Vectors: displacement, position, magnitude, direction, sums, differences, and scalar multiplication

Overview

A vector is a quantity with both size and direction. It can describe a movement, a force, a velocity, or a position from a chosen origin.

In Form IV, vectors are often introduced through movement on a plane. A movement of $3$ units to the right and $2$ units up can be written as a column vector:

$$ \begin{pmatrix}3\\2\end{pmatrix} $$

The top number describes horizontal movement. The bottom number describes vertical movement. This makes vectors a bridge between geometry and algebra: a movement can be drawn as an arrow and also calculated using numbers.

This chapter develops vectors slowly from displacement, to position, to magnitude and direction, then to vector addition, subtraction, and scalar multiplication. The aim is for learners to see each vector as an instruction: "move this far in this direction."

Prerequisites

• Mathematics Form IV - This topic belongs to the Form IV coordinate geometry, trigonometry, and vectors competence area.
• Coordinate Geometry - Vectors use the same coordinate-plane language of horizontal and vertical movement.
• Coordinate geometry: midpoint, distance, parallel lines, and perpendicular lines - Distance and gradient ideas help learners understand magnitude and direction.
• Pythagoras' theorem - Vector magnitude in two dimensions comes from the Pythagorean relationship.
• Trigonometric Ratios - Direction angles can be connected to tangent ratios.
• Directed numbers - Vector addition and subtraction require careful handling of positive and negative components.

Learning Scope

This chapter covers the meaning of a vector, displacement vectors, position vectors, column-vector notation, equal vectors, negative vectors, magnitude, direction, vector addition, vector subtraction, scalar multiplication, and basic geometric interpretation of vector operations.

This page does not develop advanced vector proofs, three-dimensional vectors, or mechanics applications. It focuses on the two-dimensional Form IV skills needed to represent and calculate movement on a plane.

Subtopics

Scalars And Vectors

A scalar has size only. Examples include mass, time, area, and temperature.

A vector has both size and direction. Examples include displacement, velocity, force, and directed movement on a coordinate plane.

The difference matters because direction changes the meaning. A movement of $5$ km east is not the same as a movement of $5$ km west, even though both have the same size.

| Quantity | Scalar or vector? | Reason | | --- | --- | --- | | $20$ seconds | Scalar | Time has size only. | | $15$ N upward | Vector | Force has size and direction. | | $8$ m east | Vector | Displacement has size and direction. | | $30^\circ$C | Scalar | Temperature has size only. |

Key insight: if changing direction changes the meaning of the quantity, the quantity is being treated as a vector.

Displacement Vectors

Displacement is a movement from one point to another. It does not describe the whole path travelled; it describes the straight movement from start to finish.

If a point moves $4$ units right and $3$ units up, the displacement vector is:

$$ \begin{pmatrix}4\\3\end{pmatrix} $$

If it moves $2$ units left and $5$ units down, the displacement vector is:

$$ \begin{pmatrix}-2\\-5\end{pmatrix} $$

The signs show direction:

| Component sign | Meaning | | --- | --- | | Top component positive | Move right | | Top component negative | Move left | | Bottom component positive | Move up | | Bottom component negative | Move down |

For points $A(x_1,y_1)$ and $B(x_2,y_2)$, the displacement from $A$ to $B$ is:

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

Example:

If $A(2,1)$ and $B(7,4)$, then:

$$ \overrightarrow{AB}= \begin{pmatrix} 7-2\\ 4-1 \end{pmatrix} = \begin{pmatrix} 5\\ 3 \end{pmatrix} $$

This means the movement from $A$ to $B$ is $5$ units right and $3$ units up.

Position Vectors

A position vector describes the position of a point from the origin.

If $P(4,-2)$, then the position vector of $P$ is:

$$ \overrightarrow{OP}= \begin{pmatrix} 4\\ -2 \end{pmatrix} $$

where $O$ is the origin.

Position vectors and displacement vectors look similar, but they answer different questions:

| Vector type | Question answered | | --- | --- | | Position vector | Where is the point from the origin? | | Displacement vector | How do you move from one point to another? |

If position vectors are known, displacement can be found by subtracting:

$$ \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA} $$

Example:

If:

$$ \overrightarrow{OA}=\begin{pmatrix}2\\1\end{pmatrix} \quad \text{and}\quad \overrightarrow{OB}=\begin{pmatrix}7\\4\end{pmatrix} $$

then:

$$ \overrightarrow{AB}= \begin{pmatrix}7\\4\end{pmatrix} - \begin{pmatrix}2\\1\end{pmatrix} = \begin{pmatrix}5\\3\end{pmatrix} $$

Equal And Negative Vectors

Vectors are equal if they have the same magnitude and direction. In column form, equal vectors have the same components.

