Vectors in two and three dimensions

A vector is a quantity with both magnitude and direction. It can describe a displacement, velocity, acceleration or force. A scalar has magnitude but no direction, such as distance, speed, time, mass or temperature.

For example,

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

represents a movement of $4$ units in the positive $x$ direction and $3$ units in the negative $y$ direction. It describes a change, not a fixed point.

This lesson develops the notation and geometric meaning needed for all later vector work.

Prerequisites

You should be able to:

• read coordinates in all four quadrants;
• subtract negative numbers;
• use the $x$, $y$ and $z$ axes;
• distinguish distance from displacement.

Review coordinate geometry or GCSE vectors if these ideas are uncertain.

Scalars and vectors

Direction is part of a vector’s identity. A speed of $12\text{ m s}^{-1}$ is a scalar, but a velocity of $12\text{ m s}^{-1}$ due east is a vector. Changing only the direction changes the velocity even when the speed stays constant.

Quantity	Scalar or vector?	Reason
$7$ km travelled	Scalar	Distance has no direction
$7$ km north	Vector	Displacement includes direction
$20^\circ\text{C}$	Scalar	Temperature has no direction
$5$ N vertically downwards	Vector	Force includes direction
$3\text{ m s}^{-2}$ towards the origin	Vector	Acceleration includes direction

Suppose a walker travels $5$ km east and then $5$ km west. The total distance is $10$ km, but the overall displacement is the zero vector because the walker finishes where they started.

Self-check 1

Classify each quantity as scalar or vector.

1. A mass of $2.4$ kg.
2. A wind velocity of $18\text{ km h}^{-1}$ south-west.
3. A journey time of $35$ minutes.
4. An acceleration of $9.8\text{ m s}^{-2}$ downwards.

Answer

Mass and time are scalars. Wind velocity and acceleration are vectors because their directions are specified.

Vector notation

A vector can be represented by an arrow. The arrow’s length represents its magnitude and its arrowhead gives its direction.

Common notation includes

$$\mathbf a, \qquad \overrightarrow{AB}, \qquad \begin{pmatrix}a_1\\a_2\end{pmatrix}.$$

• $\mathbf a$ is a named vector. Printed vectors are usually bold.
• $\overrightarrow{AB}$ is the vector from $A$ to $B$. The order of the letters gives the direction.
• A column vector gives the change in each coordinate.

In handwriting, bold type is impractical, so a vector may be written as an underlined letter, such as $\underline a$. Use one convention consistently.

An arrow needs the correct order

$\overrightarrow{AB}$ starts at $A$ and finishes at $B$. Reversing the letters reverses the direction:

$$\boxed{\overrightarrow{BA}=-\overrightarrow{AB}}.$$

Worked example 1: interpret a column vector

Describe the movement represented by

$$\mathbf p=\begin{pmatrix}-6\\2\end{pmatrix}.$$

The first component is the horizontal change. Since it is negative, the movement is $6$ units left.

The second component is the vertical change. Since it is positive, the movement is $2$ units up.

Therefore $\mathbf p$ represents $6$ units left and $2$ units up.

It does not by itself specify where the movement begins.

Free vectors and equal vectors

A vector records a movement, not a location. An arrow may be translated parallel to itself without changing the vector. Such a vector is called a free vector.

Two vectors are equal precisely when they have the same magnitude and the same direction. Their starting points may differ. In component form,

$$\begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}c\\d\end{pmatrix} \quad\Longleftrightarrow\quad a=c\text{ and }b=d.$$

So every arrow representing $3$ units right and $4$ units down is the vector

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

regardless of where the arrow is drawn.

Equal length is not enough

The vectors $\begin{pmatrix}3\\4\end{pmatrix}$ and $\begin{pmatrix}-3\\-4\end{pmatrix}$ have equal magnitudes, but they point in opposite directions and are not equal.

Worked example 2: use equality of vectors

Given

$$\begin{pmatrix}2p-1\\q+4\end{pmatrix} = \begin{pmatrix}7\\-3\end{pmatrix},$$

find $p$ and $q$.

Equal vectors have equal corresponding components, so

$$2p-1=7, \qquad q+4=-3.$$

Solving gives

$$2p=8\implies p=4, \qquad q=-7.$$

Hence

$$\boxed{p=4,\quad q=-7}.$$

The vector equation is simply two scalar equations written together.

Components in two dimensions

In two dimensions,

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

has horizontal component $x$ and vertical component $y$.

The standard unit vectors are

$$\mathbf i=\begin{pmatrix}1\\0\end{pmatrix}, \qquad \mathbf j=\begin{pmatrix}0\\1\end{pmatrix}.$$

They point one unit along the positive $x$ and $y$ axes respectively. Therefore

$$\boxed{\begin{pmatrix}x\\y\end{pmatrix}=x\mathbf i+y\mathbf j}.$$

This is called unit vector notation or $\mathbf i$, $\mathbf j$ notation.

Worked example 3: convert between notations

Write $\mathbf a=-3\mathbf i+5\mathbf j$ as a column vector, and write

$$\mathbf b=\begin{pmatrix}4\\-7\end{pmatrix}$$

in unit vector notation.

