Vectors

What you'll learn

Vectors are fundamental to understanding quantities that have both magnitude and direction, unlike scalars which have magnitude only. In CSEC Additional Mathematics, you'll work with vectors in two dimensions, performing operations and solving problems involving displacement, position, and geometric applications. This topic connects directly to coordinate geometry and forms essential groundwork for Further Mathematics and applied sciences.

Key terms and definitions

Vector — a quantity with both magnitude (size) and direction, represented by a directed line segment or in component form.

Scalar — a quantity with magnitude only, such as mass, temperature, or distance.

Magnitude — the size or length of a vector, denoted |a| or |AB|, calculated using Pythagoras' theorem in component form.

Position vector — a vector that describes the position of a point relative to the origin O, typically denoted r or $\overrightarrow{OA}$.

Unit vector — a vector with magnitude 1, used to indicate direction; i and j are unit vectors in the x and y directions respectively.

Resultant vector — the single vector obtained by adding two or more vectors, representing the combined effect.

Parallel vectors — vectors with the same or opposite direction, where one is a scalar multiple of the other.

Collinear points — three or more points that lie on the same straight line, verified using vectors.

Core concepts

Vector notation and representation

Vectors can be represented in multiple ways, and you must be comfortable with all forms:

Column vector notation: $$\mathbf{a} = \begin{pmatrix} x \ y \end{pmatrix}$$

where x is the horizontal component and y is the vertical component.

Component form using unit vectors: $$\mathbf{a} = x\mathbf{i} + y\mathbf{j}$$

Directed line segment: Written as $\overrightarrow{AB}$ or AB, representing displacement from point A to point B.

Single bold letter: Written as a or v in print, or $\underline{a}$ or $\vec{a}$ in handwriting.

Key relationships to remember:

• $\overrightarrow{AB} = -\overrightarrow{BA}$ (reversing direction changes sign)
• If A has coordinates (x₁, y₁) and B has coordinates (x₂, y₂), then $\overrightarrow{AB} = \begin{pmatrix} x_2 - x_1 \ y_2 - y_1 \end{pmatrix}$

Magnitude and direction of vectors

The magnitude of a vector $\mathbf{a} = \begin{pmatrix} x \ y \end{pmatrix}$ is:

$$|\mathbf{a}| = \sqrt{x^2 + y^2}$$

For example, if a cargo ship travels from Port of Spain to Bridgetown with displacement vector $\begin{pmatrix} 450 \ 200 \end{pmatrix}$ km, the actual distance travelled (if going in a straight line) is:

$$\sqrt{450^2 + 200^2} = \sqrt{202500 + 40000} = \sqrt{242500} \approx 492.4 \text{ km}$$

The direction of a vector is found using trigonometry:

$$\theta = \tan^{-1}\left(\frac{y}{x}\right)$$

where θ is the angle measured from the positive x-axis. Be careful with quadrants:

• Quadrant I: angle is positive as calculated
• Quadrant II: add 180° or use 180° - angle (depending on reference)
• Quadrant III: add 180° to the calculated angle
• Quadrant IV: add 360° or report as negative angle

A unit vector in the direction of a is: $\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}$

Vector operations

Addition and subtraction:

Vectors are added or subtracted component-wise:

$$\begin{pmatrix} x_1 \ y_1 \end{pmatrix} + \begin{pmatrix} x_2 \ y_2 \end{pmatrix} = \begin{pmatrix} x_1 + x_2 \ y_1 + y_2 \end{pmatrix}$$

$$\begin{pmatrix} x_1 \ y_1 \end{pmatrix} - \begin{pmatrix} x_2 \ y_2 \end{pmatrix} = \begin{pmatrix} x_1 - x_2 \ y_1 - y_2 \end{pmatrix}$$

Geometrically, use the triangle law (tip-to-tail method) or parallelogram law.

Scalar multiplication:

Multiplying a vector by a scalar k:

$$k\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} kx \ ky \end{pmatrix}$$

Properties:

• If k > 1, the vector is stretched
• If 0 < k < 1, the vector is compressed
• If k < 0, the vector is reversed and scaled by |k|
• If k = 0, the result is the zero vector

Parallel vectors:

Vectors a and b are parallel if and only if b = ka for some scalar k. This is crucial for proving collinearity.

Position vectors and geometric applications

The position vector of point A relative to origin O is $\overrightarrow{OA}$ or a.