For example:

$$ \begin{pmatrix}3\\-2\end{pmatrix} $$

is equal to any other vector that moves $3$ units right and $2$ units down, even if it starts at a different point.

A negative vector has the same magnitude but the opposite direction:

$$ -\begin{pmatrix}3\\-2\end{pmatrix} = \begin{pmatrix}-3\\2\end{pmatrix} $$

So if:

$$ \overrightarrow{AB}= \begin{pmatrix}5\\3\end{pmatrix} $$

then:

$$ \overrightarrow{BA}= \begin{pmatrix}-5\\-3\end{pmatrix} $$

Key insight: reversing the arrow changes the sign of every component.

Magnitude Of A Vector

The magnitude of a vector is its length.

For:

$$ \mathbf{a}= \begin{pmatrix}x\\y\end{pmatrix} $$

the magnitude is:

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

Why this works: the horizontal and vertical components form a right-angled triangle. The vector is the hypotenuse.

Example:

Find the magnitude of:

$$ \mathbf{a}= \begin{pmatrix}6\\8\end{pmatrix} $$

Then:

$$ |\mathbf{a}|=\sqrt{6^2+8^2} =\sqrt{36+64} =\sqrt{100} =10 $$

So the vector has magnitude $10$ units.

Checking routine:

1. Square both components.
2. Add the squares.
3. Take the square root.
4. Keep the magnitude positive.
5. Compare with the components: the magnitude should not be smaller than the larger component in absolute value.

Direction Of A Vector

Direction tells where the vector points. It may be described using words, a bearing, or an angle from the positive $x$-axis.

For a vector:

$$ \mathbf{a}= \begin{pmatrix}x\\y\end{pmatrix} $$

the direction angle $\theta$ from the positive $x$-axis often satisfies:

$$ \tan \theta=\frac{y}{x} $$

This must be used with care because the signs of $x$ and $y$ determine the quadrant.

Example:

For:

$$ \mathbf{a}= \begin{pmatrix}4\\3\end{pmatrix} $$

the vector points right and up, so it is in the first quadrant. Its direction angle satisfies:

$$ \tan \theta=\frac{3}{4} $$

Therefore:

$$ \theta \approx 36.9^\circ $$

So the direction is about $36.9^\circ$ above the positive $x$-axis.

Direction checking routine:

1. Use the signs of the components to identify the quadrant.
2. Use tangent to find a reference angle if needed.
3. Adjust the angle if the vector is not in the first quadrant.
4. State the direction clearly, including the reference line used.

Adding Vectors

Vector addition combines movements.

If:

$$ \mathbf{a}= \begin{pmatrix}2\\5\end{pmatrix} \quad \text{and}\quad \mathbf{b}= \begin{pmatrix}4\\-1\end{pmatrix} $$

then:

$$ \mathbf{a}+\mathbf{b} = \begin{pmatrix}2+4\\5+(-1)\end{pmatrix} = \begin{pmatrix}6\\4\end{pmatrix} $$

The top components are added together and the bottom components are added together.

Geometrically, addition means doing one movement after another. If you move according to $\mathbf{a}$, then from the new position move according to $\mathbf{b}$, the final displacement is $\mathbf{a}+\mathbf{b}$.

Checking routine:

1. Add horizontal components with horizontal components.
2. Add vertical components with vertical components.
3. Use brackets for negative components.
4. Interpret the result as the combined movement.

Subtracting Vectors

Vector subtraction compares movements or adds the opposite vector.

If:

$$ \mathbf{a}= \begin{pmatrix}7\\2\end{pmatrix} \quad \text{and}\quad \mathbf{b}= \begin{pmatrix}3\\5\end{pmatrix} $$

then:

$$ \mathbf{a}-\mathbf{b} = \begin{pmatrix}7-3\\2-5\end{pmatrix} = \begin{pmatrix}4\\-3\end{pmatrix} $$

Another way to read it is:

$$ \mathbf{a}-\mathbf{b}=\mathbf{a}+(-\mathbf{b}) $$

Subtraction is important for displacement from one position vector to another:

$$ \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA} $$

Warning sign: learners often subtract only the first component and forget the second. Both components must be subtracted.

Scalar Multiplication

Scalar multiplication stretches, shrinks, or reverses a vector.

