.



Subtract each part separately. Complex Number 2

Imaginary part

a  bi   c  di   a  c   b  d i Complex Number 1

Real part

EXAMPLE 3

Subtract 7  3i and 1 7i SOLUTION: ( 7  3i ) - ( 1 7i )

REAL PART

7  1  6

IMAGINARY PART

3  7i  4i

ANSWER

6  4i

Perform the subtraction (5 + 3𝑖) − (3 + 2𝑖) SOLUTION:

5  3i   3  2i  REAL PART

5  3  2

IMAGINARY PART

3  2i  i

ANSWER

2i

[CHAPTER 4: COMPLEX NUMBER]

LET’S PRACTICE 3 Solve the following operation of complex number. a.  7  9i   5  2i 

b. 8  2i   1  3i 

c. 8i  6   5  1i 

d. 2  6i    4  9i 

e. 30  4i    6  48i 

f.  70  6i   4  21i 

g.  32i  11  14  20i 

h. 91  27i    13  9i 

Ans: −12 + 7i

Ans: 7  i

Ans: 11  9i

Ans: 6  3i

Ans: 36 + 44i

Ans:  74  27i

Ans:  25  12i

Ans: 104  36i 65

[CHAPTER 4: COMPLEX NUMBER]

4.2.3 MULTIPLICATION

To multiply complex numbers, each part of the first complex number gets multiplied



by each part of the second complex number. Just use "FOIL", which stands for "Firsts, Outers, Inners, Lasts"



a  bi c  di   ac  adi  bci  bdi 2 (a  bi)(c  di)  (ac  bd )  (ad  bc)i

EXAMPLE 4

Solve the equation 3  2i 1  7i  SOLUTION:

3  2i 1  7i   (3  1)  (3  7i)  (2i  1)  (2i  7i)  (3)  (21i )  (2i )  (14i 2 )  (3)  ( 21i )  (2i )  (14)(1)

Remember:- i 2  1

 (3)  ( 21i )  ( 2i )  ( 14)

 11 23i b. Solve the equation 1  i 2 SOLUTION:

1  i 1  i   (1  1)  (1  i )  (1  i )  (i 2 )  (1)  ( 2i )  ( 1)

Remember:- i 2  1

= 2𝑖

[CHAPTER 4: COMPLEX NUMBER]

c. Solve the equation −9(4 − 2𝑖) SOLUTION: −9(4 − 2𝑖) = (−9 × 4) + [(−9) × (−2𝑖)] = −38 + 18𝑖 d. Solve the equation 3𝑖(2 + 6𝑖) SOLUTION: 3𝑖(2 + 6𝑖) = (3𝑖 × 2) + (3𝑖 × 6𝑖) = 6𝑖 + 18𝑖

Remember:- i 2  1

= 6𝑖 + 18(−1) = −18 + 6𝑖

LET’S PRACTICE 4 Solve the following operation of complex number. a. 8i  6  5  1i 

b. 8  2i 1  3i 

c.  7  9i 5  2i 

d. 2  6i  4  9i 

Ans:  22  46i

Ans: 2  26i

Ans:  53  30i

Ans:  44  6i

[CHAPTER 4: COMPLEX NUMBER]

4.2.4 DIVISION 

To perform the division of complex numbers, we use the conjugate.



For example, a  bi , the conjugate of the bottom number will be used. c  di

Bottom number is c  di , so the conjugate will be c  di . The sign for imaginary part



will be changed or the conjugate can be written as c  di  c  di . 

Then the division of



Hence,

a  bi a  bi c  di when using conjugate is written as  c  di c  di c  di

a  bi c  di a  bi   c  di  =  c  di c  di c  di   c  di 

ac  adi  bci  bdi  cc  cdi  cdi  ddi  2

; i 2  1

ac  bd   bc  ad i c2  d 2

EXAMPLE 5

1. Do the division

2  3i 4  5i

SOLUTION:

EXPLANATION:

2  3i 4  5i  4  5i 4  5i

a. Multiply top and bottom by the conjugate of



8  10i  12i  15i 16  20i  20i  25i 2

8  10i  12i  15  16  20i  20i  25



4  5i 2

 7  22i 41



b. Remember that i 2  1 c. Add like terms and notice how on the bottom 20i  20i cancels out. d. Lastly we should put the answer back into

a  bi form

7 22  i 41 41

[CHAPTER 4: COMPLEX NUMBER]

2. Write

4  3i 7 22 in the standard form 𝑎 + 𝑏𝑖.    i 2  5i 41 41

SOLUTION: 4 + 3𝑖 4 + 3𝑖 2 − 5𝑖 = × 2 + 5𝑖 2 + 5𝑖 2 − 5𝑖 =

8 − 20𝑖 + 6𝑖 − 15𝑖 4 − 10𝑖 + 10𝑖 − 25𝑖

8 − 14𝑖 − 15(−1) 4 − 25(−1)

8 + 15 − 14𝑖 4 + 25 23 − 14𝑖 = 29 23 14𝑖 = − 29 29 =

LET’S PRACTICE 5

Solve the following operation of complex number. a.

(

)

(

)

Ans:

1  3i  8  2i 

Ans: c.

5  2i 

2  26i 68

 7  9i  Ans:

 4  9i   2  6i  Ans:

 46  42i 40 69

[CHAPTER 4: COMPLEX NUMBER]

4.3

CONJUGATE OF A COMPLEX NUMBERS



A conjugate is where we change the sign in the middle. Refer the diagram below.



In other word, it is obtained by changing the sign of the imaginary part.



A conjugate is often written with a bar over it. Conjugate solving

Explanation change the sign of the imaginary part

5  3i = 5  3i Let z  3  4i ;



then the conjugate of z represented by z  3  4i .



