Slide 1

• Topic: Vectors - Some examples on vectors

Slide 2

• Introduction to vectors:
  • Definition of a vector
  • Representation of vectors
  • Types of vectors: zero vector, unit vector

Slide 3

• Vector addition:
  • Commutative property
  • Associative property
  • Geometric interpretation of vector addition

Slide 4

• Vector subtraction:
  • Definition of vector subtraction
  • Geometric interpretation of vector subtraction

Slide 5

• Scalar multiplication of vectors:
  • Definition of scalar multiplication
  • Properties of scalar multiplication

Slide 6

• Cross product of vectors:
  • Definition of cross product
  • Properties of cross product
  • Calculating the cross product of two vectors

Slide 7

• Dot product of vectors:
  • Definition of dot product
  • Properties of dot product
  • Calculating the dot product of two vectors

Slide 8

• Application of vectors in geometry:
  • Finding the area of a triangle using vector cross product
  • Finding the angle between two vectors using vector dot product

Slide 9

• Application of vectors in physics:
  • Resolving forces into components using vectors
  • Solving problems involving projectile motion using vectors

Slide 10

• Application of vectors in engineering:
  • Vector quantities in engineering calculations
  • Using vectors to analyze forces and moments in structures

Slide 11

• Example 1: Adding Vectors
  • Let’s say we have two vectors: A = (3, 4) and B = (-2, 5)
  • To add these vectors, we simply add their corresponding components: A + B = (3 + (-2), 4 + 5)
  • Simplifying: A + B = (1, 9)
• Example 2: Subtracting Vectors
  • Consider two vectors: P = (5, 2) and Q = (3, 7)
  • To subtract these vectors, we subtract their corresponding components: P - Q = (5 - 3, 2 - 7)
  • Simplifying: P - Q = (2, -5)
• Example 3: Scalar Multiplication
  • Let’s take a vector A = (2, -3) and a scalar k = 4
  • To multiply the vector by the scalar, we multiply each component by the scalar: kA = (4 * 2, 4 * (-3))
  • Simplifying: kA = (8, -12)
• Example 4: Cross Product
  • Consider two vectors A = (1, 2, -3) and B = (-2, 1, 4)
  • To find the cross product, we use the formula: A x B = [(2 * 4) - (1 * (-3)), (-3 * (-2)) - (1 * 4), (1 * 1) - (2 * (-2))]
  • Simplifying: A x B = (11, -14, 5)
• Example 5: Dot Product
  • Let’s consider two vectors A = (3, -2) and B = (-4, 5)
  • To find the dot product, we use the formula: A · B = (3 * (-4)) + (-2 * 5)
  • Simplifying: A · B = (-12) + (-10) = -22

Slide 12

• Properties of Vectors:
  • Zero Vector: A vector with all components equal to zero, denoted as 0 or O
  • Unit Vector: A vector with magnitude equal to 1
  • Commutative Property of Addition: A + B = B + A
  • Associative Property of Addition: (A + B) + C = A + (B + C)
  • Distributive Property: k(A + B) = kA + kB, where k is a scalar
• Geometric interpretation of vectors:
  • Vectors can be represented as directed line segments
  • The length of the line segment represents the magnitude of the vector
  • The direction of the line segment represents the direction of the vector

Slide 13

• Area of a Triangle:
  • Given two vectors A and B, the area of the triangle formed by these vectors can be found using the cross product
  • Area = 1/2 |A x B|
• Angle between two vectors:
  • The angle between two vectors A and B can be found using the dot product
  • Angle θ = cos^(-1)(A · B / |A| |B|)
• Example:
  • Suppose we have two vectors A = (2, 3) and B = (4, -1)
  • To find the area of the triangle formed by A and B, we calculate the cross product: |A x B| = |(2 * (-1)) - (3 * 4)|
  • The area of the triangle is 10 square units
  • To find the angle between A and B, we calculate the dot product: A · B = (2 * 4) + (3 * (-1))
  • Using the formula, we find θ = cos^(-1)((2 * 4) + (3 * (-1))) / √((2^2 + 3^2)(4^2 + (-1)^2))

Slide 14

• Resolving Forces:
  • In physics, vectors are used to represent forces
  • Vectors can be resolved into components along different axes (x, y, z)
  • This makes it easier to analyze forces in different directions
• Projectile Motion:
  • Vectors are used to analyze the motion of projectiles, such as objects thrown or launched into the air
  • By breaking down the motion into horizontal and vertical components, various properties (range, time of flight, maximum height) can be calculated
• Example:
  • Consider a ball being thrown at an angle of 30 degrees above the horizontal with a velocity of 20 m/s
  • By resolving the velocity into horizontal and vertical components, we can analyze its motion
  • The horizontal component remains constant while the vertical component is affected by gravity
  • This allows us to determine the range, time of flight, and maximum height of the ball