If:

$$ \mathbf{a}= \begin{pmatrix}3\\-2\end{pmatrix} $$

then:

$$ 2\mathbf{a} = 2\begin{pmatrix}3\\-2\end{pmatrix} = \begin{pmatrix}6\\-4\end{pmatrix} $$

and:

$$ -\mathbf{a} = \begin{pmatrix}-3\\2\end{pmatrix} $$

Effects of a scalar $k$:

| Scalar | Effect on vector | | --- | --- | | $k>1$ | Same direction, longer vector | | $0<k<1$ | Same direction, shorter vector | | $k=0$ | Zero vector | | $k<0$ | Opposite direction, length multiplied by $|k|$ |

Example:

If:

$$ \mathbf{p}= \begin{pmatrix}-2\\5\end{pmatrix} $$

then:

$$ -3\mathbf{p} = \begin{pmatrix}6\\-15\end{pmatrix} $$

The vector is three times as long and points in the opposite direction.

Vector Expressions

Many exam-style vector questions combine addition, subtraction, and scalar multiplication.

Example:

If:

$$ \mathbf{a}=\begin{pmatrix}2\\-1\end{pmatrix} \quad \text{and}\quad \mathbf{b}=\begin{pmatrix}-3\\4\end{pmatrix} $$

find $2\mathbf{a}-3\mathbf{b}$.

First multiply:

$$ 2\mathbf{a}=2\begin{pmatrix}2\\-1\end{pmatrix} =\begin{pmatrix}4\\-2\end{pmatrix} $$

$$ 3\mathbf{b}=3\begin{pmatrix}-3\\4\end{pmatrix} =\begin{pmatrix}-9\\12\end{pmatrix} $$

Then subtract:

$$ 2\mathbf{a}-3\mathbf{b} = \begin{pmatrix}4\\-2\end{pmatrix} - \begin{pmatrix}-9\\12\end{pmatrix} = \begin{pmatrix}13\\-14\end{pmatrix} $$

Check each component separately:

Top component:

$$ 2(2)-3(-3)=4+9=13 $$

Bottom component:

$$ 2(-1)-3(4)=-2-12=-14 $$

Key Terms

| Term | Meaning | | --- | --- | | Scalar | A quantity with size only. | | Vector | A quantity with size and direction. | | Displacement | Directed movement from one point to another. | | Position vector | Vector from the origin to a point. | | Component | One part of a vector, usually horizontal or vertical. | | Column vector | Vector written in vertical component form. | | Magnitude | Length or size of a vector. | | Direction | The way a vector points. | | Equal vectors | Vectors with the same magnitude and direction. | | Negative vector | Vector with the same magnitude but opposite direction. | | Zero vector | Vector with zero magnitude, often written $\begin{pmatrix}0\\0\end{pmatrix}$. | | Scalar multiplication | Multiplying a vector by a number. | | Resultant vector | The single vector produced by adding vectors. |

Worked Examples

Example 1: Displacement From Coordinates

Find $\overrightarrow{AB}$ if $A(-1,3)$ and $B(5,-2)$.

Use:

$$ \overrightarrow{AB}= \begin{pmatrix} x_B-x_A\\ y_B-y_A \end{pmatrix} $$

Substitute:

$$ \overrightarrow{AB}= \begin{pmatrix} 5-(-1)\\ -2-3 \end{pmatrix} = \begin{pmatrix} 6\\ -5 \end{pmatrix} $$

Answer:

$$ \overrightarrow{AB}=\begin{pmatrix}6\\-5\end{pmatrix} $$

This means move $6$ units right and $5$ units down.

Example 2: Position Vectors To Displacement

Given:

$$ \overrightarrow{OP}=\begin{pmatrix}4\\7\end{pmatrix} \quad \text{and}\quad \overrightarrow{OQ}=\begin{pmatrix}-2\\1\end{pmatrix} $$

find $\overrightarrow{PQ}$.

Use:

$$ \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP} $$

Then:

$$ \overrightarrow{PQ} = \begin{pmatrix}-2\\1\end{pmatrix} - \begin{pmatrix}4\\7\end{pmatrix} = \begin{pmatrix}-6\\-6\end{pmatrix} $$

Answer:

$$ \overrightarrow{PQ}=\begin{pmatrix}-6\\-6\end{pmatrix} $$

Example 3: Magnitude

Find the magnitude of:

$$ \mathbf{v}= \begin{pmatrix}-5\\12\end{pmatrix} $$

Use:

$$ |\mathbf{v}|=\sqrt{(-5)^2+12^2} $$

So:

$$ |\mathbf{v}|=\sqrt{25+144} =\sqrt{169} =13 $$

Answer: the magnitude is $13$ units.

Check: the negative component does not make the magnitude negative because it is squared.

