Vectors in two dimensions (KCSE Mathematics Form 3)

TL;DR

Vectors are quantities with both magnitude (size) and direction, represented by arrows. You'll learn to add, subtract, and multiply vectors by scalars, and use them to describe positions and movements in a flat plane. Mastering these operations helps you solve geometry problems and understand relative positions.

1. The Mental Model

Think of vectors as instructions for movement: "go 5 steps east" or "move 3 units up and 2 units right." They tell you how far to go and in what direction, making them perfect for describing paths and positions.

2. The Core Material

What is a Vector?

A vector is a quantity that has both magnitude (size or length) and direction. Examples include displacement, velocity, and force.
A scalar is a quantity that only has magnitude. Examples include distance, speed, mass, and time.

You can represent a vector in a few ways:
1. Column Vector: $\begin{pmatrix} x \ y \end{pmatrix}$, where $x$ is the horizontal component and $y$ is the vertical component.
2. Position Vector: If a point P has coordinates $(x, y)$, its position vector from the origin O is $\vec{OP} = \begin{pmatrix} x \ y \end{pmatrix}$.
3. Directed Line Segment: An arrow from point A to point B, written as $\vec{AB}$. The length of the arrow is the magnitude, and the arrowhead shows the direction.

Magnitude of a Vector

The magnitude (or length) of a vector $\mathbf{v} = \begin{pmatrix} x \ y \end{pmatrix}$ is denoted by $|\mathbf{v}|$ or $||\mathbf{v}||$. You calculate it using Pythagoras' theorem:
$|\mathbf{v}| = \sqrt{x^2 + y^2}$.

Vector Addition

To add two vectors, you add their corresponding components.
If $\mathbf{a} = \begin{pmatrix} a_x \ a_y \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} b_x \ b_y \end{pmatrix}$, then $\mathbf{a} + \mathbf{b} = \begin{pmatrix} a_x + b_x \ a_y + b_y \end{pmatrix}$.
Geometrically, you place the tail of the second vector at the head of the first vector. The resultant vector goes from the tail of the first to the head of the second (Triangle Law).

Vector Subtraction

Subtracting a vector is the same as adding its negative. The negative of a vector $\mathbf{b}$ is $-\mathbf{b}$, which has the same magnitude but opposite direction.
So, $\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b})$.
If $\mathbf{a} = \begin{pmatrix} a_x \ a_y \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} b_x \ b_y \end{pmatrix}$, then $\mathbf{a} - \mathbf{b} = \begin{pmatrix} a_x - b_x \ a_y - b_y \end{pmatrix}$.

Scalar Multiplication

Multiplying a vector by a scalar (a number) changes its magnitude but not its direction (unless the scalar is negative, which reverses the direction).
If $\mathbf{v} = \begin{pmatrix} x \ y \end{pmatrix}$ and $k$ is a scalar, then $k\mathbf{v} = \begin{pmatrix} kx \ ky \end{pmatrix}$.

Parallel Vectors

Two vectors $\mathbf{a}$ and $\mathbf{b}$ are parallel if one is a scalar multiple of the other. That is, $\mathbf{a} = k\mathbf{b}$ for some scalar $k$.
If $k > 0$, they are in the same direction. If $k < 0$, they are in opposite directions.

Position Vectors and Relative Vectors

If O is the origin, and A and B are points with position vectors $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$ respectively, then the vector from A to B is:
$\vec{AB} = \vec{AO} + \vec{OB} = -\vec{OA} + \vec{OB} = \mathbf{b} - \mathbf{a}$.
This is a very important concept! Remember: "end minus start".

Collinear Points

Three points A, B, and C are collinear (lie on the same straight line) if the vector $\vec{AB}$ is parallel to $\vec{BC}$ (or $\vec{AC}$), and they share a common point (B in this case).
So, if $\vec{AB} = k \vec{BC}$ (or $\vec{AB} = k \vec{AC}$), then A, B, and C are collinear.

Vector Operations Flow

graph TD
    A[Start] --> B{Choose Operation};
    B --> C{Addition/Subtraction};
    C --> D[Add/Subtract Corresponding Components];
    D --> E[Resultant Vector];
    B --> F{Scalar Multiplication};
    F --> G[Multiply Each Component by Scalar];
    G --> E;
    B --> H{Magnitude};
    H --> I[Use Pythagoras: sqrt(x^2 + y^2)];
    I --> J[Magnitude Value];
    B --> K{Relative Vector (e.g., AB)};
    K --> L[End Position Vector - Start Position Vector];
    L --> E;
    E --> M[End];
    J --> M;

3. Worked Example

Given points A(2, 1), B(5, 7), and C(11, 19).
1. Find the position vectors $\vec{OA}$, $\vec{OB}$, and $\vec{OC}$.
2. Find the vectors $\vec{AB}$ and $\vec{BC}$.
3. Show that A, B, and C are collinear.
4. Find the magnitude of $\vec{AB}$.