Slide 15

• Vectors in Engineering:
  • Vectors play a crucial role in engineering applications, ranging from structural analysis to fluid dynamics
  • They are used to represent and analyze various quantities such as forces, velocities, accelerations, and displacements
• Vector Quantities:
  • Various physical quantities such as force, velocity, and acceleration have both magnitude and direction
  • Vectors provide a concise and accurate representation of these quantities, enabling engineers to analyze and solve problems effectively
• Forces and Moments:
  • Vectors are used to analyze forces and moments in structures
  • By breaking them down into components, engineers can determine the equilibrium conditions and design structures accordingly

Slide 16

• Example:
  • Let’s consider a simple truss structure with two forces acting on it: F1 = 10N at an angle of 30 degrees above the horizontal and F2 = 5N at an angle of 60 degrees below the horizontal
  • To analyze the forces, we can resolve them into horizontal and vertical components
  • F1’s horizontal component = 10N * cos(30 degrees), F1’s vertical component = 10N * sin(30 degrees)
  • F2’s horizontal component = 5N * cos(120 degrees), F2’s vertical component = 5N * sin(120 degrees)
  • By summing the horizontal and vertical components separately, we can determine the equilibrium conditions of the truss structure
• Fluid Dynamics:
  • Vectors are used to represent and analyze fluid flow, including velocity, pressure, and direction of flow
  • This helps engineers design efficient systems for fluid transportation and control

Slide 17

• Summary:
  • Vectors are a fundamental concept in mathematics and find widespread applications in various fields including physics, engineering, and geometry
  • They are represented by directed line segments and have both magnitude and direction
  • Addition, subtraction, scalar multiplication, dot product, and cross product are some operations that can be performed on vectors
  • Geometric interpretation of vectors helps in visualizing vector operations and solving real-world problems
• Key Points:
  • Vectors have magnitude and direction
  • Addition, subtraction, scalar multiplication, dot product, and cross product can be performed on vectors
  • Vectors are used in geometry, physics, and engineering for various applications
• Questions?

Slide 18

• Exercise:
  1. Add the following vectors: a = (2, 5), b = (1, -3), and c = (4, 0)
  2. Subtract the vector d = (6, -2) from the vector e = (-3, 7)
  3. Multiply the vector f = (3, -4) by the scalar k = 2
  4. Calculate the cross product of the vectors g = (1, 2, -3) and h = (-2, 1, 4)
  5. Find the dot product of the vectors i = (3, -2) and j = (-4, 5)
• Solution:
  1. a + b + c = (2 + 1 + 4, 5 + (-3) + 0)
  2. e - d = (-3 - 6, 7 - (-2))
  3. k * f = (2 * 3, 2 * (-4))
  4. g x h = [(2 * 4) - (1 * (-3)), (-3 * (-2)) - (1 * 4), (1 * 1) - (2 * (-2))]
  5. i · j = (3 * (-4)) + (-2 * 5)

Slide 19

• Review:
  • Vectors have both magnitude and direction
  • Addition, subtraction, scalar multiplication, dot product, and cross product can be performed on vectors
  • Vectors find applications in various fields including geometry, physics, and engineering
• Homework:
  • Solve the exercise problems given in the previous slide
  • Research and find real-world examples where vectors are used
  • Prepare for the next class on matrices
• Thank you and see you in the next class!

Slide 21

• Example 1:
  • Add the following vectors:
    • A = (4, -2)
    • B = (3, 6)
    • C = (-1, -3)
  • Solution:
    • A + B + C = (4 + 3 + (-1), -2 + 6 + (-3))
    • A + B + C = (6, 1)
• Example 2:
  • Subtract the vector D = (6, 4) from the vector E = (10, 8)
  • Solution:
    • E - D = (10 - 6, 8 - 4)
    • E - D = (4, 4)
• Example 3:
  • Multiply the vector F = (5, -3) by the scalar k = 3
  • Solution:
    • kF = (3 * 5, 3 * (-3))
    • kF = (15, -9)
• Example 4:
  • Calculate the cross product of the vectors G = (1, 2, 3) and H = (-2, 0, 4)
  • Solution:
    • G x H = [(2 * 4) - (3 * 0), (3 * (-2)) - (1 * 4), (1 * 0) - (2 * (-2))]
    • G x H = [8, -10, 4]
• Example 5:
  • Find the dot product of the vectors I = (4, 5) and J = (-2, 3)
  • Solution:
    • I · J = (4 * (-2)) + (5 * 3)
    • I · J = (-8) + 15
    • I · J = 7

