6- Vectors in 2-dimensions

# 6- Vectors in 2-dimensions ## Definitions, representation of vectors, magnitude of a vector, direction of a vector and operations with vectors ### Introduction Welcome to our lesson on vectors in two dimensions! In mathematics and physics, quantities can be classified into two main types: scalars and vectors. While scalar quantities only have a magnitude (a numerical value), vector quantities have both a magnitude and a direction. Understanding vectors is crucial for describing movements, forces, and velocities in a plane. In this lesson, we will define what vectors are, learn how to represent them, calculate their magnitude and direction, and perform basic operations like addition, subtraction, and scalar multiplication. ### Key Concepts #### What is a Vector? * **Scalar Quantity:** A quantity that has only magnitude. Examples include mass, temperature, time, and speed. * **Vector Quantity:** A quantity that has both magnitude and direction. Examples include displacement, velocity, acceleration, and force. #### Representation of Vectors Vectors can be represented in two main ways: * **Geometrical Representation:** A vector is represented by a directed line segment, which is essentially an arrow. The length of the arrow represents the **magnitude** of the vector, and the arrowhead indicates its **direction**. * The starting point of the arrow is called the **initial point** or **tail**. * The ending point with the arrowhead is called the **terminal point** or **head**. * Notation: If a vector starts at point A and ends at point B, it is denoted as \( \vec{AB} \). Alternatively, a single bold lowercase letter can be used, e.g., \( \mathbf{a} \) or \( \underline{a} \). * **Component Form (Column Vector):** In a Cartesian coordinate system, a vector can be represented by its components along the x-axis and y-axis. If a vector starts at the origin \( O(0,0) \) and ends at a point \( P(x,y) \), it can be written as: \[ \mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix} \] Here, \( x \) is the horizontal component and \( y \) is the vertical component. If a vector starts at \( A(x_1, y_1) \) and ends at \( B(x_2, y_2) \), its component form is: \[ \vec{AB} = \begin{pmatrix} x_2-x_1 \\ y_2-y_1 \end{pmatrix} \] #### Magnitude of a Vector The magnitude (or length) of a vector is denoted by \( |\mathbf{v}| \) or \( ||\mathbf{v}|| \). It represents the size of the vector quantity. * For a vector \( \mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix} \), its magnitude is calculated using the Pythagorean theorem: \[ |\mathbf{v}| = \sqrt{x^2 + y^2} \] * For a vector \( \vec{AB} \) with initial point \( A(x_1, y_1) \) and terminal point \( B(x_2, y_2) \), its magnitude is: \[ |\vec{AB}| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \] #### Direction of a Vector The direction of a vector in 2-dimensions is usually given by the angle \( \theta \) it makes with the positive x-axis, measured anticlockwise. * For a vector \( \mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix} \), the angle \( \theta \) can be found using trigonometry: \[ \tan\theta = \frac{y}{x} \] It is important to determine the correct quadrant for \( \theta \) based on the signs of \( x \) and \( y \). * **Unit Vector:** A unit vector is a vector with a magnitude of 1. If \( \mathbf{v} \) is a vector, its unit vector in the same direction, denoted as \( \hat{\mathbf{v}} \), is given by: \[ \hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|} \] #### Operations with Vectors ##### 1. Addition of Vectors * **Geometrical (Triangle Rule):** To add \( \mathbf{a} \) and \( \mathbf{b} \), place the tail of \( \mathbf{b} \) at the head of \( \mathbf{a} \). The resultant vector \( \mathbf{a} + \mathbf{b} \) is drawn from the tail of \( \mathbf{a} \) to the head of \( \mathbf{b} \). * **Geometrical (Parallelogram Rule):** If vectors \( \mathbf{a} \) and \( \mathbf{b} \) start from the same point, their sum \( \mathbf{a} + \mathbf{b} \) is the diagonal of the parallelogram formed by \( \mathbf{a} \) and \( \mathbf{b} \) starting from that same point. * **Component Form:** If \( \mathbf{a} = \begin{pmatrix} a_x \\ a_y \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} b_x \\ b_y \end{pmatrix} \), then: \[ \mathbf{a} + \mathbf{b} = \begin{pmatrix} a_x + b_x \\ a_y + b_y \end{pmatrix} \] ##### 2. Subtraction of Vectors Subtracting a vector \( \mathbf{b} \) from \( \mathbf{a} \) is equivalent to adding the negative of \( \mathbf{b} \), i.e., \( \mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b}) \). The vector \( -\mathbf{b} \) has the same magnitude as \( \mathbf{b} \) but points in the opposite direction. * **Component Form:** If \( \mathbf{a} = \begin{pmatrix} a_x \\ a_y \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} b_x \\ b_y \end{pmatrix} \), then: \[ \mathbf{a} - \mathbf{b} = \begin{pmatrix} a_x - b_x \\ a_y - b_y \end{pmatrix} \] ##### 3. Scalar Multiplication Multiplying a vector by a scalar \( k \) changes its magnitude but not its direction (unless \( k \) is negative, in which case the direction is reversed). * If \( k > 0 \), \( k\mathbf{a} \) has magnitude \( k|\mathbf{a}| \) and the same direction as \( \mathbf{a} \). * If \( k < 0 \), \( k\mathbf{a} \) has magnitude \( |k||\mathbf{a}| \) and the opposite direction to \( \mathbf{a} \). * **Component Form:** If \( \mathbf{a} = \begin{pmatrix} a_x \\ a_y \end{pmatrix} \) and \( k \) is a scalar, then: \[ k\mathbf{a} = k\begin{pmatrix} a_x \\ a_y \end{pmatrix} = \begin{pmatrix} ka_x \\ ka_y \end{pmatrix} \] ##### 4. Parallel Vectors Two non-zero vectors \( \mathbf{a} \) and \( \mathbf{b} \) are parallel if one is a scalar multiple of the other. That is, if \( \mathbf{a} = k\mathbf{b} \) for some non-zero scalar \( k \). ##### 5. Position Vectors A position vector is a vector that starts from the origin \( O(0,0) \) and points to a specific point \( P(x,y) \). It is denoted as \( \vec{OP} \) or simply \( \mathbf{p} \). If \( P \) has coordinates \( (x,y) \), then \( \vec{OP} = \begin{pmatrix} x \\ y \end{pmatrix} \). * If you have two points \( A \) and \( B \) with position vectors \( \vec{OA} \) and \( \vec{OB} \) respectively, then the vector \( \vec{AB} \) can be found using: \[ \vec{AB} = \vec{OB} - \vec{OA} \] This is sometimes called the **triangle law for position vectors**. ### Vocabulary * **Vector:** A quantity with both magnitude and direction. * **Scalar:** A quantity with only magnitude. * **Magnitude:** The length or size of a vector. * **Direction:** The orientation of a vector, usually specified by an angle. * **Component Form:** Representing a vector by its horizontal (x) and vertical (y) parts, e.g., \( \begin{pmatrix} x \\ y \end{pmatrix} \). * **Column Vector:** Another name for the component form of a vector. * **Position Vector:** A vector that starts from the origin \( (0,0) \) and points to a specific point. * **Initial Point (Tail):** The starting point of a vector's arrow. * **Terminal Point (Head):** The ending point of a vector's arrow. * **Unit Vector:** A vector with a magnitude of 1. * **Parallel Vectors:** Vectors that point in the same or opposite direction; one is a scalar multiple of the other. ### Examples **Example 1: Representation and Magnitude** Given points \( P(1, 2) \) and \( Q(5, 5) \). (a) Write the vector \( \vec{PQ} \) in component form. (b) Find the magnitude of \( \vec{PQ} \). *Solution:* (a) \( \vec{PQ} = \begin{pmatrix} 5-1 \\ 5-2 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix} \) (b) \( |\vec{PQ}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \) units. **Example 2: Direction of a Vector** Find the direction of the vector \( \mathbf{v} = \begin{pmatrix} -3 \\ 4 \end{pmatrix} \). *Solution:* Let \( \theta \) be the angle with the positive x-axis. \( \tan\alpha = \left|\frac{4}{-3}\right| = \frac{4}{3} \) \( \alpha = \arctan\left(\frac{4}{3}\right) \approx 53.13^\circ \) Since the x-component is negative and the y-component is positive, the vector is in the second quadrant. Therefore, \( \theta = 180^\circ - \alpha = 180^\circ - 53.13^\circ = 126.87^\circ \). **Example 