Pure · Vectors · Year 1

Introduction to Vectors

A vector carries both size and direction. Add them component by component, scale them by a number, and spot when two are parallel — the foundations everything else builds on.

9 min read 3 worked examples

What a vector is

A vector has two properties: magnitude (size) and direction. This makes it fundamentally different from a scalar, which is just a number — a size with no direction. Speed is a scalar; velocity (speed in a direction) is a vector. A displacement of "3 km north-east" is a vector; a distance of "3 km" is a scalar.

Because a vector is defined by magnitude and direction alone — not by where it starts — two arrows that are the same length and point the same way represent the same vector, even if they're drawn in different places.

Several arrows on a grid. Two arrows of equal length pointing the same direction are labelled as the same vector a, even though they start at different points. A third arrow pointing the opposite way is labelled negative a. — A vector is fixed by its length and direction, not its position — so equal, parallel arrows represent the same vector. Reversing the direction gives −a.

Two ways to write a vector

You'll meet vectors written in two equivalent notations. Column form stacks the horizontal and vertical components: $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$ means "3 right, 2 up". i–j form uses the unit vectors $\mathbf{i}$ (one unit right) and $\mathbf{j}$ (one unit up): the same vector is $3\mathbf{i} + 2\mathbf{j}$. They carry identical information — $\begin{pmatrix} 3 \\ 2 \end{pmatrix} = 3\mathbf{i} + 2\mathbf{j}$ — so switch between them freely. In print, vectors are bold ($\mathbf{a}$); handwritten, they're underlined ($\underline{a}$).

Adding and subtracting vectors

To add vectors, add their components. Geometrically, $\mathbf{a} + \mathbf{b}$ means "follow $\mathbf{a}$, then follow $\mathbf{b}$" — placing them nose-to-tail, the sum is the single vector from the very start to the very end (the triangle law).

Addition: $\begin{pmatrix} a_1 \\ a_2 \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \end{pmatrix}$ — add component by component.

Subtraction: $\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b})$ — subtracting is adding the reverse.

Two vectors a and b placed nose to tail. Vector a goes from the start point to a middle point, then vector b continues from there to the end point. The resultant a plus b is drawn as a single arrow from the start directly to the end, forming a triangle. — The triangle law: place b's tail at a's nose, and a + b is the arrow closing the triangle from start to finish.

Multiplying by a scalar

Multiplying a vector by a scalar (an ordinary number) scales its length but keeps its line of direction. So $2\mathbf{a}$ is twice as long as $\mathbf{a}$, same direction; $-\mathbf{a}$ is the same length, opposite direction; $\tfrac12\mathbf{a}$ is half as long. In components, multiply each one: $k\begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} ka_1 \\ ka_2 \end{pmatrix}$.

Parallel vectors — the key test

Two non-zero vectors are parallel if and only if one is a scalar multiple of the other: $\mathbf{a}$ and $\mathbf{b}$ are parallel exactly when $\mathbf{b} = k\mathbf{a}$ for some scalar $k$. This is one of the most-used facts in the whole topic. For example $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} 6 \\ 9 \end{pmatrix}$ are parallel because $\begin{pmatrix} 6 \\ 9 \end{pmatrix} = 3\begin{pmatrix} 2 \\ 3 \end{pmatrix}$. If $k > 0$ they point the same way; if $k < 0$, opposite ways.

Worked example 1

Add and subtract in component form

Given $\mathbf{a} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -2 \\ 5 \end{pmatrix}$, find $\mathbf{a} + \mathbf{b}$, $\mathbf{a} - \mathbf{b}$, and $3\mathbf{a}$.

MethodWork component by component

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

Add the components.

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

Subtract the components — watch the double negative.

$3\mathbf{a} = 3\begin{pmatrix} 4 \\ 1 \end{pmatrix} = \begin{pmatrix} 12 \\ 3 \end{pmatrix}$.

Multiply each component by 3.

▪ solved

Examiner's eye

The one place marks slip is the double negative in subtraction: $4 - (-2) = 6$, not $2$. Treat each component as its own little sum and take the signs slowly. Component arithmetic is the entire mechanic here — get comfortable with it and the rest of the topic follows.

