Vectors in Two Dimensions – A Level Maths Year 1

1 What a Vector Is, and How to Write One

A scalar has size only (mass $5$ kg, speed $12$ m s$^{-1}$). A vector has size and direction (a force of $5$ N upwards, a velocity of $12$ m s$^{-1}$ due east). This page is DfE content area J · Vectors; in Year 1 every vector is two-dimensional.

Notation	Example	Notes
Bold lower case (printed)	$\mathbf{a}$	Handwritten as $\underline{a}$ — underline it, or you lose the distinction from a number.
Arrow between points	$\overrightarrow{AB}$	The displacement from $A$ to $B$.
$\mathbf{i}, \mathbf{j}$ form	$3\mathbf{i} + 4\mathbf{j}$	$\mathbf{i}$ and $\mathbf{j}$ are unit vectors along the $x$- and $y$-axes.
Column vector	$\begin{pmatrix} 3 \\ 4 \end{pmatrix}$	Exactly the same vector as $3\mathbf{i} + 4\mathbf{j}$.

A vector has no fixed position. Any arrow 3 across and 4 up represents $3\mathbf{i} + 4\mathbf{j}$, wherever it is drawn. Two vectors are equal when both components are equal.

2 Magnitude and Direction

Magnitude and direction of $\mathbf{a} = x\mathbf{i} + y\mathbf{j}$

$|\mathbf{a}| = \sqrt{x^2 + y^2}$ angle with $\mathbf{i}$: $\tan\theta = \dfrac{y}{x}$ (then check the quadrant)

Worked Example 1 — Magnitude and direction

Find the magnitude of $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}$ and the angle it makes with $\mathbf{i}$.

①$|\mathbf{a}| = \sqrt{3^2 + 4^2} = 5$

②$\tan\theta = \tfrac43$, so $\theta = 53.1°$ above the positive $x$-axis.

Worked Example 2 — A vector in the second quadrant

Find the magnitude of $\mathbf{b} = -2\mathbf{i} + 5\mathbf{j}$ and the angle it makes with the positive $x$-direction.

①$|\mathbf{b}| = \sqrt{4 + 25} = \sqrt{29} \approx 5.39$

②Sketch: $\mathbf{b}$ points up and to the left. The angle it makes with the negative $x$-axis is $\tan^{-1}\tfrac52 = 68.2°$.

③So the angle from the positive $x$-direction is $180° - 68.2° = 111.8°$.

Typing $\tan^{-1}\left(\tfrac{5}{-2}\right)$ gives $-68.2°$, which points into the fourth quadrant — the wrong way. Always draw the vector first.

Worked Example 3 — From magnitude and direction to components

(a) A force of $10$ N acts at $30°$ above the positive $x$-axis. Write it in $\mathbf{i}, \mathbf{j}$ form. (b) A boat moves at $12$ km h$^{-1}$ on a bearing of $060°$. With $\mathbf{i}$ east and $\mathbf{j}$ north, write its velocity as a vector.

①(a) $10\cos 30°\,\mathbf{i} + 10\sin 30°\,\mathbf{j} = 5\sqrt3\,\mathbf{i} + 5\mathbf{j}$ N

②(b) A bearing is measured from north, so the angle to $\mathbf{j}$ is $60°$: east component $12\sin 60° = 6\sqrt3$, north component $12\cos 60° = 6$.

③$\mathbf{v} = 6\sqrt3\,\mathbf{i} + 6\mathbf{j}$ km h$^{-1}$

Resolving

Magnitude $r$ at angle $\theta$ to $\mathbf{i}$: $r\cos\theta\,\mathbf{i} + r\sin\theta\,\mathbf{j}$

3 Adding, Scaling and Unit Vectors

Component by component

$(a_1\mathbf{i} + a_2\mathbf{j}) + (b_1\mathbf{i} + b_2\mathbf{j}) = (a_1 + b_1)\mathbf{i} + (a_2 + b_2)\mathbf{j}$ $\lambda(a_1\mathbf{i} + a_2\mathbf{j}) = \lambda a_1\mathbf{i} + \lambda a_2\mathbf{j}$

Geometrically, $\mathbf{a} + \mathbf{b}$ means "do $\mathbf{a}$, then do $\mathbf{b}$" (the triangle law). Multiplying by a scalar $\lambda$ stretches the vector by factor $|\lambda|$, and reverses it if $\lambda < 0$. So $\lambda\mathbf{a}$ is always parallel to $\mathbf{a}$.

Worked Example 4 — A linear combination

$\mathbf{a} = 2\mathbf{i} + \mathbf{j}$ and $\mathbf{b} = -\mathbf{i} + 3\mathbf{j}$. Find $3\mathbf{a} - 2\mathbf{b}$ and its magnitude.