If z  5  2i , then its conjugate z  5  2i .



The conjugate is used to solve the division of complex numbers.



It is important to ease the complex number with denominator.



The trick is to multiply both top and bottom by the conjugate of the bottom.

4.3.1 THE PROPERTIES OF THE COMPLEX CONJUGATE 

For any complex numbers 𝑧 , 𝑧 , 𝑧 , the algebraic properties of the conjugate operation:

𝑧 +𝑧 = 𝑧 + 𝑧 𝑧 𝑧 =𝑧 ∙𝑧

𝑧 −𝑧 =𝑧 − 𝑧 𝑧 𝑧

𝑧 , 𝑧

𝑧 ≠0

[CHAPTER 4: COMPLEX NUMBER]

EXAMPLE 6 Given that 𝑧 = 2 − i and 𝑧 = 3 + 3i. Write each expression in the standard form a + bi a. 𝑧 − 𝑧 b. 𝑧 + 𝑧 SOLUTION: a.

𝑧 = 3 − 3𝑖 Therefore; 𝑧 − 𝑧 = (2 − 𝑖) − (3 − 3𝑖) 𝑧 − 𝑧 = 2 − 𝑖 − 3 + 3𝑖 𝒛𝟏 − 𝒛𝟐 = −𝟏 + 𝟐𝒊

b. 𝑧 + 𝑧 = 2 − 𝑖 + 3 + 3𝑖 𝑧 + 𝑧 = 5 + 2𝑖 Therefore; 𝒛𝟏 + 𝒛𝟐 = 𝟓 − 𝟐𝒊

LET’S PRACTICE 6 Given that z = 8 + 3i and w = 3 − 4i. Write each expression in the standard form a + bi. a.

𝑧 + 𝑧̅

Ans: 16 b.

𝑤− 𝑤

Ans: −8𝑖 c.

𝑧̅ + 𝑤

Ans: 11 + 𝑖

[CHAPTER 4: COMPLEX NUMBER]

4.4

GRAPHICAL REPRESENTATION OF A COMPLEX NUMBER THROUGH ARGAND DIAGRAM

Complex number can be represented by argand diagram. If a complex number, z  a  bi ; a → real part b → imaginary part.

 

4.4.1 ARGAND’S DIAGRAM TO REPRESENT A COMPLEX NUMBER Based on the following diagram,



z  a  bi is the reflection image of z  a  bi Where;

a  bi  (a  bi)  2a and

(a  bi )  (a  bi )  2bi

EXAMPLE 7 1.

Show the following complex number on Argands Diagram. a. x  yi b. 3  5i

c. 4  7i

2  5i

[CHAPTER 4: COMPLEX NUMBER]

LET’S PRACTICE 7 Represent the following complex number on Argand Diagram

a. 4  9i

Ans:

b.  3  6i

Ans:

c.  7  2i

Ans:

d. 8  5i

Ans: 4.4.2 MODULUS AND ARGUMENT 

Refer to argand diagram below:

 Modulus of the complex number z, donated by

z where it is a positive value which is equal to the length of the segment OZ. |𝑧| =

𝑥 +𝑦

 Argument of z is angle  ;

tan  

y. x

Hence, the argument, donated by; 𝜃 = tan

𝑦 𝑥 73

[CHAPTER 4: COMPLEX NUMBER]

EXAMPLE 8 Find the modulus and the argument for the following complex numbers. Then show them on argand diagram. z  3  4i z  4  7i a. b. Modulus, z = 32  42

Modulus, z = 42   7 

= 9  16 = 25 =5

= 16  49 = 65 = 8.06

4 3 = 53.13º

Argument, θ = tan 1

7 4 = 60.25 

Argument, θ = tan 1

θ in Quadrant 4;

  360  60.25 = 299.75º Argand diagram:

Argand diagram:

θ θ

[CHAPTER 4: COMPLEX NUMBER]

LET’S PRACTICE 8 Find the modulus and the argument for the following complex numbers. 1.

z  8  4i

Ans: z  8.94 ,   333.43 2.

z  6  9i

Ans: z  10.8 ,   56.3 3.

z  10i  12

Ans: z  15 .62 ,   320.19 3.

z  10i  12

Ans: z  15 .62 ,   320.19 4.5

COMPLEX NUMBER IN OTHER FORM

Complex number can be expressed as shown in the following table. Complex Number

The form of

r  a  bi

Cartesian

r 

 can be in degree( ) r  cos   i sin  

 can be in degree( ) re i

 must be in radian (rad)

Polar

Trigonometric

Exponent

[CHAPTER 4: COMPLEX NUMBER]

4.5.1 COMPLEX NUMBER IN POLAR FORM EXAMPLE 9 Given a complex number z  4  9i . State z in polar form, z  SOLUTION: Modulus; z  4 2  9 2

z  9.85 Argument; 𝜃 = tan 9 𝜃 = tan 4 𝜃 = 66.04° In Polar Form; |𝒛|∠𝛉 = 𝟗. 𝟖𝟓∠𝟔𝟔. 𝟎𝟒°

LET’S PRACTICE 9 Express for the following complex numbers in polar form 1.

y  9  5i

Ans: 10.333.1 2.

p  12  7i

Ans: 13.8930.26 3.

r  9(cos 120  i sin 120)

Ans: 9120 4. x  6.4e1.12i Ans: 6.464.17 76

[CHAPTER 4: COMPLEX NUMBER]

4.5.2 COMPLEX NUMBER IN TRIGONOMETRIC FORM EXAMPLE 10 If z  530 , write z in trigonometric form { r  cos   i sin   } SOLUTION: z  530 → in polar form ( r  )



z  5 cos 30   i sin 30 



LET’S PRACTICE 10 Express for the following complex numbers in trigonometric form 1.