3: Vector Addition and Scalar Multiplication** Given \( \mathbf{a} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} -1 \\ 3 \end{pmatrix} \). (a) Find \( \mathbf{a} + \mathbf{b} \). (b) Find \( 3\mathbf{a} - 2\mathbf{b} \). *Solution:* (a) \( \mathbf{a} + \mathbf{b} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 + (-1) \\ 1 + 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix} \) (b) \( 3\mathbf{a} - 2\mathbf{b} = 3\begin{pmatrix} 2 \\ 1 \end{pmatrix} - 2\begin{pmatrix} -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \end{pmatrix} - \begin{pmatrix} -2 \\ 6 \end{pmatrix} = \begin{pmatrix} 6 - (-2) \\ 3 - 6 \end{pmatrix} = \begin{pmatrix} 8 \\ -3 \end{pmatrix} \) **Example 4: Position Vectors** Given the position vectors \( \vec{OA} = \begin{pmatrix} 4 \\ -2 \end{pmatrix} \) and \( \vec{OB} = \begin{pmatrix} 1 \\ 5 \end{pmatrix} \). Find the vector \( \vec{AB} \). *Solution:* \( \vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 1 \\ 5 \end{pmatrix} - \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 - 4 \\ 5 - (-2) \end{pmatrix} = \begin{pmatrix} -3 \\ 7 \end{pmatrix} \) ### Exercises 1. Given points \( R(-2, 1) \) and \( S(3, 13) \): (a) Express \( \vec{RS} \) as a column vector. (b) Calculate the magnitude of \( \vec{RS} \). (c) Find the unit vector in the direction of \( \vec{RS} \). 2. A vector \( \mathbf{u} = \begin{pmatrix} 6 \\ -8 \end{pmatrix} \). Calculate its magnitude and direction. 3. Given \( \mathbf{p} = \begin{pmatrix} 5 \\ -3 \end{pmatrix} \) and \( \mathbf{q} = \begin{pmatrix} -2 \\ 4 \end{pmatrix} \). Find: (a) \( \mathbf{p} + \mathbf{q} \) (b) \( \mathbf{p} - \mathbf{q} \) (c) \( 2\mathbf{p} + 3\mathbf{q} \) 4. The position vectors of points A and B are \( \mathbf{a} = \begin{pmatrix} 7 \\ 2 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} -1 \\ 6 \end{pmatrix} \) respectively. Find the vector \( \vec{BA} \). 5. Determine if the vectors \( \mathbf{v} = \begin{pmatrix} -4 \\ 6 \end{pmatrix} \) and \( \mathbf{w} = \begin{pmatrix} 6 \\ -9 \end{pmatrix} \) are parallel. ### Assignment 1. Points P, Q, and R have coordinates \( (1, -1) \), \( (4, 5) \) and \( (-2, 3) \) respectively. (a) Find the column vectors \( \vec{PQ} \) and \( \vec{QR} \). (b) Show that P, Q, and R are not collinear (do not lie on the same straight line). (c) Calculate the magnitude of \( \vec{PR} \). 2. A vector \( \mathbf{f} \) has a magnitude of 10 units and makes an angle of \( 210^\circ \) with the positive x-axis. Express \( \mathbf{f} \) in column vector form. 3. Given \( \mathbf{u} = \begin{pmatrix} m \\ 5 \end{pmatrix} \) and \( \mathbf{v} = \begin{pmatrix} -6 \\ 10 \end{pmatrix} \). If \( \mathbf{u} \) and \( \mathbf{v} \) are parallel, find the value of \( m \). 4. If \( \mathbf{x} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} \) and \( \mathbf{y} = \begin{pmatrix} 1 \\ 4 \end{pmatrix} \), find a scalar \( k \) such that \( |\mathbf{x} + k\mathbf{y}| = 5 \). ### Summary In this lesson, we explored the fundamentals of vectors in two dimensions. We learned that vectors are quantities with both magnitude and direction, distinguishing them from scalars which only have magnitude. We covered: * **Representation:** Vectors can be represented geometrically as directed line segments or algebraically as column vectors \( \begin{pmatrix} x \\ y \end{pmatrix} \). * **Magnitude:** The length of a vector is calculated using the distance formula, \( |\mathbf{v}| = \sqrt{x^2 + y^2} \). * **Direction:** The angle a vector makes with the positive x-axis is found using \( \tan\theta = \frac{y}{x} \), paying attention to the quadrant. * **Operations:** We performed vector addition and subtraction by adding or subtracting corresponding components. Scalar multiplication changes the magnitude of a vector and can reverse its direction. * **Position Vectors:** Vectors from the origin to a point, useful for finding vectors between two points: \( \vec{AB} = \vec{OB} - \vec{OA} \). These concepts form the building blocks for further study in vector geometry and their applications in