①$3\mathbf{a} = 6\mathbf{i} + 3\mathbf{j}$, $2\mathbf{b} = -2\mathbf{i} + 6\mathbf{j}$

②$3\mathbf{a} - 2\mathbf{b} = 8\mathbf{i} - 3\mathbf{j}$

③$|8\mathbf{i} - 3\mathbf{j}| = \sqrt{64 + 9} = \sqrt{73} \approx 8.54$

Worked Example 5 — Finding unknown scalars

With $\mathbf{a}$ and $\mathbf{b}$ as above, find $p$ and $q$ such that $p\mathbf{a} + q\mathbf{b} = 7\mathbf{i} + 7\mathbf{j}$.

①$\mathbf{i}$: $2p - q = 7$ $\mathbf{j}$: $p + 3q = 7$

②From the first, $q = 2p - 7$. Then $p + 6p - 21 = 7$, so $p = 4$ and $q = 1$.

③Check: $4(2\mathbf{i} + \mathbf{j}) + (-\mathbf{i} + 3\mathbf{j}) = 7\mathbf{i} + 7\mathbf{j}$ ✓

Unit vector in the direction of $\mathbf{a}$

$\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}$ — magnitude $1$, same direction

Worked Example 6 — Unit vector and a given magnitude

Find the unit vector in the direction of $3\mathbf{i} - 4\mathbf{j}$, and a vector of magnitude $15$ in the same direction.

①$|3\mathbf{i} - 4\mathbf{j}| = 5$, so the unit vector is $\tfrac15(3\mathbf{i} - 4\mathbf{j}) = 0.6\mathbf{i} - 0.8\mathbf{j}$.

②Magnitude $15$: $15(0.6\mathbf{i} - 0.8\mathbf{j}) = 9\mathbf{i} - 12\mathbf{j}$.

Worked Example 7 — Parallel vectors

$(k + 1)\mathbf{i} + 4\mathbf{j}$ is parallel to $3\mathbf{i} + 6\mathbf{j}$. Find $k$.

①Parallel means one is a multiple of the other: $(k + 1)\mathbf{i} + 4\mathbf{j} = \lambda(3\mathbf{i} + 6\mathbf{j})$.

②$\mathbf{j}$: $4 = 6\lambda$, so $\lambda = \tfrac23$.

③$\mathbf{i}$: $k + 1 = 3 \times \tfrac23 = 2$, so $k = 1$.

4 Position Vectors and Distances

The position vector of a point $A$ is $\overrightarrow{OA}$, the vector from the origin to $A$, usually written $\mathbf{a}$. The point $A(2, 5)$ has position vector $\mathbf{a} = 2\mathbf{i} + 5\mathbf{j}$.

The key result

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ ("end minus start") distance $AB = |\mathbf{b} - \mathbf{a}|$

Worked Example 8 — Displacement and distance

$A$ and $B$ have position vectors $2\mathbf{i} + 5\mathbf{j}$ and $6\mathbf{i} - \mathbf{j}$. Find $\overrightarrow{AB}$, the exact distance $AB$, and the position vector of the midpoint $M$.

①$\overrightarrow{AB} = (6 - 2)\mathbf{i} + (-1 - 5)\mathbf{j} = 4\mathbf{i} - 6\mathbf{j}$

②$AB = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$

③$\overrightarrow{OM} = \mathbf{a} + \tfrac12\overrightarrow{AB} = (2\mathbf{i} + 5\mathbf{j}) + (2\mathbf{i} - 3\mathbf{j}) = 4\mathbf{i} + 2\mathbf{j}$

In general $\overrightarrow{OM} = \tfrac12(\mathbf{a} + \mathbf{b})$ — the vector form of the midpoint formula from Straight Lines.

5 Quick Reference

Notation

$x\mathbf{i} + y\mathbf{j} = \begin{pmatrix} x \\ y \end{pmatrix}$; underline by hand.

Magnitude

$\sqrt{x^2 + y^2}$

Direction

$\tan\theta = \dfrac{y}{x}$ — sketch for the quadrant.

Resolving

$r\cos\theta\,\mathbf{i} + r\sin\theta\,\mathbf{j}$

Unit vector

$\dfrac{\mathbf{a}}{|\mathbf{a}|}$

Parallel

$\mathbf{a} = \lambda\mathbf{b}$

Equal vectors

Equate $\mathbf{i}$ and $\mathbf{j}$ components.

Position vector

$\overrightarrow{OA} = \mathbf{a}$

Displacement

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

Midpoint

$\tfrac12(\mathbf{a} + \mathbf{b})$

6 Practice Questions

Question 1

Find the magnitude of $5\mathbf{i} - 12\mathbf{j}$.