k  7  10i

Ans: 12.2(cos 305  i sin 305) 2.

p  9  12i

Ans: 15(cos 53.13  i sin 53.13) 3.

y  40270

Ans: 40(cos 270  i sin 270) 4. x  21.3e

2.13i

Ans:, 21.3(cos122  i sin 122)

[CHAPTER 4: COMPLEX NUMBER]

4.5.3 COMPLEX NUMBER IN EXPONENTIAL FORM EXAMPLE 11 Given z  6  4i . State z in the form of re i SOLUTION: Modulus; |𝑧| =

6 + (−4) = √52

|𝑧| = 7.21

Argument; 𝜃 = tan 𝜃 = tan

4 6

𝜃 = 33.69° Since 𝑧 is in Quadrant 4; 𝜃 = 360° − 33.69° = 326.31° 𝜋 𝜃 = 326.31° × = 5.695𝑟𝑎𝑑 180° In exponential form; 𝒛 = 𝒓𝒆𝜽𝒊 = 𝟕. 𝟐𝟏𝒆𝟓.𝟔𝟗𝟓𝒊

LET’S PRACTICE 11 Express for the following complex numbers in exponential form. 1.

a  6i  10

Ans: 11.7e 5.742i

[CHAPTER 4: COMPLEX NUMBER]

r  11 52i

Ans: 53.2e1.36i 3.

p  32275

Ans: 32e 4.8i 4. x  38(cos 86  i sin 86)

Ans: 38e1.5i 4.5.4 COMPLEX NUMBER IN CARTESIAN FORM

EXAMPLE 12 If z  245 , express z in Cartesian form. SOLUTION: Write z in trigonometric form:

z = 2  cos 45  i sin 45  In Cartesian form: 𝑧 = 2(0.7071 + 0.7071𝑖) 𝒛 = 𝟏. 𝟒𝟏𝟒𝟐 + 𝟏. 𝟒𝟏𝟒𝟐𝒊

[CHAPTER 4: COMPLEX NUMBER]

LET’S PRACTICE 12 Express for the following complex numbers in Cartesian form. a. z  42(cos 56  i sin 56)

Ans: 23.49  34.82i b. y  34e

1.48i

Ans: 30.83  33.86i y  28130

Ans:  18  21.45i 4.6

MULTIPLICATION AND DIVISION OF COMPLEX NUMBERS IN POLAR FORM Complex numbers in polar form are especially easy to multiply and divide. The rules



are: i.

Multiplication rule: To form the product, multiply the magnitudes and add the angles.

ii.

Division rule: To form the quotient, divide the magnitudes and subtract the angles.

EXAMPLE 13 1. Given z1  335 and z 2  547 . Calculate: a.

𝑧 ×𝑧 SOLUTION: 𝑧 × 𝑧 = (3∠35° )(5∠47° ) 𝑧 × 𝑧 = (3 × 5)∠(35° + 47° ) 𝑧 × 𝑧 = 15∠82°

[CHAPTER 4: COMPLEX NUMBER] b.

SOLUTION: =

∠

𝑧 5 = ∠(47° − 35° ) 𝑧 3 𝑧 = 1.67∠12° 𝑧

2. Given a  414 , b  6  71.5 , c  745 . Solve: a. a  b  c SOLUTION:

a  b  c  414  6  71.5  745 a  b  c  (4  6  7)(14  71.5  45)

a  b  c  168  12.5

a c b

SOLUTION:

a 414 c   745 b 6  71.5 a 4  c   714  (71.5)  45 b 6

a  c  4.67130.5 b

[CHAPTER 4: COMPLEX NUMBER]

bc a

SOLUTION:

b  c 6  75  745  a 414 bc 67   75  45  14 a 4

bc  10.5  44 a d.

c2 ab SOLUTION:

c2 (745) 2  a  b 414  6  71.5

c2 7 2 2  45  a  b 4  6(14  71.5) c2 49  (90  61) a  b 24

c2  2.04151 ab

LET’S PRACTICE 13 1. Solve the following expression in an exponential form. 25(cos 180  i sin 180)  8(cos 12  i sin 12) 20(cos 50  i sin 50)

Ans: 10e 2.48i

[CHAPTER 4: COMPLEX NUMBER]

2. Given that Z 1  8(cos 34  i sin 34) and Z 2  6040 . Solve

Z2 in trigonometry form Z1

Ans: 7.5(cos 6  i sin 6) 3. Given z1  3  2i and z 2  3(cos 30  i sin 30) a.

Calculate z1  z 2 and express the answer in the form of a  bi and exponential form.

Ans: 4.8  9.7i & 10.8e1.11i b. Calculate

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

Ans: 1.2  0.077i & 1.2e 0.064i 4.

Given z1  8145 , z 2  2460 a. Calculate z1  z 2 and express the answer in the form of a  bi

Ans:  174  81.14i b. State

2z 2 in trigonometric form. Illustrate the answer on Argand diagram. z1



Ans: 6 cos 85 0  i sin  85 0



[CHAPTER 5: MATRIX]

MATRIX OBJECTIVES: At the end of this topic, students should be able to understand about: i. ii. iii. iv. v.

matrix definition dimension or order of matrix types of matrix operation of matrix (addition, subtraction, multiplication and division) simultaneous equation using matrix (inverse matrix method and Cramer’s rule) 5.1

INTRODUCTION OF MATRIX



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 matrix 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

[CHAPTER 5: MATRIX]

MATRIX 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 MATRIX 

The number of ROWS in matrix 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 B   3 1  0 4

3  2 -

3 rows & 2 columns

[CHAPTER 5: MATRIX]

EXAMPLE 1

34  5 31  A  22 15 55  52 13  9 i.