For any two points A and B with position vectors a and b:

$$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$$

This is the fundamental relationship for solving geometric problems.

Midpoint formula:

If M is the midpoint of AB, then:

$$\mathbf{m} = \frac{\mathbf{a} + \mathbf{b}}{2}$$

Section formula:

If point P divides AB in the ratio m:n, then:

$$\mathbf{p} = \frac{n\mathbf{a} + m\mathbf{b}}{m + n}$$

For example, if a hurricane is tracked at point A and moves toward point B, and we need to know its position when it's travelled 2/5 of the distance, we use the section formula with m = 2, n = 3.

Collinearity of points

Three points A, B, and C are collinear if they lie on the same straight line.

Method to prove collinearity:

1. Express $\overrightarrow{AB}$ and $\overrightarrow{AC}$ (or $\overrightarrow{BC}$) in terms of position vectors
2. Show that one is a scalar multiple of the other: $\overrightarrow{AC} = k\overrightarrow{AB}$
3. Verify they share a common point (point A in this case)

If $\overrightarrow{AC} = k\overrightarrow{AB}$ where k is a positive scalar, then C lies on the line through A and B, beyond B if k > 1, or between A and B if 0 < k < 1.

Solving vector equations

Vector equations often require you to find unknown scalars or position vectors.

Key techniques:

• Equate corresponding components (i and j components must be equal separately)
• Use the property that if p = q, then their i-components are equal AND their j-components are equal
• For equations like ka + mb = 0 where a and b are non-parallel, both k = 0 and m = 0

Example application:

In navigation problems common to the Caribbean maritime industry, if a boat's displacement from port A to point P is $\mathbf{p} = \lambda\mathbf{a} + \mu\mathbf{b}$, you find λ and μ by:

1. Substituting known vectors
2. Equating i and j components
3. Solving simultaneous equations

Worked examples

Example 1: Magnitude, direction and unit vectors

Points A and B have position vectors a = 3i + 4j and b = 7i + j respectively.

(a) Find $\overrightarrow{AB}$ in the form xi + yj. (2 marks)

(b) Calculate the magnitude of $\overrightarrow{AB}$. (2 marks)

(c) Find a unit vector in the direction of $\overrightarrow{AB}$. (2 marks)

Solution:

(a) $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$

$= (7\mathbf{i} + \mathbf{j}) - (3\mathbf{i} + 4\mathbf{j})$

$= 4\mathbf{i} - 3\mathbf{j}$

(b) $|\overrightarrow{AB}| = \sqrt{4^2 + (-3)^2}$

$= \sqrt{16 + 9}$

$= \sqrt{25} = 5$ units

(c) Unit vector $= \frac{\overrightarrow{AB}}{|\overrightarrow{AB}|}$

$= \frac{4\mathbf{i} - 3\mathbf{j}}{5}$

$= \frac{4}{5}\mathbf{i} - \frac{3}{5}\mathbf{j}$ or $\begin{pmatrix} 0.8 \ -0.6 \end{pmatrix}$

Example 2: Section formula and collinearity

The position vectors of points P, Q and R are p = $\begin{pmatrix} 2 \ 5 \end{pmatrix}$, q = $\begin{pmatrix} 6 \ 9 \end{pmatrix}$ and r = $\begin{pmatrix} 8 \ 11 \end{pmatrix}$ respectively.

(a) Show that P, Q and R are collinear. (4 marks)

(b) Find the ratio in which Q divides PR. (2 marks)

Solution:

(a) $\overrightarrow{PQ} = \mathbf{q} - \mathbf{p} = \begin{pmatrix} 6 \ 9 \end{pmatrix} - \begin{pmatrix} 2 \ 5 \end{pmatrix} = \begin{pmatrix} 4 \ 4 \end{pmatrix}$

$\overrightarrow{PR} = \mathbf{r} - \mathbf{p} = \begin{pmatrix} 8 \ 11 \end{pmatrix} - \begin{pmatrix} 2 \ 5 \end{pmatrix} = \begin{pmatrix} 6 \ 6 \end{pmatrix}$

Check if $\overrightarrow{PR} = k\overrightarrow{PQ}$:

$\begin{pmatrix} 6 \ 6 \end{pmatrix} = k\begin{pmatrix} 4 \ 4 \end{pmatrix}$

From i-component: 6 = 4k, so k = 1.5

From j-component: 6 = 4k, so k = 1.5

Since both components give the same k value and P is common to both vectors, P, Q and R are collinear.