▶ Show solution

$\sqrt{25 + 144} = \sqrt{169} = 13$

Question 2

$\mathbf{a} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}$. Find $2\mathbf{a} + \mathbf{b}$ and $\mathbf{a} - 3\mathbf{b}$.

▶ Show solution

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

$\mathbf{a} - 3\mathbf{b} = \begin{pmatrix} 3 + 6 \\ 1 - 12 \end{pmatrix} = \begin{pmatrix} 9 \\ -11 \end{pmatrix}$

Question 3

Find the unit vector in the direction of $6\mathbf{i} + 8\mathbf{j}$.

▶ Show solution

$|6\mathbf{i} + 8\mathbf{j}| = 10$, so the unit vector is $0.6\mathbf{i} + 0.8\mathbf{j}$.

Question 4

Find the angle between $2\mathbf{i} - 5\mathbf{j}$ and the positive $x$-direction.

▶ Show solution

The vector points right and down (fourth quadrant).

$\tan^{-1}\tfrac52 = 68.2°$, so it is $68.2°$ below the positive $x$-axis (equivalently $291.8°$ measured anticlockwise).

Question 5

$A(-1, 4)$ and $B(5, -4)$. Find $\overrightarrow{AB}$ and the distance $AB$.

▶ Show solution

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = 6\mathbf{i} - 8\mathbf{j}$

$AB = \sqrt{36 + 64} = 10$

Question 6

Find a vector of magnitude $26$ in the same direction as $5\mathbf{i} - 12\mathbf{j}$.

▶ Show solution

$|5\mathbf{i} - 12\mathbf{j}| = 13$, so multiply by $\tfrac{26}{13} = 2$: $10\mathbf{i} - 24\mathbf{j}$.

Question 7

Find $p$ and $q$ such that $p(\mathbf{i} + 2\mathbf{j}) + q(3\mathbf{i} - \mathbf{j}) = 11\mathbf{i} + \mathbf{j}$.

▶ Show solution

$\mathbf{i}$: $p + 3q = 11$; $\mathbf{j}$: $2p - q = 1$, so $q = 2p - 1$.

$p + 6p - 3 = 11 \;\Rightarrow\; p = 2$, $q = 3$.

Question 8

The vector $k\mathbf{i} + 3\mathbf{j}$ has magnitude $5$. Find the possible values of $k$.

▶ Show solution

$k^2 + 9 = 25 \;\Rightarrow\; k^2 = 16 \;\Rightarrow\; k = \pm 4$.

Question 9

$ABCD$ is a parallelogram with $A(1, 2)$, $B(5, 3)$ and $C(7, 8)$. Find the position vector of $D$.

▶ Show solution

In a parallelogram $\overrightarrow{AD} = \overrightarrow{BC} = \mathbf{c} - \mathbf{b} = 2\mathbf{i} + 5\mathbf{j}$.

$\mathbf{d} = \mathbf{a} + \overrightarrow{AD} = (\mathbf{i} + 2\mathbf{j}) + (2\mathbf{i} + 5\mathbf{j}) = 3\mathbf{i} + 7\mathbf{j}$.

Check: $\overrightarrow{DC} = 4\mathbf{i} + \mathbf{j} = \overrightarrow{AB}$ ✓

Question 10

A drone takes off from $O$. It flies on a displacement of $(3\mathbf{i} + 4\mathbf{j})$ km to $A$, then $(5\mathbf{i} - 10\mathbf{j})$ km to $B$, where $\mathbf{i}$ is east and $\mathbf{j}$ is north.

(a) Find the position vector of $B$. (b) Find the straight-line distance $OB$. (c) Find the bearing of $B$ from $O$. (d) Find the total distance flown, and how much further this is than flying directly from $O$ to $B$.

▶ Show solution

(a) $\overrightarrow{OB} = (3 + 5)\mathbf{i} + (4 - 10)\mathbf{j} = 8\mathbf{i} - 6\mathbf{j}$ km.

(b) $|8\mathbf{i} - 6\mathbf{j}| = \sqrt{64 + 36} = 10$ km.

(c) $B$ is east and south of $O$. The angle below east is $\tan^{-1}\tfrac68 = 36.9°$.

Bearing $= 090° + 36.9° = 126.9°$, about $127°$.

(d) $|3\mathbf{i} + 4\mathbf{j}| = 5$ and $|5\mathbf{i} - 10\mathbf{j}| = \sqrt{125} = 5\sqrt5$.

Total $= 5 + 5\sqrt5 \approx 16.18$ km, which is $16.18 - 10 = 6.18$ km further than the direct route.

Vectors in Two Dimensions · A Level Maths Year 1 · Created with MathJax