Example 4: Vector Sum And Difference

Let:

$$ \mathbf{a}=\begin{pmatrix}3\\-4\end{pmatrix} \quad \text{and}\quad \mathbf{b}=\begin{pmatrix}-1\\6\end{pmatrix} $$

Find $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$.

Sum:

$$ \mathbf{a}+\mathbf{b} = \begin{pmatrix}3+(-1)\\-4+6\end{pmatrix} = \begin{pmatrix}2\\2\end{pmatrix} $$

Difference:

$$ \mathbf{a}-\mathbf{b} = \begin{pmatrix}3-(-1)\\-4-6\end{pmatrix} = \begin{pmatrix}4\\-10\end{pmatrix} $$

Answers:

$$ \mathbf{a}+\mathbf{b}=\begin{pmatrix}2\\2\end{pmatrix} $$

and:

$$ \mathbf{a}-\mathbf{b}=\begin{pmatrix}4\\-10\end{pmatrix} $$

Example 5: Scalar Multiplication And Expression

If:

$$ \mathbf{x}=\begin{pmatrix}1\\-3\end{pmatrix} \quad \text{and}\quad \mathbf{y}=\begin{pmatrix}4\\2\end{pmatrix} $$

find $5\mathbf{x}-2\mathbf{y}$.

First:

$$ 5\mathbf{x} = \begin{pmatrix}5\\-15\end{pmatrix} $$

and:

$$ 2\mathbf{y} = \begin{pmatrix}8\\4\end{pmatrix} $$

Then:

$$ 5\mathbf{x}-2\mathbf{y} = \begin{pmatrix}5\\-15\end{pmatrix} - \begin{pmatrix}8\\4\end{pmatrix} = \begin{pmatrix}-3\\-19\end{pmatrix} $$

Answer:

$$ 5\mathbf{x}-2\mathbf{y}=\begin{pmatrix}-3\\-19\end{pmatrix} $$

Example 6: Direction Angle

Find the direction angle of:

$$ \mathbf{u}=\begin{pmatrix}3\\3\end{pmatrix} $$

The vector points right and up, so it is in the first quadrant.

$$ \tan \theta=\frac{3}{3}=1 $$

Therefore:

$$ \theta=45^\circ $$

Answer: the vector points at $45^\circ$ above the positive $x$-axis.

Common Mistakes

• Treating all quantities as scalars. Correction: ask whether direction is part of the meaning. Warning sign: "5 km east" is shortened to only "5 km."
• Reversing displacement direction. Correction: $\overrightarrow{AB}$ means final point $B$ minus starting point $A$. Warning sign: the answer for $\overrightarrow{AB}$ is actually $\overrightarrow{BA}$.
• Confusing a position vector with a displacement vector. Correction: position starts at the origin; displacement starts at one named point and ends at another.
• Adding top and bottom components together. Correction: top combines with top, bottom combines with bottom. Warning sign: a two-component vector becomes a single number.
• Forgetting brackets around negative components. Correction: write $5-(-1)$ before simplifying. Warning sign: subtraction of a negative becomes subtraction of a positive.
• Making magnitude negative. Correction: magnitude is length, so it is always zero or positive.
• Forgetting to multiply every component by the scalar. Correction: apply the scalar to both the top and bottom components. Warning sign: $3\begin{pmatrix}2\\5\end{pmatrix}$ becomes $\begin{pmatrix}6\\5\end{pmatrix}$.
• Using $\tan \theta=\frac{y}{x}$ without checking the quadrant. Correction: first inspect the signs of the components.
• Assuming equal magnitude means equal vectors. Correction: equal vectors must have the same magnitude and the same direction.

Practice Tasks

Foundation

1. State whether each quantity is a scalar or vector: $12$ kg, $8$ m north, $20$ seconds, $5$ N downward.
2. Write the column vector for a movement of $6$ units right and $4$ units down.
3. Find the negative of $\begin{pmatrix}3\\-7\end{pmatrix}$.
4. Find the magnitude of $\begin{pmatrix}5\\12\end{pmatrix}$.
5. Add $\begin{pmatrix}2\\1\end{pmatrix}$ and $\begin{pmatrix}4\\-3\end{pmatrix}$.

Skill-Building

1. If $A(3,-2)$ and $B(-1,5)$, find $\overrightarrow{AB}$.
2. If $\overrightarrow{OP}=\begin{pmatrix}6\\-4\end{pmatrix}$ and $\overrightarrow{OQ}=\begin{pmatrix}1\\3\end{pmatrix}$, find $\overrightarrow{PQ}$.
3. Find $3\begin{pmatrix}-2\\5\end{pmatrix}$.
4. If $\mathbf{a}=\begin{pmatrix}4\\-1\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}-3\\2\end{pmatrix}$, find $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$.
5. Find the magnitude of $\begin{pmatrix}-8\\15\end{pmatrix}$.