What are the dimensions of a matrix above?

ii.

Identify entry A23

SOLUTION: i.

Dimension of a matrix; 3 3

ii.

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

7  12  45  2 V   3 15   4  13

21 31 11 14 27 19  36 71 26  55 34 15 

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 86

[CHAPTER 5: MATRIX]

TYPES OF MATRIX a.

RECTANGULAR MATRIX  A matrix in which number of rows is not

1 3 A 4  1

equal to number of columns. b.

ROW MATRIX

c. COLUMN MATRIX

 A matrix with single

 A matrix with

 A matrix with equal

single column and

number of rows and

of columns.

any number of

columns.

5

rows.



1   2 A   3    4

DIAGONAL MATRIX which all the

in which all the

elements except

diagonal elements

those on the leading

are equal.

diagonal are zero.

4 0 0 A  0 4 0 0 0 4

IDENTITY MATRIX /UNIT MATRIX 



 A diagonal matrix

 2 0 0 A  0 5 0 0 0 7 

When the diagonal elements

m  n  1 2 1 A  2 1 2 1 2 1

f. SCALAR MATRIX

 Is a square matrix in

SQUARE MATRIX

row and any number

A  1 2 3 4

2 3 2 1  3 2  2 3

ZERO MATRIX  A matrix in which every element is zero.

0 0 0 A  0 0 0 0 0 0

TRANPOSE MATRIX  A matrix which is formed by turning

are one and nondiagonal

all the rows of a given matrix into

elements are zero.

columns and vice versa.

A unit matrix is always a square matrix. 1 0 0  A  0 1 0 0 0 1

 The transpose of matrix A is written as AT 1 4  T 1 2 3   4 5 6   2 5     3 6  87

[CHAPTER 5: MATRIX]

5.2 

Addition



Subtraction



Multiplication -

By scalar

By matrices

OPERATION OF MATRIX

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  4 6  1  9  5  3      

3 47 4 1  5

80  8 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  4 6  1  9   3 15      

3  4  1 4 1 3

80  8 6   9  15

[CHAPTER 5: MATRIX] LET’S PRACTICE 2

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

 1  8 Ans:    1 0  2.

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

 3 2 Ans:    7 4 3.

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

 2  1 6 Ans:    6 2 9 4.

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

4  9 2 Ans:   4 6 3

[CHAPTER 5: MATRIX]

5.2.3 MULTIPLICATION MULTIPLICATION BY SCALAR 

To multiply a matrix by a single number is easy. EXAMPLE 4

These are the calculations:

 4 0  8 0  2     1  9 2  18    

2 4  8 2 1  2

20  0 2   9  18

MULTIPLYING A MATRIX BY ANOTHER MATRIX 

To multiply a matrix by another matrix, 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

1(7) + 2(9) + 3(11) = 58

[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 

4(7) + 5(9) + 6(11) = 139

Second rows, second columns:

4,5,6  8,10,12  154 

1(8) + 2(10) + 3(12) = 64

4(8) + 5(10) + 6(12) = 154

THE ANSWER:

7 8   1 2 3     58 64   9 10  4 5 6       11 12 139 154  

REMEMBER 

In arithmetic:

3  5  5  3 [Commutative Law] 

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

[CHAPTER 5: MATRIX]

EXAMPLE 6

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

1 2 2 0 A B     3 4 1 2  4 4 A B    10 8

2 0 1 2 B A     1 2 3 4 2 4  B A    7 10

The answer is DIFFERENT [PROVED]

LET’S PRACTICE 3 1.

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

2 4 0  Ans:   0 2  6  2.

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

 4 0 2 Ans:    2 6 4 3.

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

 7 14  Ans:   10 22 92

[CHAPTER 5: MATRIX]

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

  3 11 Ans:    19 13 5.

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

 1 0 Ans:    8 16 6.

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

5  7  8  Ans: 14  3 20    13  16 24 

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

 5 7 6 Ans:  3 19 10    3 10 2  93

[CHAPTER 5: MATRIX]

5.3

DETERMINANT

DETERMINANT OF A MATRIX 

Determinant of a matrix 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 matrix may be negative or positive.



Example:

means the determinant of the matrix A

CALCULATING THE DETERMINANT



First of all the matrices must be square. 

have the same number of rows as columns.

FOR A 2  2 MATRIX 

For a 2 2 matrix (2 rows and 2 columns)



Formula:

a b  A  c d 

The determinant is;

A  ad  bc

“BUTTERFLY” rule for matrix 2 2

[CHAPTER 5: MATRIX]

EXAMPLE 7

4 6 Given B    . Find the determinant of matrix B. 3 8 SOLUTION |B| = (4 × 8) − (6 × 3) = 32 − 18 |B| = 14

ii.

FOR A 3 3 MATRIX



For a 3 3 matrix (3 rows and 3 columns)



Formula a A   d  g

b e h

c f  i 

The determinant is;

A  aei  fh  bdi  fg   cdh  eg 

EXAMPLE 8

6 1 1  C  4  2 5  2 8 7  Given;

Find the determinant of matrix C. SOLUTION:

C  6 14  40  128  10  132  4 C  306

[CHAPTER 5: MATRIX]

LET’S PRACTICE 4 1.

Find the determinant of matrix A.

2 5  A  1  3 Ans:  11 2.

Find the determinant of matrix B.

3 2 B  1 4 Ans: 10 3.

Find the determinant of matrix M. 0  1 3  M   2  5 4   3 1 3 

Ans:  44 4.