Slide 22

• Properties of Vectors:
  • Zero Vector: A vector with all components equal to zero, denoted as 0 or O
  • Unit Vector: A vector with magnitude equal to 1
  • Commutative Property of Addition: A + B = B + A
  • Associative Property of Addition: (A + B) + C = A + (B + C)
  • Distributive Property: k(A + B) = kA + kB, where k is a scalar
• Geometric interpretation of vectors:
  • Vectors can be represented as directed line segments
  • The length of the line segment represents the magnitude of the vector
  • The direction of the line segment represents the direction of the vector
• Example:
  • Suppose A = (3, -4) and B = (2, 1)
  • Zero Vector: 0 = (0, 0)
  • Unit Vector: A/|A| = (3/5, -4/5)
  • Commutative Property: A + B = B + A
  • Associative Property: (A + B) + C = A + (B + C)
  • Distributive Property: 2(A + B) = 2A + 2B

Slide 23

• Area of a Triangle:
  • Given two vectors A and B, the area of the triangle formed by these vectors can be found using the cross product
  • Area = 1/2 |A x B|
• Example:
  • Consider two vectors A = (2, 3) and B = (4, -1)
  • To find the area of the triangle formed by A and B, we calculate the cross product: |A x B| = |(2 * (-1)) - (3 * 4)|
  • The area of the triangle is 10 square units
• Angle between two vectors:
  • The angle between two vectors A and B can be found using the dot product
  • Angle θ = cos^(-1)(A · B / |A| |B|)
• Example:
  • Let A = (3, -2) and B = (-4, 5)
  • To find the angle between A and B, we calculate the dot product: A · B = (3 * (-4)) + (-2 * 5)
  • Using the formula, we find θ = cos^(-1)((3 * (-4)) + (-2 * 5)) / √((3^2 + (-2)^2)(-4^2 + 5^2))

Slide 24

• Resolving Forces:
  • In physics, vectors are used to represent forces
  • Vectors can be resolved into components along different axes (x, y, z)
  • This makes it easier to analyze forces in different directions
• Projectile Motion:
  • Vectors are used to analyze the motion of projectiles, such as objects thrown or launched into the air
  • By breaking down the motion into horizontal and vertical components, various properties (range, time of flight, maximum height) can be calculated
• Example:
  • Consider a ball being thrown at an angle of 30 degrees above the horizontal with a velocity of 20 m/s
  • By resolving the velocity into horizontal and vertical components, we can analyze its motion
  • The horizontal component remains constant while the vertical component is affected by gravity
  • This allows us to determine the range, time of flight, and maximum height of the ball

Slide 25

• Vectors in Engineering:
  • Vectors play a crucial role in engineering applications, ranging from structural analysis to fluid dynamics
  • They are used to represent and analyze various quantities such as forces, velocities, accelerations, and displacements
• Vector Quantities:
  • Various physical quantities such as force, velocity, and acceleration have both magnitude and direction
  • Vectors provide a concise and accurate representation of these quantities, enabling engineers to analyze and solve problems effectively
• Forces and Moments:
  • Vectors are used to analyze forces and moments in structures
  • By breaking them down into components, engineers can determine the equilibrium conditions and design structures accordingly

Slide 26

• Example:
  • Let’s consider a simple truss structure with two forces acting on it: F1 = 10N at an angle of 30 degrees above the horizontal and F2 = 5N at an angle of 60 degrees below the horizontal
  • To analyze the forces, we can resolve them into horizontal and vertical components
  • F1’s horizontal component = 10N * cos(30 degrees), F1’s vertical component = 10N * sin(30 degrees)
  • F2’s horizontal component = 5N * cos(120 degrees), F2’s vertical component = 5N * sin(120 degrees)
  • By summing the horizontal and vertical components separately, we can determine the equilibrium conditions of the truss structure
• Fluid Dynamics:
  • Vectors are used to represent and analyze fluid flow, including velocity, pressure, and direction of flow
  • This helps engineers design efficient systems for fluid transportation and control

Slide 27

• Summary:
  • Vectors are a fundamental concept in mathematics and find widespread applications in various fields including physics, engineering, and geometry
  • They are represented by directed line segments and have both magnitude and direction
  • Addition, subtraction, scalar multiplication, dot product, and cross product are some operations that can be performed on vectors
  • Geometric interpretation of vectors helps in visualizing vector operations and solving real-world problems
• Key Points:
  • Vectors have magnitude and direction
  • Addition, subtraction, scalar multiplication, dot product, and cross product can be performed on vectors
  • Vectors are used in geometry, physics, and engineering for various applications
• Questions?

Slide 28

• Exercise:
  1. Add the following vectors: a = (2,