Exam-Style

1. Points $A(-2,1)$ and $B(4,9)$ are given. Find $\overrightarrow{AB}$ and its magnitude.
2. Given $\mathbf{p}=\begin{pmatrix}2\\-3\end{pmatrix}$ and $\mathbf{q}=\begin{pmatrix}-5\\1\end{pmatrix}$, find $4\mathbf{p}-2\mathbf{q}$.
3. The position vectors of $A$ and $B$ are $\begin{pmatrix}3\\7\end{pmatrix}$ and $\begin{pmatrix}-1\\2\end{pmatrix}$ respectively. Find $\overrightarrow{AB}$ and $\overrightarrow{BA}$.
4. A vector has components $\begin{pmatrix}6\\6\end{pmatrix}$. Find its magnitude and direction angle from the positive $x$-axis.
5. Show that $\begin{pmatrix}2\\-4\end{pmatrix}$ and $\begin{pmatrix}-1\\2\end{pmatrix}$ are parallel but point in opposite directions.

Challenge

1. Create two different vectors with magnitude $10$ and explain how you checked them.
2. A vector $\mathbf{a}$ is multiplied by $-\frac{1}{2}$. Describe what happens to its magnitude and direction.
3. Points $A$, $B$, and $C$ have position vectors $\begin{pmatrix}1\\2\end{pmatrix}$, $\begin{pmatrix}5\\4\end{pmatrix}$, and $\begin{pmatrix}9\\6\end{pmatrix}$. Use vectors to decide whether the points lie on one straight line.
4. If $\mathbf{a}+\mathbf{b}=\begin{pmatrix}7\\-1\end{pmatrix}$ and $\mathbf{a}=\begin{pmatrix}3\\4\end{pmatrix}$, find $\mathbf{b}$ and describe the movement it represents.

Generated Question Layer

• Scalar-vector classification questions: decide whether direction is part of the quantity.
• Displacement questions: find $\overrightarrow{AB}$ from coordinates or position vectors.
• Magnitude questions: use Pythagoras' theorem on vector components.
• Direction questions: interpret component signs and calculate a direction angle when required.
• Addition and subtraction questions: combine or compare vector movements component by component.
• Scalar multiplication questions: stretch, shrink, or reverse vectors using a scalar.
• Vector-expression questions: simplify expressions such as $2\mathbf{a}-3\mathbf{b}$.
• Geometric interpretation questions: connect equal, negative, or parallel vectors to arrows on a plane.

Learner Aid Opportunities

• diagram: Show a column vector as an arrow with labelled horizontal and vertical components.
• chart: Compare scalar quantities and vector quantities using familiar daily-life examples.
• graph: Plot position vectors from the origin and displacement vectors between two non-origin points.
• animation: Show vector addition by placing the second arrow at the head of the first arrow.
• interactive: Let learners adjust vector components and observe changes in magnitude, direction, and quadrant.
• video: Demonstrate how to find $\overrightarrow{AB}$ from both coordinates and position vectors.
• LLM tutor: Prompt learners to explain the starting point, ending point, component signs, and final interpretation of a vector answer.

Exam-Derived Signals

These signals are assessment leads, not verified official past-question links. They should be checked against original papers and marking schemes before being used as final learner-facing references.

| Source | Current Signal | Review Status | Use Carefully As | | --- | --- | --- | --- | | data/exam_format_topic_crosswalk_2022.jsonl | Official 2022 format grouping includes vectors within the broader Form IV competence area for coordinate geometry, trigonometry, and vectors. | Official format mapping; topic-page use still unreviewed. | Evidence that vector skills belong in the assessed Form IV skill cluster. | | data/topic_frequency_2021_2025.json | Frequency data may contain direct or indirect vector coverage, but no item has been reviewed and promoted on this page. | Unreviewed aggregate. | A prompt for reviewers to search for displacement, magnitude, direction, and vector-operation tasks. | | data/question_map_2021_2025.jsonl | No reviewed 2021-2025 question-map item is promoted here yet. | No reviewed links. | Future review should check whether coordinate geometry, trigonometry, or transformation questions include vector reasoning. |

Source And Review Notes

• Topic registry status: official in data/curriculum_map.json.
• Learner expansion status: original prose drafted from the official syllabus topic and local assessment-signal files.
• Exam mapping status: unreviewed except for official format crosswalk records.
• Review risk: vector notation varies across resources, so future reviewers should confirm notation choices against local textbook and examination conventions before final publication.
• Media status: no media assets were added; learner-aid entries are planning markers only.