Find the determinant of matrix N. 1 1 1  N  2 5 7  2 1  1

Ans:  4 5. Find the determinant of matrix Z.   7 5 4 Z   4 2 6  2  3 4

Ans:  266

[CHAPTER 5: MATRIX]

5.4

INVERSE MATRIX USING MINOR, COFACTOR AND ADJOIN



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 matrix 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

0 m11   1 0 m21   1 0 m31   0

 2  0   21  2 1  2  0  2  2 1  2  00  0  2

2 m12   0 3 m22   0 3 m32   2

 2  21   2 0   2 1  2  31  20   3 1  2   3 2   22   10  2

2 m13   0 3 m23   0

0  21  0  2 1 0  31  0  3 1

3 0 m33     30  0  0 2 0

2 2  2  M   2 3 3  0  10 0 97

[CHAPTER 5: MATRIX]

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 matrix. The elements of row will be elements of columns. 2 0 2  AdjA   2 3 10  2  3 0 

STEP 5: INVERSE MATRIX 

Formula;

A 1 

𝐴

1  AdjA A

2 1 2 = −2 3 10 2 −3

1 ⎡ 5 0 10 = ⎢⎢− 1 5 ⎢ 1 0 ⎣ 5

1 3

10 − 3 10

0⎤ 1⎥⎥ ⎥ 0⎦

LET’S PRACTICE 5 1. Find the inverse matrix 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 98

[CHAPTER 5: MATRIX]

2. Find the inverse matrix  2 1 0  M   1 3  1  3 0 1  Using minor, cofactor and adjoin.

3 4 1 Ans:  2 9  4

1 4 1 2 3 4

1 4 1  2 7 4 

3. Find the inverse matrix  3 1  6 B   2 0 4    1 2  3 Using minor, cofactor and adjoin.

 4  9 2 2   3 Ans:  1 0 2   5 2  1  2  4. Find the inverse matrix   3 7  5 F   4  1 12  9  1  2 Using minor, cofactor and adjoin.

  107  327  28 Ans:   327  38  327

 38 327 13 327 41 327

79  327  16   327   25  327  99

[CHAPTER 5: MATRIX]

5.5

SIMULTANEOUS LINEAR EQUATION USING MATRIX

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 matrix exist precisely when its determinant is non-zero.

5.5.1 SIMULTANEOUS EQUATION USING INVERSE MATRIX METHOD EXAMPLE 10

Given;

 x  3y  z  1

2x  5 y  3 3 x  y  2 z  2

Solve the simultaneous equation by using inverse matrix method. SOLUTION: STEP 1: Rewrite the system using matrix multiplication  1 3 1   x   1   2 5 0   y   3        3 1  2  z   2

STEP 2: Writing the coefficient matrix as A  x  1  A y    3   z   2

Where

 1 3 1  A   2 5 0   3 1  2

100

[CHAPTER 5: MATRIX]

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 m13  2 1  53  13 m11  5 2   0 1  10 m12  2 2  0 3  4 m21  3 2  11  7 m22   1 2   13  1 m 23   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 Minor    7  1  10   5  2  11

STEP 6: Cofactor of A

 10  4  13      10 4  13 Cofactor    7  1  10       7  1 10    5  2  11       5 2  11

101

[CHAPTER 5: MATRIX]

STEP 7: Adjoin of A  10 7  5  AdjA   4  1 2    13 10  11

STEP 8: Inverse matrix of A

A 1 

1 AdjA A

  10 7  5  1 A   4  1 2  9   13 10  11 1

STEP 9: Solve the equation  x  y     z   x  y     z   x  y     z 

 10 7  5   1  1 4  1 2   3  9   13 10  11  2   10  21  10  1 4  3  4  9  13  30  22  21  1  3 9   39 

STEP 10: Find the x, y and z   7  x  3   y    1     3  z   13     3 

102

[CHAPTER 5: MATRIX]

LET’S PRACTICE 6 Solve the simultaneous equation below by using inverse matrix method. x  y  z  3

2 x  3 y  4 z  23  3x  y  2 z  15

Ans: 2,1,4

x  2y  z  7

2 x  3 y  4 z  3 x yz 0

Ans:  1,3,2 

4 x  2 y  2 z  10

2 x  8 y  4 z  32 30 x  12 y  4 z  24

Ans:  2,6,3

3 x  2 y  z  24

2 x  2 y  2 z  12 x  5 y  2 z  31

Ans: 3,4,7 

5x  2 y  4x  0

2 x  3 y  5z  8 3 x  4 y  3z  11

Ans:  2,1,3 103

[CHAPTER 5: MATRIX]

5.5.2 SIMULTANEOUS EQUATION USING CRAMER’s RULE EXAMPLE 11

x  2 y  3 z  5

Given 3 x  y  3 z  4  3 x  4 y  7 z  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 1  3  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  Ax   4 1  3   7 4 7 

 1 5 3  Ay   3 4  3  3  7 7 

 1 2  5 Az   3 1 4    3 4  7 

STEP 4: Find the determinant of each matrices  5 2 3  Ax   4 1  3  517    34  247    3 7   344  1 7   95  14  69  40  7 4 7   1 5 3  Ay   3 4  3  147    3 7   537    3 3  33 7  4 3  7  60  27  40  3  7 7 

104

[CHAPTER 5: MATRIX]  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 y z

Ax A AY A Az A



 40  1 40



40 1 40



 80  2 40

LET’S PRACTICE 7

Solve the simultaneous equation below by using Cramer’s Rule  2 x  2 y  4 z  22

 3 x  5 y  2 z  35  6 x  3 y  3 z  21

Ans: 2,9,2  4 x  y  2 z  1

 2 x  y  6 z  3 4 x  6 y  3 z  7

Ans:  1,1,1  5 x  9 y  3 z  40

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

Ans:  2,3,1 105

[CHAPTER 5: MATRIX]

4 x  2 y  2 z  4

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

Ans: 2,4,2  4 x  7 y  3 z  37

 2 x  3 y  6 z  61 2 x  2 y  4 z  44

Ans:  5,5,6 

106

[CHAPTER 6: 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

understand area of parallelogram

6.1 

INTRODUCTION OF 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 .