(b) Since $\overrightarrow{PR} = 1.5\overrightarrow{PQ}$, Q divides PR in the ratio PQ:QR.

$\overrightarrow{QR} = \overrightarrow{PR} - \overrightarrow{PQ} = \begin{pmatrix} 6 \ 6 \end{pmatrix} - \begin{pmatrix} 4 \ 4 \end{pmatrix} = \begin{pmatrix} 2 \ 2 \end{pmatrix}$

$|\overrightarrow{PQ}| : |\overrightarrow{QR}| = 4 : 2 = 2 : 1$

Example 3: Vector equation problem

Given that a = 2i - 3j and b = i + 4j, find the values of λ and μ such that:

$$\lambda\mathbf{a} + \mu\mathbf{b} = 9\mathbf{i} + 5\mathbf{j}$$

(5 marks)

Solution:

Substitute the vectors:

$\lambda(2\mathbf{i} - 3\mathbf{j}) + \mu(\mathbf{i} + 4\mathbf{j}) = 9\mathbf{i} + 5\mathbf{j}$

$2\lambda\mathbf{i} - 3\lambda\mathbf{j} + \mu\mathbf{i} + 4\mu\mathbf{j} = 9\mathbf{i} + 5\mathbf{j}$

$(2\lambda + \mu)\mathbf{i} + (-3\lambda + 4\mu)\mathbf{j} = 9\mathbf{i} + 5\mathbf{j}$

Equating i-components: $2\lambda + \mu = 9$ ... (1)

Equating j-components: $-3\lambda + 4\mu = 5$ ... (2)

From equation (1): $\mu = 9 - 2\lambda$

Substitute into equation (2):

$-3\lambda + 4(9 - 2\lambda) = 5$

$-3\lambda + 36 - 8\lambda = 5$

$-11\lambda = -31$

$\lambda = \frac{31}{11}$

Substitute back: $\mu = 9 - 2\left(\frac{31}{11}\right) = 9 - \frac{62}{11} = \frac{99 - 62}{11} = \frac{37}{11}$

Therefore: $\lambda = \frac{31}{11}$ and $\mu = \frac{37}{11}$

Common mistakes and how to avoid them

• Confusing position vectors with displacement vectors — Remember: $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ (final position minus initial position), not a - b. The order matters.

• Forgetting to use square root when finding magnitude — The magnitude formula is $\sqrt{x^2 + y^2}$, not just $x^2 + y^2$. Always complete the calculation.

• Incorrect handling of negative components — When a component is negative, be careful with signs during addition/subtraction. Write out each step clearly, especially with nested brackets.

• Claiming collinearity without checking the common point — Showing $\overrightarrow{AC} = k\overrightarrow{AB}$ isn't enough unless you also state that A is common to both vectors. Without a common point, the lines could be parallel but not coincident.

• Using the wrong formula for section divisions — Internal division (point between A and B) uses ratio m:n differently than external division. For internal division in ratio m:n from A toward B, use $\frac{n\mathbf{a} + m\mathbf{b}}{m+n}$.

• Not equating both components in vector equations — When solving vector equations, you must equate i-components AND j-components separately to form simultaneous equations. One equation alone is insufficient.

Exam technique for "Vectors"

• "Show that" questions require complete working — When asked to "show" or "prove" collinearity or parallelism, you must write out every step. State explicitly that both components give the same scalar k value, and identify the common point.

• Use clear notation consistently — Choose either column vector or component form (i, j) at the start of your solution and maintain it throughout. Don't switch mid-calculation unless necessary. Underline handwritten vectors: $\underline{a}$ not just a.

• Express answers in the required form — If the question asks for the answer "in the form xi + yj", don't leave it as a column vector. Read the command words carefully: "find," "calculate," "express," "hence," and "show" all require different response styles.

• Check magnitudes are positive — Since magnitude represents length or distance, your final answer must be positive. If you get a negative value, you've made an error in calculation or taken a square root incorrectly.

Quick revision summary

Vectors have magnitude and direction, represented in column form or using unit vectors i and j. Calculate magnitude using $|\mathbf{v}| = \sqrt{x^2 + y^2}$. The displacement from A to B is $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$. Add/subtract vectors component-wise; multiply by scalars to stretch or compress. Points are collinear when one displacement vector is a scalar multiple of another and they share a common point. Use section formula for dividing line segments in given ratios. Always equate corresponding components when solving vector equations.