Worked example 2

Convert notation and combine

Given $\mathbf{p} = 5\mathbf{i} - 2\mathbf{j}$ and $\mathbf{q} = \mathbf{i} + 4\mathbf{j}$, find $2\mathbf{p} - \mathbf{q}$ in i–j form.

MethodScale, then subtract term by term

$2\mathbf{p} = 2(5\mathbf{i} - 2\mathbf{j}) = 10\mathbf{i} - 4\mathbf{j}$.

Distribute the 2.

$2\mathbf{p} - \mathbf{q} = (10\mathbf{i} - 4\mathbf{j}) - (\mathbf{i} + 4\mathbf{j})$.

Set up the subtraction.

Collect $\mathbf{i}$ terms: $10\mathbf{i} - \mathbf{i} = 9\mathbf{i}$. Collect $\mathbf{j}$ terms: $-4\mathbf{j} - 4\mathbf{j} = -8\mathbf{j}$.

$\mathbf{i}$ and $\mathbf{j}$ are separate.

So $2\mathbf{p} - \mathbf{q} = 9\mathbf{i} - 8\mathbf{j}$.

Combine.

▪ solved

The key insight

The $\mathbf{i}$ and $\mathbf{j}$ parts never mix — they behave like two separate tracks, exactly like the top and bottom of a column vector. Working in i–j form is really just column arithmetic written on one line. If you prefer, rewrite as columns, compute, and convert back; the answer is identical.

Worked example 3

Test for parallel vectors

Determine whether $\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ is parallel to $\mathbf{b} = \begin{pmatrix} -9 \\ 6 \end{pmatrix}$. If so, state the scalar and whether they point the same or opposite ways.

MethodCheck whether one is a scalar multiple of the other

Look for $k$ with $\mathbf{b} = k\mathbf{a}$. From the first components: $-9 = 3k \Rightarrow k = -3$.

Solve using one component.

Check the second components with $k = -3$: $3 \times (-3) \times ... $ — test $k\mathbf{a} = -3\begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} -9 \\ 6 \end{pmatrix}$.

Both components must match.

This equals $\mathbf{b}$ exactly, so $\mathbf{b} = -3\mathbf{a}$ — they are parallel.

Confirmed for both components.

Since $k = -3 < 0$, they point in opposite directions.

Negative scalar reverses direction.

▪ parallel

Look deeper — you must check both components

Finding a $k$ that works for the first component isn't enough — it must give the same $k$ for the second too. If the two components demanded different values of $k$, the vectors would not be parallel. Here both give $k = -3$, so the test passes. The sign of $k$ then tells you the relative direction: negative means antiparallel (opposite ways).

Common mistake

Don't confuse a vector with a scalar: adding vectors means adding components separately, never adding them into one number. And keep the $\mathbf{i}$ and $\mathbf{j}$ parts apart — $9\mathbf{i} - 8\mathbf{j}$ cannot be simplified to $1\mathbf{j}$ or any single term; the two directions are independent. Finally, for the parallel test, a scalar multiple must work for all components at once.

Check yourself

Given $\mathbf{a} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$, find $\mathbf{a} + 2\mathbf{b}$. Then determine whether $\begin{pmatrix} 4 \\ -6 \end{pmatrix}$ is parallel to $\mathbf{a}$.
(Hint: scale b first; for the parallel test, look for a single scalar k.)

Method

$2\mathbf{b} = \begin{pmatrix} 2 \\ 8 \end{pmatrix}$, so $\mathbf{a} + 2\mathbf{b} = \begin{pmatrix} 2 + 2 \\ -3 + 8 \end{pmatrix} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}$.

Scale then add.

Parallel test: is $\begin{pmatrix} 4 \\ -6 \end{pmatrix} = k\begin{pmatrix} 2 \\ -3 \end{pmatrix}$? First components: $4 = 2k \Rightarrow k = 2$.

Find k from one component.

Check: $2 \times (-3) = -6$ ✓ — matches the second component.

Both components agree.

So $\begin{pmatrix} 4 \\ -6 \end{pmatrix} = 2\mathbf{a}$: yes, parallel, same direction (since $k = 2 > 0$).

Positive k ⇒ same way.

▪ solved

The scalar $k = 2$ worked for both components, confirming the vectors are parallel — and being positive, they point the same way.

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