107

[CHAPTER 6: VECTOR AND SCALAR]

VECTOR REPRESENTATION 

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

Figure 1 

In the Figure 1, the line OA represents a vector OA .



You can write:

 4 OA     2 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,

 4 OA     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 .

108

[CHAPTER 6: VECTOR AND SCALAR]

EXAMPLE 1 

By referring Figure 2;

Figure 2 The magnitude of a vector;

Angle;

 y x

OA  x 2  y 2

  tan 1  

OA  4 2  2 2

  tan 1  

OA  20units

2 4   26.57 0

LET’S PRACTICE 1 1. Find the magnitude of each of these vectors. a. 4 i  3 j

  9 c.    7 

b. 2i  2 j  k

 5     7  3    Ans: 5 , 3 , 11.40 & 9.11

109

[CHAPTER 6: VECTOR AND SCALAR]

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.

UNIT VECTOR 

A unit vector is a vector of length 1.  u unit vector, uˆ   u

EXAMPLE 2 1. Find the unit vector in the direction of v = 5i – 2j +4k. SOLUTION: v  52   2   42  45 2

magnitude of v,

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

vˆ  vˆ 

5i  2 j  4k 45 5 45

i

2 45

j

4 45

110

[CHAPTER 6: VECTOR AND SCALAR]

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  5 k .

Ans:

3i  2 j  5k 38

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

Ans:

 7i  9 j 130

  3   4. Find a unit vector in the direction of the vector  12    4  

Ans:

 3i  12 j  4k 13

111

[CHAPTER 6: VECTOR AND SCALAR]

POSITION VECTORS 

By referring to the Figure 3, the 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 fixed point.



It can be write as; OP  P



Then,

PQ  PO  OQ PQ   P  Q



PQ  Q  P

Figure 3

EXAMPLE 3

7    5     1. Given that a   3  and b   2  , find a  b and a  b . Hence, find unit vector   2   3  in the direction of a  b . SOLUTION: → → 𝑎+𝑏 =

7 2 −5 3 + 2 = 5 −2 1 3

→ → 𝑎+𝑏 =

Unit vector,

2 + 5 + 1 = √30     ab ab    ab   2i  5 j  k 2 5 1 ab   i j k 30 30 30 30

112

[CHAPTER 6: VECTOR AND SCALAR]

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 XY   4    3   7   5  1    6 XY 

 4 2  7 2   6 2

Unit vector, XY 



101

XY XY

 4i  7 j  6k 101 4 7 6 XY   i j k 101 101 101 XY 

LET’S PRACTICE 3

1   1. If given B  3 , find the unit vector of B . 2

Ans:

1 14

i

3 14

j

2 14

113

[CHAPTER 6: VECTOR AND SCALAR]

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

2A  B

c. Unit vector of 2 A  B

Ans: 5i  5 j  14 k ,

246 &

5 246

i

5 246

j

14 246

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

Ans: 

6.2

2 125

i

11 125

THE OPERATION OF VECTOR

6.2.1 VECTOR ADDITION

Figure 4

By referring the Figure 4 above; i)

OR  OP  PR OR  A  B  C 

ii)

OR  OQ  QR OR   A  B   C

From i) and ii)

OR  A  B  C    A  B   C 114

[CHAPTER 6: 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

115

[CHAPTER 6: 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 116

[CHAPTER 6: 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.

ii.

iii.

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

117

[CHAPTER 6: VECTOR AND SCALAR]

LET’S PRACTICE 4 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

CA BC Ans: BA

OB  DO Ans: DB

118

[CHAPTER 6: VECTOR AND SCALAR]

LET’S PRACTICE 5

O, A, B, C and D are five points where OA  a , OB  b , OC  a  2b and OD  2a  b . Express AB, BC, CD, AC and BD in terms of 𝑎 and 𝑏. Ans:

AB  b  a BC  a  b CD  a  3b AC  2b BD  2a  2b LET’S PRACTICE 6 𝑉𝑊𝑋𝑌 is a parallelogram with VW  a and WX  b . Express the vectors below in terms of 𝑎 and 𝑏. i. VX Ans: a  b ii. XV Ans:  b  a iii. WY

Ans: b  a iv. YW

Ans: a  b 119

[CHAPTER 6: VECTOR AND SCALAR]

6.3

APPLY SCALAR (DOT) PRODUCT OF TWO VECTORS

Definition of scalar product 

In Figure, 𝑂𝐴⃗ = 𝑎 and 𝑂𝐵⃗ = 𝑏.



The angle between 𝑎 and 𝑏 is defined as the angle between 𝑂𝐴⃗ and 𝑂𝐵⃗. The scalar product of a vector 𝑎 and 𝑏 is



represented by 𝑎 ∙ 𝑏, and 𝑎 ∙ 𝑏 = |𝑎||𝑏| cos 𝜃, with 𝜃 as the angle



between 𝑎 and 𝑏. 

𝜃 as the angle between 𝑎 and 𝑏, then 𝑎 ∙ 𝑏 = |𝑎||𝑏| cos 𝜃.



Therefore, cos 𝜃 = | cos 𝜃 =

Figure

∙ || |

𝑥 𝑥 +𝑦 𝑦 (𝑥 ) +(𝑦 )

(𝑥 ) +(𝑦 )

Properties of scalar product i.

   A B  B  A

ii.

For non zero vector A and B ;

iii. iv.