The coefficient of $\mathbf i$ is the horizontal component and the coefficient of $\mathbf j$ is the vertical component. Thus

$$\boxed{\mathbf a=\begin{pmatrix}-3\\5\end{pmatrix}}.$$

Conversely,

$$\boxed{\mathbf b=4\mathbf i-7\mathbf j}.$$

The minus sign belongs to the $\mathbf j$ component. Writing $4\mathbf i+7\mathbf j$ would reverse the vertical movement.

Components in three dimensions

Three dimensional vectors need one further component:

$$\mathbf a= \begin{pmatrix}x\\y\\z\end{pmatrix} =x\mathbf i+y\mathbf j+z\mathbf k,$$

where

$$\mathbf i=\begin{pmatrix}1\\0\\0\end{pmatrix}, \qquad \mathbf j=\begin{pmatrix}0\\1\\0\end{pmatrix}, \qquad \mathbf k=\begin{pmatrix}0\\0\\1\end{pmatrix}.$$

The $\mathbf k$ direction is the positive $z$ direction. The order is always $x$, then $y$, then $z$.

Worked example 4: interpret a 3D vector

Write

$$\mathbf r=2\mathbf i-\mathbf j+6\mathbf k$$

as a column vector and state its components.

An omitted numerical coefficient is $1$, so $-\mathbf j=-1\mathbf j$. Therefore

$$\boxed{\mathbf r=\begin{pmatrix}2\\-1\\6\end{pmatrix}}.$$

Its $x$, $y$ and $z$ components are $2$, $-1$ and $6$ respectively.

Do not mix dimensions

A two dimensional vector and a three dimensional vector cannot be equal. Even

$$\begin{pmatrix}2\\3\end{pmatrix} \quad\text{and}\quad \begin{pmatrix}2\\3\\0\end{pmatrix}$$

belong to different spaces, although the second lies in the $xy$ plane of three dimensional space.

Self-check 2

1. Write $5\mathbf i-2\mathbf j$ as a column vector.
2. Write $\begin{pmatrix}-4\\0\\7\end{pmatrix}$ in $\mathbf i$, $\mathbf j$, $\mathbf k$ notation.
3. If $\begin{pmatrix}m+1\\2n\\3\end{pmatrix}=\begin{pmatrix}6\\-8\\3\end{pmatrix}$, find $m$ and $n$.

Answer

1. $\begin{pmatrix}5\\-2\end{pmatrix}$.
2. $-4\mathbf i+7\mathbf k$. The zero $\mathbf j$ term may be omitted.
3. $m+1=6$ and $2n=-8$, so $\boxed{m=5}$ and $\boxed{n=-4}$.

The zero vector and opposite vectors

The zero vector has every component equal to zero:

$$\mathbf 0=\begin{pmatrix}0\\0\end{pmatrix} \quad\text{or}\quad \mathbf 0=\begin{pmatrix}0\\0\\0\end{pmatrix}.$$

It represents no change in position. Its magnitude is zero, so it has no defined direction.

The vector opposite to $\mathbf a$ is $-\mathbf a$. Every component changes sign. For example,

$$-\begin{pmatrix}-2\\5\\1\end{pmatrix} =\begin{pmatrix}2\\-5\\-1\end{pmatrix}.$$

The two vectors have the same magnitude but opposite directions.

Worked example 5: reverse a journey

Suppose

$$\overrightarrow{PQ}=3\mathbf i-4\mathbf j+2\mathbf k.$$

Find $\overrightarrow{QP}$.

Reversing the journey negates the vector:

$$\begin{aligned} \overrightarrow{QP} &=-\overrightarrow{PQ}\\ &=-(3\mathbf i-4\mathbf j+2\mathbf k)\\ &=\boxed{-3\mathbf i+4\mathbf j-2\mathbf k}. \end{aligned}$$

Notice that all three signs change.

Finding a vector between two points

For points $A(x_1,y_1)$ and $B(x_2,y_2)$,

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

In words, calculate finish minus start in each coordinate.

In three dimensions, if $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$, then

$$\boxed{\overrightarrow{AB} =\begin{pmatrix}x_2-x_1\\y_2-y_1\\z_2-z_1\end{pmatrix}}.$$

This works because each component measures the change needed to get from $A$ to $B$.

Worked example 6: displacement in two dimensions

Let $A(-2,5)$ and $B(4,-1)$. Find $\overrightarrow{AB}$ and $\overrightarrow{BA}$.

For $A$ to $B$, subtract the coordinates of $A$ from those of $B$:

$$\begin{aligned} \overrightarrow{AB} &=\begin{pmatrix}4-(-2)\\-1-5\end{pmatrix}\\ &=\boxed{\begin{pmatrix}6\\-6\end{pmatrix}}. \end{aligned}$$

The result makes geometric sense: move $6$ units right and $6$ units down.

For the reverse journey,

$$\overrightarrow{BA} =\begin{pmatrix}-2-4\\5-(-1)\end{pmatrix} =\boxed{\begin{pmatrix}-6\\6\end{pmatrix}} =-\overrightarrow{AB}.$$

The final equality provides a useful check.