physics and engineering. ## Angle between two vectors, Perpendicular and parallel vectors ### Introduction Welcome to this lesson on vectors in two dimensions! In our previous lessons, we learned about vector addition, subtraction, and scalar multiplication. Today, we will explore more advanced concepts: how to find the angle between two vectors, and how to determine if vectors are perpendicular or parallel to each other. These concepts are crucial for understanding forces, velocities, and many other physical quantities in engineering and physics. ### Key Concepts #### The Dot Product (Scalar Product) Before we can find the angle between two vectors, we need to understand the **dot product**, also known as the scalar product. The dot product of two vectors results in a scalar (a single number), not another vector. Let \( \mathbf{a} \) and \( \mathbf{b} \) be two vectors. The dot product is defined in two ways: 1. **Geometric Definition:** The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by: \[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \] where: * \( |\mathbf{a}| \) is the magnitude (length) of vector \( \mathbf{a} \). * \( |\mathbf{b}| \) is the magnitude (length) of vector \( \mathbf{b} \). * \( \theta \) is the angle between the two vectors \( \mathbf{a} \) and \( \mathbf{b} \) (\( 0^\circ \le \theta \le 180^\circ \)). 2. **Component Definition (in 2-dimensions):** If \( \mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} \), then their dot product is: \[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 \] #### Angle Between Two Vectors By combining the two definitions of the dot product, we can derive a formula to find the angle \( \theta \) between two non-zero vectors \( \mathbf{a} \) and \( \mathbf{b} \): From \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \), we can rearrange to solve for \( \cos \theta \): \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \] To find the angle \( \theta \), you would then use the inverse cosine function: \[ \theta = \cos^{-1}\left( \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \right) \] **Steps to find the angle between two vectors:** 1. Calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \) using the component definition. 2. Calculate the magnitude of vector \( \mathbf{a} \), which is \( |\mathbf{a}| = \sqrt{a_1^2 + a_2^2} \). 3. Calculate the magnitude of vector \( \mathbf{b} \), which is \( |\mathbf{b}| = \sqrt{b_1^2 + b_2^2} \). 4. Substitute these values into the formula for \( \cos \theta \). 5. Calculate \( \theta \) using the \( \cos^{-1} \) function. #### Perpendicular Vectors (Orthogonal Vectors) Two non-zero vectors \( \mathbf{a} \) and \( \mathbf{b} \) are **perpendicular** (or **orthogonal**) if the angle between them is \( 90^\circ \) (or \( \frac{\pi}{2} \) radians). If \( \theta = 90^\circ \), then \( \cos 90^\circ = 0 \). Using the dot product formula, \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \): If \( \mathbf{a} \) is perpendicular to \( \mathbf{b} \), then: \[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos 90^\circ = |\mathbf{a}| |\mathbf{b}| (0) = 0 \] Therefore, the condition for two vectors \( \mathbf{a} \) and \( \mathbf{b} \) to be perpendicular is that their dot product is zero: \[ \mathbf{a} \cdot \mathbf{b} = 0 \] In component form, if \( \mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} \), then: \[ a_1 b_1 + a_2 b_2 = 0 \] #### Parallel Vectors Two non-zero vectors \( \mathbf{a} \) and \( \mathbf{b} \) are **parallel** if they are in the same or opposite direction. This means the angle between them is \( 0^\circ \) or \( 180^\circ \). If \( \theta = 0^\circ \), then \( \cos 0^\circ = 1 \). If \( \theta = 180^\circ \), then \( \cos 180^\circ = -1 \). Mathematically, two vectors \( \mathbf{a} \) and \( \mathbf{b} \) are parallel if one is a scalar multiple of the other. That is, there exists a scalar \( k \) (where \( k \neq 0 \)) such that: \[ \mathbf{a} = k \mathbf{b} \] In component form, if \( \mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} \): \[ \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = k \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} \implies \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} k b_1 \\ k b_2 \end{pmatrix} \] This means that \( a_1 = k b_1 \) and \( a_2 = k b_2 \). Consequently, the ratio of their corresponding components must be equal: \[ \frac{a_1}{b_1} = \frac{a_2}{b_2} = k \] **Note:** If \( k > 0 \), the vectors point in the same direction. If \( k < 0 \), they point in opposite directions. ### Vocabulary * **Dot Product (Scalar Product):** An operation that takes two vectors and returns a scalar quantity. Its value depends on the magnitudes of the vectors and the angle between them. * **Magnitude:** The length or size of a vector, denoted as \( |\mathbf{v}| \). * **Perpendicular Vectors (Orthogonal Vectors):** Two vectors that meet at a right angle (\( 90^\circ \)). Their dot product is zero. * **Parallel Vectors:** Two vectors that point in the same or exactly opposite directions. One vector is a scalar multiple of the other. * **Scalar:** A quantity that has magnitude only (e.g., mass, temperature). In this context, it's a real number that scales a vector. * **Vector:** A quantity that has both magnitude and direction (e.g., force, velocity). ### Examples #### Example 1: Finding the Angle Between Two Vectors Given vectors \( \mathbf{p} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} \) and \( \mathbf{q} = \begin{pmatrix} -1 \\ 2 \end{pmatrix} \). Find the angle \( \theta \) between \( \mathbf{p} \) and \( \mathbf{q} \). **Solution:** 1. **Calculate the dot product:** \( \mathbf{p} \cdot \mathbf{q} = (3)(-1) + (4)(2) = -3 + 8 = 5 \) 2. **Calculate magnitudes:** \( |\mathbf{p}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \) \( |\mathbf{q}| = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \) 3. **Apply the angle formula:** \[ \cos \theta = \frac{\mathbf{p} \cdot \mathbf{q}}{|\mathbf{p}| |\mathbf{q}|} = \frac{5}{5 \sqrt{5}} = \frac{1}{\sqrt{5}} \] 4. **Find \( \theta \):** \( \theta = \cos^{-1}\left( \frac{1}{\sqrt{5}} \right) \approx \cos^{-1}(0.4472) \approx 63.43^\circ \) #### Example 2: Determining if Vectors are Perpendicular Are vectors \( \mathbf{a} = \begin{pmatrix} 2 \\ -3 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} 6 \\ 4 \end{pmatrix} \) perpendicular? **Solution:** To check for perpendicularity, we calculate their dot product. If \( \mathbf{a} \cdot \mathbf{b} = 0 \), they are perpendicular. \( \mathbf{a} \cdot \mathbf{b} = (2)(6) + (-3)(4) = 12 - 12 = 0 \) Since the dot product is 0, vectors \( \mathbf{a} \) and \( \mathbf{b} \) are perpendicular. #### Example 3: Determining if Vectors are Parallel Are vectors \( \mathbf{u} = \begin{pmatrix} -4 \\ 6 \end{pmatrix} \) and \( \mathbf{v} = \begin{pmatrix} 2 \\ -3 \end{pmatrix} \) parallel? **Solution:** To check for parallelism, we see if one is a scalar multiple of the other, i.e., \( \mathbf{u} = k \mathbf{v} \). Let's compare the components: \( -4 = k(2) \implies k = -2 \) \( 6 = k(-3) \implies k = -2 \) Since we found a consistent scalar \( k = -2 \) such that \( \mathbf{u} = -2 \mathbf{v} \), the vectors \( \mathbf{u} \) and \( \mathbf{v} \) are parallel. They point in opposite directions because \( k \) is negative. #### Example 4: Finding an Unknown Component for Perpendicular Vectors Given vectors \( \mathbf{r} = \begin{pmatrix} x \\ 5 \end{pmatrix} \) and \( \mathbf{s} = \begin{pmatrix} -2 \\ 4 \end{pmatrix} \). If \( \mathbf{r} \) and \( \mathbf{s} \) are perpendicular, find the value of \( x \). **Solution:** For perpendicular vectors, their dot product must be zero: \( \mathbf{r} \cdot \mathbf{s} = 0 \). \( (x)(-2) + (5)(4) = 0 \) \( -2x + 20 = 0 \) \( -2x = -20 \) \( x = 10 \) ### Exercises 1. Find the angle between the following pairs of vectors, giving your answer to one decimal place: a. \( \mathbf{a} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \) b. \( \mathbf{p} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} \) and \( \mathbf{q} = \begin{pmatrix} 4 \\ 5 \end{pmatrix} \) c. \( \mathbf{u} = \begin{pmatrix} -5 \\ 12 \end{pmatrix} \) and \( \mathbf{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \) 2. Determine if the following pairs of vectors are perpendicular, parallel, or neither: a. \( \mathbf{c} = \begin{pmatrix} 7 \\ 1 \end{pmatrix} \) and \( \mathbf{d} = \begin{pmatrix} -2 \\ 14 \end{pmatrix} \) b. \( \mathbf{e} = \begin{pmatrix} -3 \\ 9 \end{pmatrix} \) and \( \mathbf{f} = \begin{pmatrix} 1 \\ -3 \end{pmatrix} \) c. \( \mathbf{g} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} \) and \( \mathbf{h} = \begin{pmatrix} 4 \\ -5 \end{pmatrix} \) 3. Find the value of \( y \) such that the vectors \( \mathbf{m} = \begin{pmatrix} -3 \\ y \end{pmatrix} \) and \( \mathbf{n} = \begin{pmatrix} 6 \\ 2 \end{pmatrix} \) are perpendicular. 4. Given that vectors \( \mathbf{r} = \begin{pmatrix} 2 \\ x \end{pmatrix} \) and \( \mathbf{s} = \begin{pmatrix} -4 \\ 6 \end{pmatrix} \) are parallel, find the value of \( x \). ### Assignment 1. Given the points A(1, 2), B(4, 6), and C(7, 2). Find: a. The vector \( \vec{AB} \). b. The vector \( \vec{BC} \). c. The angle \( \angle ABC \) (the angle between \( \vec{BA} \) and \( \vec{BC} \)). 2. A vector \( \mathbf{u} = \begin{pmatrix} p \\ q \end{pmatrix} \) is perpendicular to \( \mathbf{v} = \begin{pmatrix} 3 \\ -5 \end{pmatrix} \). Find a possible pair of values for \( p \) and \( q \). 3. Show that the points P(-2, 1), Q(1, 7), and R(4, 13) are collinear (lie on the same straight line) by using vectors. 4. If \( \mathbf{a} = \begin{pmatrix} k \\ 2 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} 3 \\ k \end{pmatrix} \), and \( \mathbf{a} \) is perpendicular to \( \mathbf{b} \), find the value(s) of \( k \). ### Summary In this lesson, we explored key relationships between vectors in 2-dimensions: * **Dot Product:** A scalar quantity defined as \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \) or \( a_1 b_1 + a_2 b_2 \). * **Angle Between Vectors:** Calculated using the formula \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \). * **Perpendicular Vectors:** Two non-zero vectors \( \mathbf{a} \) and \( \mathbf{b} \) are perpendicular if and only if their dot product is zero (\( \mathbf{a} \cdot \mathbf{b} = 0 \)). * **Parallel Vectors:** Two non-zero vectors \( \mathbf{a} \) and \( \mathbf{b} \) are parallel if and only if one is a scalar multiple of the other (\( \mathbf{a} = k \mathbf{b} \) for some scalar \( k \neq 0 \)). This implies their corresponding components are proportional. Understanding these concepts is fundamental for further studies in mathematics and various scientific fields. Practice the examples and exercises to master these skills! ## Vector Geometry, the Midpoint Theorem and Proportional division of a vector ### Introduction Welcome to this lesson on Vector Geometry! In Form Four, we build on our understanding of vectors to solve more complex geometric problems. Today, we will explore how vectors can be used to determine the midpoint of a line segment and how a point can divide a line segment in a specific ratio, both internally and externally. These concepts are fundamental in coordinate geometry and have wide applications in physics and engineering. ### Key Concepts #### Review of Position and Displacement Vectors Before diving into new concepts, let's quickly recap what we know about vectors: * **Position Vector:** A vector that starts from the origin (O) to a point A is called the position vector of A, denoted as \( \vec{OA} \) or \( \mathbf{a} \). * **Displacement Vector:** The vector representing the displacement from point A to point B is denoted as \( \vec{AB} \). It can be found using the position vectors of A and B: \[ \vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a} \] #### The Midpoint Theorem (Vector Form) The midpoint of a line segment is the point that divides the segment into two equal parts. In vector form, we can find the position vector of the midpoint M of a line segment AB if we know the position vectors of A and B. * Let A have position vector \( \mathbf{a} \) and B have