   B  A  0 ; If and only if A is perpendicular to B        A B  C  A B  A C       A  kB  kA  B  k A  B

If vectors A and B are given in term of their component with respect to the

      







standard vectors i, j and k as

  A  a1i  a 2 j  a3 k and B  b1i  b2 j  b3 k Then;

  A  B  a1b1  a 2 b2  a3 b3

120

[CHAPTER 6: VECTOR AND SCALAR]

EXAMPLE 6 1.

If 𝑅 = 2𝑖 + 𝑗 and 𝑆 = 𝑖 − 3𝑗. Find a. 𝑅 ∙ 𝑆 b. 𝑅 ∙ (𝑆 − 𝑅) c. (10𝑅 + 𝑆) ∙ 𝑆 d. (3𝑅 ∙ 𝑆) ∙ 𝑅 SOLUTION: a. 𝑹 ∙ 𝑺 = (2 × 1) + [1 × (−3)] = 2 − 3 𝑹 ∙ 𝑺 = −𝟏 b. 𝑅 ∙ (𝑆 − 𝑅) = (2𝑖 + 𝑗 ) ∙ [(𝑖 − 3𝑗) − (2𝑖 + 𝑗)] 𝑅 ∙ (𝑆 − 𝑅) = (2𝑖 + 𝑗 ) ∙ (−𝑖 − 4𝑗) 𝑅 ∙ (𝑆 − 𝑅) = [2 × (−1)] + [1 × (−4)] 𝑹 ∙ (𝑺 − 𝑹) = −𝟔 c. (10𝑅 + 𝑆) ∙ 𝑆 = [10(2𝑖 + 𝑗) + (𝑖 − 3𝑗)] ∙ (𝑖 − 3𝑗) (10𝑅 + 𝑆) ∙ 𝑆 = [(20𝑖 + 10𝑗) + (𝑖 − 3𝑗)] ∙ (𝑖 − 3𝑗) (10𝑅 + 𝑆) ∙ 𝑆 = (21𝑖 + 7𝑗) ∙ (𝑖 − 3𝑗) (10𝑅 + 𝑆) ∙ 𝑆 = (21 × 1) + [7 × (−3)] (𝟏𝟎𝑹 + 𝑺) ∙ 𝑺 = 𝟎

d. (3𝑅 ∙ 𝑆) ∙ 𝑅 = [3(2𝑖 + 𝑗)] ∙ (𝑖 − 3𝑗) ∙ (2𝑖 + 𝑗) (3𝑅 ∙ 𝑆) ∙ 𝑅 = [(6𝑖 + 3𝑗) ∙ (𝑖 − 3𝑗)] ∙ (2𝑖 + 𝑗) (3𝑅 ∙ 𝑆) ∙ 𝑅 = [(6 × 1) + 3(−3)] ∙ (2𝑖 + 𝑗) (3𝑅 ∙ 𝑆) ∙ 𝑅 = (−3) ∙ (2𝑖 + 𝑗) (𝟑𝑹 ∙ 𝑺) ∙ 𝑹 = −𝟔𝒊 − 𝟑𝒋

121

[CHAPTER 6: VECTOR AND SCALAR]

A, B and C are three points with 𝐴𝐵⃗ = 2𝑖 + 𝑗 and 𝐵𝐶⃗ = 𝑖 + 3𝑗. Find angle 𝐴𝐵𝐶. SOLUTION: 𝐵𝐴⃗ = −𝐴𝐵⃗ = −2𝑖 − 𝑗 cos 𝐴𝐵𝐶 = cos 𝐴𝐵𝐶 = cos 𝐴𝐵𝐶 =

𝐵𝐴⃗ ∙ 𝐵𝐶⃗ 𝐵𝐴⃗ 𝐵𝐶⃗ (−2𝑖 − 𝑗) ∙ (𝑖 + 3𝑗) (−2) +(−1)

(1) +(3)

−2 − 3

√5√10 −2 − 3 ∠ 𝐴𝐵𝐶 = cos √5√10 ∠ 𝑨𝑩𝑪 = 𝟏𝟑𝟓°

The position vectors of points A, B and C, relative to the origin O, are −4𝑎 0 0 , 𝑎𝑛𝑑 respectively, with a > 0. Calculate: 𝑎 −2𝑎 2𝑎 a. The scalar product of 𝐴𝐵⃗. 𝐴𝐶⃗ b. Angle 𝐵𝐴𝐶 SOLUTION: a. The scalar product of 𝐴𝐵⃗ . 𝐴𝐶⃗ 𝐴𝐵⃗ = 𝑂𝐵⃗ − 𝑂𝐴⃗ 𝐴𝐵⃗ =

0 −4𝑎 4𝑎 − = −2𝑎 𝑎 −3𝑎

𝐴𝐶⃗ = 𝑂𝐶⃗ − 𝑂𝐴⃗ 0 −4𝑎 4𝑎 𝐴𝐶⃗ = − = 2𝑎 𝑎 𝑎 𝐴𝐵⃗ . 𝐴𝐶⃗ = (4𝑎)(4𝑎) + (−3𝑎)(𝑎) 𝐴𝐵⃗ . 𝐴𝐶⃗ = 16a

122

[CHAPTER 6: VECTOR AND SCALAR]

b. Angle 𝐵𝐴𝐶 cos 𝐵𝐴𝐶 =

𝐴𝐵⃗ . 𝐴𝐶⃗ 𝐴𝐵⃗ 𝐴𝐶⃗ 16a

cos 𝐵𝐴𝐶 = cos 𝐵𝐴𝐶 = cos 𝐵𝐴𝐶 =

(4𝑎) +(−3𝑎)

(4𝑎) +(𝑎)

16a √25𝑎 √17𝑎 16a (5𝑎)(4.123𝑎)

∠ 𝑩𝑨𝑪 = cos

0.6306 = 𝟓𝟎. 𝟗𝟏°

LET’S PRACTICE 7



1. If P  2i  4 j  6k and U  2i  4 j  3k . Find a.

  P U

P  U P

   2. Given A4,4,8 , B5,1,8 and C 2,4,3 . Find: i.

  A B

ii.