Solution:

1. Position vectors:
   $\vec{OA} = \begin{pmatrix} 2 \ 1 \end{pmatrix}$
   $\vec{OB} = \begin{pmatrix} 5 \ 7 \end{pmatrix}$
   $\vec{OC} = \begin{pmatrix} 11 \ 19 \end{pmatrix}$

2. Vectors $\vec{AB}$ and $\vec{BC}$:
   $\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 5 \ 7 \end{pmatrix} - \begin{pmatrix} 2 \ 1 \end{pmatrix} = \begin{pmatrix} 5-2 \ 7-1 \end{pmatrix} = \begin{pmatrix} 3 \ 6 \end{pmatrix}$
   $\vec{BC} = \vec{OC} - \vec{OB} = \begin{pmatrix} 11 \ 19 \end{pmatrix} - \begin{pmatrix} 5 \ 7 \end{pmatrix} = \begin{pmatrix} 11-5 \ 19-7 \end{pmatrix} = \begin{pmatrix} 6 \ 12 \end{pmatrix}$

3. Show A, B, C are collinear:
   We need to check if $\vec{BC}$ is a scalar multiple of $\vec{AB}$.
   $\vec{BC} = \begin{pmatrix} 6 \ 12 \end{pmatrix}$
   $\vec{AB} = \begin{pmatrix} 3 \ 6 \end{pmatrix}$
   Notice that $\begin{pmatrix} 6 \ 12 \end{pmatrix} = 2 \begin{pmatrix} 3 \ 6 \end{pmatrix}$.
   So, $\vec{BC} = 2 \vec{AB}$.
   Since $\vec{BC}$ is a scalar multiple of $\vec{AB}$ (scalar $k=2$), and they share a common point B, points A, B, and C are collinear.

4. Magnitude of $\vec{AB}$:
   $|\vec{AB}| = \left| \begin{pmatrix} 3 \ 6 \end{pmatrix} \right| = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45}$
   $|\vec{AB}| = \sqrt{9 \times 5} = 3\sqrt{5}$ units.

4. Key Takeaways

• Vectors have both magnitude and direction, unlike scalars which only have magnitude.
• You can represent vectors as column vectors $\begin{pmatrix} x \ y \end{pmatrix}$ or directed line segments $\vec{AB}$.
• Vector addition and subtraction involve adding or subtracting corresponding components.
• Scalar multiplication scales the magnitude of a vector; a negative scalar reverses its direction.
• The magnitude of a vector $\begin{pmatrix} x \ y \end{pmatrix}$ is found using Pythagoras: $\sqrt{x^2 + y^2}$.
• A relative vector $\vec{AB}$ is calculated as $\vec{OB} - \vec{OA}$ ("end minus start").
• Points A, B, C are collinear if $\vec{AB} = k \vec{BC}$ (or $\vec{AC}$) and they share a common point.

Common Mistakes to Avoid:
- Confusing scalars with vectors; remember vectors have direction.
- Incorrectly applying Pythagoras' theorem for magnitude (e.g., forgetting to square or square root).
- Calculating $\vec{AB}$ as $\vec{OA} - \vec{OB}$ instead of $\vec{OB} - \vec{OA}$.
- Forgetting that for collinearity, vectors must be parallel AND share a common point.
- Adding or subtracting components incorrectly (e.g., adding $x$ to $y$).

5. Now Try It

Given points P(1, 2), Q(4, 6), and R(10, 14).
1. Calculate the vectors $\vec{PQ}$ and $\vec{QR}$.
2. Determine if P, Q, and R are collinear. If they are, state the scalar relationship between $\vec{PQ}$ and $\vec{QR}$.
3. Find the magnitude of $\vec{PR}$.

What success looks like: You should be able to correctly calculate the component vectors, show the scalar relationship for collinearity, and find the magnitude using the formula. Your final answer for collinearity should be a clear "Yes, they are collinear because..." with the scalar value, and the magnitude should be a simplified surd or decimal.