position vector \( \mathbf{b} \). * If M is the midpoint of AB, then \( \vec{AM} = \vec{MB} \) and \( \vec{AM} = \frac{1}{2}\vec{AB} \). * Using displacement vectors: \( \vec{OM} = \vec{OA} + \vec{AM} \) \( \mathbf{m} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) \) \( \mathbf{m} = \mathbf{a} + \frac{1}{2}\mathbf{b} - \frac{1}{2}\mathbf{a} \) \( \mathbf{m} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} \) * Therefore, the position vector of the midpoint M is: \[ \mathbf{m} = \frac{1}{2}(\mathbf{a} + \mathbf{b}) \] #### Proportional Division of a Vector Sometimes, a point divides a line segment in a specific ratio that is not necessarily 1:1 (midpoint). This division can be internal or external. ##### Internal Division of a Vector A point P divides a line segment AB internally in the ratio \( m:n \) if P lies between A and B, and \( AP : PB = m : n \). * Let A have position vector \( \mathbf{a} \) and B have position vector \( \mathbf{b} \). * Since P divides AB in the ratio \( m:n \), we have \( n\vec{AP} = m\vec{PB} \). * Expressing in terms of position vectors: \( n(\mathbf{p} - \mathbf{a}) = m(\mathbf{b} - \mathbf{p}) \) \( n\mathbf{p} - n\mathbf{a} = m\mathbf{b} - m\mathbf{p} \) \( n\mathbf{p} + m\mathbf{p} = m\mathbf{b} + n\mathbf{a} \) \( (m+n)\mathbf{p} = n\mathbf{a} + m\mathbf{b} \) * Therefore, the position vector of P is: \[ \mathbf{p} = \frac{n\mathbf{a} + m\mathbf{b}}{m+n} \] ##### External Division of a Vector A point P divides a line segment AB externally in the ratio \( m:n \) if P lies on the line containing AB but outside the segment AB, such that \( AP : PB = m : n \). In this case, one of the ratios (AP or PB) will be considered negative in terms of direction relative to the segment. * If P divides AB externally in the ratio \( m:n \), it means that \( \vec{AP} : \vec{PB} = m : (-n) \) (or similar interpretation depending on which side P lies). * Using the internal division formula as a base, if we consider the ratio to be \( m:(-n) \) where the division is external and P is further from B than A (i.e., beyond B), then the formula becomes: \( (-n)(\mathbf{p} - \mathbf{a}) = m(\mathbf{b} - \mathbf{p}) \) (from \( (-n)\vec{AP} = m\vec{PB} \) is not right) * A more consistent derivation for \( P \) dividing \( AB \) externally in ratio \( m:n \) (i.e., \( \vec{AP} = \frac{m}{n} \vec{BP} \) no not quite). * The formula for external division, where P divides AB externally in the ratio \( m:n \), is given by: \[ \mathbf{p} = \frac{m\mathbf{b} - n\mathbf{a}}{m-n} \] (Note: This formula assumes that P is on the side of B if \( m>n \), or on the side of A if \( n>m \). Also, \( m \neq n \)). ### Vocabulary * **Position Vector:** A vector from the origin to a point. * **Displacement Vector:** A vector representing the shift from one point to another. * **Midpoint:** The point that divides a line segment into two equal parts. * **Internal Division:** A point divides a line segment when it lies between the two endpoints. * **External Division:** A point divides a line segment when it lies on the line containing the segment but outside the segment itself. * **Ratio (\( m:n \)):** The proportional relationship in which a point divides a line segment. ### Examples #### Example 1: Midpoint Given points A with position vector \( \mathbf{a} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} \) and B with position vector \( \mathbf{b} = \begin{pmatrix} 8 \\ -1 \end{pmatrix} \). Find the position vector of the midpoint M of AB. **Solution:** Using the midpoint formula: \( \mathbf{m} = \frac{1}{2}(\mathbf{a} + \mathbf{b}) \) \( \mathbf{m} = \frac{1}{2}\left( \begin{pmatrix} 2 \\ 5 \end{pmatrix} + \begin{pmatrix} 8 \\ -1 \end{pmatrix} \right) \) \( \mathbf{m} = \frac{1}{2}\begin{pmatrix} 2+8 \\ 5-1 \end{pmatrix} \) \( \mathbf{m} = \frac{1}{2}\begin{pmatrix} 10 \\ 4 \end{pmatrix} \) \( \mathbf{m} = \begin{pmatrix} 5 \\ 2 \end{pmatrix} \) The position vector of the midpoint M is \( \begin{pmatrix} 5 \\ 2 \end{pmatrix} \). #### Example 2: Internal Division Points P and Q have position vectors \( \mathbf{p} = \begin{pmatrix} 1 \\ 4 \end{pmatrix} \) and \( \mathbf{q} = \begin{pmatrix} 7 \\ -2 \end{pmatrix} \) respectively. Find the position vector of point R that divides PQ internally in the ratio 2:1. **Solution:** Here, \( \mathbf{a} = \mathbf{p} \), \( \mathbf{b} = \mathbf{q} \), and the ratio \( m:n = 2:1 \) (so \( m=2 \) and \( n=1 \)). Using the internal division formula: \( \mathbf{r} = \frac{n\mathbf{p} + m\mathbf{q}}{m+n} \) \( \mathbf{r} = \frac{1\begin{pmatrix} 1 \\ 4 \end{pmatrix} + 2\begin{pmatrix} 7 \\ -2 \end{pmatrix}}{2+1} \) \( \mathbf{r} = \frac{1}{3}\left( \begin{pmatrix} 1 \\ 4 \end{pmatrix} + \begin{pmatrix} 14 \\ -4 \end{pmatrix} \right) \) \( \mathbf{r} = \frac{1}{3}\begin{pmatrix} 1+14 \\ 4-4 \end{pmatrix} \) \( \mathbf{r} = \frac{1}{3}\begin{pmatrix} 15 \\ 0 \end{pmatrix} \) \( \mathbf{r} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} \) The position vector of point R is \( \begin{pmatrix} 5 \\ 0 \end{pmatrix} \). #### Example 3: External Division Points C and D have position vectors \( \mathbf{c} = \begin{pmatrix} -3 \\ 1 \end{pmatrix} \) and \( \mathbf{d} = \begin{pmatrix} 5 \\ 7 \end{pmatrix} \) respectively. Find the position vector of point E that divides CD externally in the ratio 3:1. **Solution:** Here, \( \mathbf{a} = \mathbf{c} \), \( \mathbf{b} = \mathbf{d} \), and the ratio \( m:n = 3:1 \) (so \( m=3 \) and \( n=1 \)). Using the external division formula: \( \mathbf{e} = \frac{m\mathbf{d} - n\mathbf{c}}{m-n} \) \( \mathbf{e} = \frac{3\begin{pmatrix} 5 \\ 7 \end{pmatrix} - 1\begin{pmatrix} -3 \\ 1 \end{pmatrix}}{3-1} \) \( \mathbf{e} = \frac{1}{2}\left( \begin{pmatrix} 15 \\ 21 \end{pmatrix} - \begin{pmatrix} -3 \\ 1 \end{pmatrix} \right) \) \( \mathbf{e} = \frac{1}{2}\begin{pmatrix} 15 - (-3) \\ 21 - 1 \end{pmatrix} \) \( \mathbf{e} = \frac{1}{2}\begin{pmatrix} 18 \\ 20 \end{pmatrix} \) \( \mathbf{e} = \begin{pmatrix} 9 \\ 10 \end{pmatrix} \) The position vector of point E is \( \begin{pmatrix} 9 \\ 10 \end{pmatrix} \). ### Exercises 1. Points P and Q have coordinates P(-1, 6) and Q(7, 2). Find the position vector of the midpoint M of PQ. 2. Points A and B have position vectors \( \mathbf{a} = \begin{pmatrix} 4 \\ 0 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} -2 \\ 6 \end{pmatrix} \). Find the position vector of point C that divides AB internally in the ratio 1:2. 3. If X has position vector \( \mathbf{x} = \begin{pmatrix} 2 \\ -5 \end{pmatrix} \) and Y has position vector \( \mathbf{y} = \begin{pmatrix} 10 \\ 3 \end{pmatrix} \). Find the position vector of Z that divides XY externally in the ratio 3:2. ### Assignment 1. The position vectors of points P and Q are \( \mathbf{p} = \begin{pmatrix} -2 \\ 3 \end{pmatrix} \) and \( \mathbf{q} = \begin{pmatrix} 6 \\ -1 \end{pmatrix} \). a) Find the position vector of M, the midpoint of PQ. b) Find the position vector of R, which divides PQ internally in the ratio 3:1. c) Find the position vector of S, which divides PQ externally in the ratio 2:3. 2. Point A(3, -4) and B(-5, 8). Point C divides AB such that \( \vec{AC} = \frac{3}{4}\vec{AB} \). Find the coordinates of C. 3. Given points M(1, 2) and N(7, 10). Find the position vector of a point P on the line MN such that P divides MN in the ratio \( \vec{MP} : \vec{PN} = 5 : -2 \). ### Summary In this lesson, we have explored fundamental concepts in vector geometry related to dividing line segments. Key takeaways include: * The **midpoint** M of a line segment AB with position vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by \( \mathbf{m} = \frac{1}{2}(\mathbf{a} + \mathbf{b}) \). * A point P **internally dividing** a line segment AB in the ratio \( m:n \) has position vector \( \mathbf{p} = \frac{n\mathbf{a} + m\mathbf{b}}{m+n} \). * A point P **externally dividing** a line segment AB in the ratio \( m:n \) has position vector \( \mathbf{p} = \frac{m\mathbf{b} - n\mathbf{a}}{m-n} \). Understanding these formulas allows us to precisely locate points on a line segment using vector operations, which is crucial for solving various geometric problems.