  AC

iii.

  BC



 

Ans: 38 , 76i  152 j  228k

Ans: 80

Ans: 32

Ans: 10 123

[CHAPTER 6: VECTOR AND SCALAR]

6.4

APPLY VECTOR (CROSS) PRODUCT OF TWO VECTORS

Properties of vector product If A, B and C are vectors and d is a scalar, then

A  B  B  A

ii.

dA  B  d  A  B   A  dB 

iii.

A  B  C   A  B  A  C

iv.

A  B C  A  C  B  C

A  B  C    A  B   C

vi.

A  B  C    A  C B   A  B C

EXAMPLE 7



1. Find the vector product for vector A  2i  2 j  2k and B  2i  2 j  3k . Find

SOLUTION:

j k  i  A  B  2 2  2 2  2 3   2 3   2  2 i  2 3   2 2  j  2  2   2 2 k  2i  10 j  8k

2. Find the unit vector of u  v . Given u  2i  2 j  3k and v  i  3 j  k .

SOLUTION:

i j k  u  v  2 2  3  11i  5 j  4k 1 3 1 

124

[CHAPTER 6: VECTOR AND SCALAR]

112   52  4 2

uv 

Vector unit of u  v is; 

 162

11 162

i

5 162

j

4 162

LET PRACTICE 8

1. Given vector OA  2i  j  3k , OB  3i  2 j  4k and OC  i  3 j  2k . Determine a.





Ans: i  3 j  7k

b. OA OB  OC

OA OB OC

  2. If P  8i  5 j  4k and Q  2i  7 j  4k , find:   a. Q  P

  b. P  Q

Ans: 39

Ans:  55i  11 j  11k

Ans:  48i  24 j  66k

Ans: 48i  24 j  66k 3. If a  3i  2 j  5k and b  i  4 j  6k , find: a. a  b Ans:  8i  13 j  10k b. b  a

Ans: 8i  13 j  10 k 125

[CHAPTER 6: VECTOR AND SCALAR]

6.4.1 APPLICATION OF THE VECTOR (CROSS) PRODUCT

Area of parallelogram

 AB AC sin   AB  BC Area of triangle ABC 1  AB BC 2

EXAMPLE 8

   1. Find the area of the parallelogram with vertices A0,5 , B2,0 , C 8,1 and  D6,4 . SOLUTION:

AB  2,5  2i  5 j  0k

BC  6,1  6i  j  0k Area of parallelogram Area  AB  BC i

AB  BC  2  5 0 6 1 0 AB  BC  0i  0 j  28k AB  BC  02  02  282  28

2. Calculate the area of the parallelogram spanned by the vectors 3 4       P    3 and Q  9  1  2

126

[CHAPTER 6: VECTOR AND SCALAR]

SOLUTION:

P  3,3,1  3i  3 j  k

Q  4,9,2  4i  9 j  2k Area of parallelogram Area  P  Q

PQ  3 3 1 4 9 2 P  Q   6  9 i  6  4  j  27  12k P  Q  15i  2 j  39k PQ 

 152   22  39 2

 41.83

LET PRACTICE 9



1. Find the area of parallelogram with vertices A0,0  , B 2,1 , C 3,6 and D 1,5 .

Ans: 9 2. Find the area of parallelogram with U  i  j  3k and V  6 j  5k .

3. Find the area of triangle with vertices P1,1,0 , Q 2,0,1 and R0,2,3 .

Ans:

230

Ans: 4.899 127

[CHAPTER 6: VECTOR AND SCALAR] 4. Find the area of parallelogram with vertices P1,5,3 , Q0,0,0 and R3,5,1 .

Ans: 24.5 5. Find the area of parallelogram with vertices x2,0,3 , y1,4,5 , and z 7,2,9 .

Ans: 64.9

128

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Bird, J. (2010). Engineering Mathematics (6th edition).UK: Newnes.(ISBN : 978-0-08096562-8)

Croft A. and Davison R. (2019). Mathematics for Engineers (5th edition), London, UK: Pearson.

Larson, R. and Edwards, B. (2018). Calculus. (11th Edition). Boston, MA. Cencage Learning.

P.M. Cohn (2017). Algebraic Numbers and Algebraic Functions. Taylor & Francis Ltd.

Siti Aishah Sheikh Abdullah, Ch’ng Pei Eng, Teoh Sian Hoon, Munirah Hamat dan Noor’ Aina Abdul Razak (2006). First Engineering Mathematics, Second Edition. McGraw-Hill (Malaysia) Sdn Bhd.

Abd. Wahib Md Raji, Hamisan Rahmat, Ismail Kamis, Mohammad Mohd Talib, Mohd Nor Mohamad, Ong Chee Tiong, (1999) Matematik Asas (Jilid 2), Jabatan Matematik, Fakulti Sains UTM.

Ong Beng Sim, Lee Khaik Yong, Yong Zulina Zubairi, Maheran Nurudin, Che Noorlia Noor, Miskiah Dzakaria, (2012) Mathematics for Matriculation for Semester 1 (Fourth Edition), Oxford Fajar

https://www.mathportal.org/algebra/solving-system-of-linear-equations/inverse-matrixmethod.php.

10.

http://www.coolmath.com/crunchers/algebra-problems-systems-equations-cramers-rule3x3

11.

https://www.mathisfun.com/numbers/complex-numbers.html

12.

https://tutorial.math.lamar.edu/Classes/Alg/ComplexNumbers.aspx

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