Worked example 7: displacement in three dimensions

Let $P(3,-2,5)$ and $Q(-1,4,2)$. Find $\overrightarrow{PQ}$ in column form and unit vector notation.

Use finish minus start:

$$\begin{aligned} \overrightarrow{PQ} &=\begin{pmatrix}-1-3\\4-(-2)\\2-5\end{pmatrix}\\ &=\begin{pmatrix}-4\\6\\-3\end{pmatrix}. \end{aligned}$$

Therefore

$$\boxed{\overrightarrow{PQ}=-4\mathbf i+6\mathbf j-3\mathbf k}.$$

The middle calculation is $4-(-2)=6$, not $2$. Brackets reduce sign errors.

Worked example 8: find an unknown coordinate

Points $R(2,a,-1)$ and $S(7,3,4)$ satisfy

$$\overrightarrow{RS}=\begin{pmatrix}5\\-2\\5\end{pmatrix}.$$

Find $a$.

From the coordinates,

$$\overrightarrow{RS} =\begin{pmatrix}7-2\\3-a\\4-(-1)\end{pmatrix} =\begin{pmatrix}5\\3-a\\5\end{pmatrix}.$$

Compare the second components:

$$3-a=-2.$$

Hence $-a=-5$, so

$$\boxed{a=5}.$$

Checking: the change in the $y$ coordinate is $3-5=-2$, as required.

Coordinates are not displacement vectors

If $B(4,-1)$, then $\begin{pmatrix}4\\-1\end{pmatrix}$ gives the displacement from the origin to $B$. It does not generally give $\overrightarrow{AB}$. For a journey between two arbitrary points, use finish minus start.

Common misconceptions

A vector is not just its magnitude

The vectors $\begin{pmatrix}1\\2\end{pmatrix}$ and $\begin{pmatrix}2\\1\end{pmatrix}$ do not point in the same direction. Having the same component values in a different order does not make vectors equal.

Negative components do not mean negative length

In $\begin{pmatrix}-3\\4\end{pmatrix}$, the negative sign means movement in the negative $x$ direction. A magnitude is never negative.

Points and vectors are different objects

The point $(3,2)$ is a location. The vector $\begin{pmatrix}3\\2\end{pmatrix}$ is a movement. They use related numbers, but brackets and context communicate different meanings.

The zero vector has no direction

It is tempting to say that $\mathbf 0$ points nowhere. More precisely, its direction is undefined because an arrow of zero length has no orientation.

Mixed self-check

Try these without looking back.

1. Explain why two arrows at different locations can represent the same vector.
2. Write $\mathbf a=3\mathbf i-5\mathbf j+2\mathbf k$ as a column vector and find $-\mathbf a$.
3. Points $C(-4,1)$ and $D(2,7)$ are given. Find $\overrightarrow{CD}$ and $\overrightarrow{DC}$.
4. Points $E(1,-3,2)$ and $F(p,4,-5)$ satisfy $\overrightarrow{EF}=\begin{pmatrix}6\\7\\-7\end{pmatrix}$. Find $p$.
5. A particle moves from $A$ to $B$ and returns to $A$. State its overall displacement vector.

Answers

1. A vector is determined by magnitude and direction, not its starting position. Translating an arrow parallel to itself does not change the vector.

$$\mathbf a=\begin{pmatrix}3\\-5\\2\end{pmatrix}, \qquad -\mathbf a=\begin{pmatrix}-3\\5\\-2\end{pmatrix}.$$

$$\overrightarrow{CD} =\begin{pmatrix}2-(-4)\\7-1\end{pmatrix} =\begin{pmatrix}6\\6\end{pmatrix}, \qquad \overrightarrow{DC}=\begin{pmatrix}-6\\-6\end{pmatrix}.$$

4. Comparing the first component gives $p-1=6$, so $\boxed{p=7}$.

5. The particle finishes where it started, so its overall displacement is $\boxed{\mathbf 0}$.

Key facts

$$\begin{aligned} \text{2D vector:}\quad &\begin{pmatrix}x\\y\end{pmatrix}=x\mathbf i+y\mathbf j,\\[4pt] \text{3D vector:}\quad &\begin{pmatrix}x\\y\\z\end{pmatrix}=x\mathbf i+y\mathbf j+z\mathbf k,\\[4pt] \text{reverse direction:}\quad &\overrightarrow{BA}=-\overrightarrow{AB},\\[4pt] \text{between points:}\quad &\overrightarrow{AB}=\text{position of }B-\text{position of }A. \end{aligned}$$

Equal vectors have equal corresponding components. The zero vector has all components zero, and a vector can be moved parallel to itself without changing its value.

Where this leads

You can now interpret vectors in two and three dimensions, move between column and unit vector notation, and find a displacement from coordinates.

Next, learn how to combine and scale vectors in vector arithmetic, then calculate length and orientation in vector magnitude and direction. Position vectors use these ideas to solve geometric problems involving points, distances and division of line segments.