A vector is a quantity with both magnitude and direction. Writing vectors as $x\mathbf{i} + y\mathbf{j}$ lets you add, scale and compare them with ordinary arithmetic — and that turns geometric facts ("these lines are parallel", "these points are in a line") into simple algebra.
Position vectors and distances · dividing a line in a ratio
Vectors in mechanics: resultant forces, equilibrium and velocities
| Word | Meaning | Example |
|---|---|---|
| Scalar | A quantity with size only. | Speed $5$ m s$^{-1}$, mass $2$ kg |
| Vector | A quantity with size and direction. | Velocity $3\mathbf{i} + 4\mathbf{j}$ m s$^{-1}$ |
| Magnitude | The length (size) of a vector, $|\mathbf{a}|$. | $|3\mathbf{i} + 4\mathbf{j}| = 5$ |
| Unit vector | A vector of magnitude $1$. | $\mathbf{i}$, $\mathbf{j}$, $0.6\mathbf{i} + 0.8\mathbf{j}$ |
| Position vector | The vector from the origin to a point. | $A(2, 5)$ has $\mathbf{a} = 2\mathbf{i} + 5\mathbf{j}$ |
| Displacement | The vector from one point to another. | $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ |
| Collinear | Lying on a single straight line. | $\overrightarrow{AB} = 2\overrightarrow{BC}$ |
| Resultant | The single vector equivalent to several added together. | The sum of all forces on a particle |
$A$ and $B$ have position vectors $\mathbf{i} + 2\mathbf{j}$ and $7\mathbf{i} + 5\mathbf{j}$. $P$ lies on $AB$ with $AP : PB = 2 : 1$. Find the position vector of $P$, the exact distance $OP$, and a unit vector in the direction of $\overrightarrow{OP}$.
The Year 1 content of DfE area J, and the page that covers it.
- JVectors in Two Dimensions$\mathbf{i}$, $\mathbf{j}$ and column notation; magnitude and direction; unit vectors; position vectors and distances; addition and scalar multiples.
- JVector GeometryParallel vectors and collinear points; dividing a line in a ratio; geometric proof; forces and velocities.
Notation
$x\mathbf{i} + y\mathbf{j} = \begin{pmatrix} x \\ y \end{pmatrix}$
Magnitude
$\sqrt{x^2 + y^2}$
Unit vector
$\mathbf{a} \div |\mathbf{a}|$
Displacement
$\mathbf{b} - \mathbf{a}$
Midpoint
$\tfrac12(\mathbf{a} + \mathbf{b})$
Parallel
$\mathbf{a} = \lambda\mathbf{b}$
Collinear
Parallel plus a common point.
Resolving
$r\cos\theta\,\mathbf{i} + r\sin\theta\,\mathbf{j}$
Equilibrium
Forces sum to $\mathbf{0}$.
Ten questions drawing on the whole Vectors topic. Work each one through before opening the solution.
Find $|-6\mathbf{i} + 8\mathbf{j}|$.
▶ Show solution
$\sqrt{36 + 64} = 10$
$\mathbf{a} = 2\mathbf{i} - \mathbf{j}$ and $\mathbf{b} = -3\mathbf{i} + 4\mathbf{j}$. Find $3\mathbf{a} - \mathbf{b}$.
▶ Show solution
$(6\mathbf{i} - 3\mathbf{j}) - (-3\mathbf{i} + 4\mathbf{j}) = 9\mathbf{i} - 7\mathbf{j}$
$A(3, -2)$ and $B(-1, 1)$. Find $\overrightarrow{AB}$ and the distance $AB$.
▶ Show solution
$\overrightarrow{AB} = -4\mathbf{i} + 3\mathbf{j}$; $AB = \sqrt{16 + 9} = 5$.
Find the unit vector in the direction of $5\mathbf{i} + 12\mathbf{j}$.
▶ Show solution
Magnitude $13$, so $\tfrac{1}{13}(5\mathbf{i} + 12\mathbf{j}) = \tfrac{5}{13}\mathbf{i} + \tfrac{12}{13}\mathbf{j}$.
Write a velocity of $8$ m s$^{-1}$ on a bearing of $150°$ in the form $x\mathbf{i} + y\mathbf{j}$, where $\mathbf{i}$ is east and $\mathbf{j}$ is north. Give exact values.
▶ Show solution
$150°$ is $30°$ east of south.
East component $8\sin 30° = 4$; south component $8\cos 30° = 4\sqrt3$.
$\mathbf{v} = 4\mathbf{i} - 4\sqrt3\,\mathbf{j}$ m s$^{-1}$.
$2\mathbf{i} + k\mathbf{j}$ is parallel to $6\mathbf{i} + 15\mathbf{j}$. Find $k$.
▶ Show solution
$2\mathbf{i} + k\mathbf{j} = \tfrac13(6\mathbf{i} + 15\mathbf{j})$, so $k = 5$.
Show that $(1, 1)$, $(3, 4)$ and $(7, 10)$ are collinear.
▶ Show solution
Call them $A$, $B$, $C$. $\overrightarrow{AB} = 2\mathbf{i} + 3\mathbf{j}$, $\overrightarrow{BC} = 4\mathbf{i} + 6\mathbf{j} = 2\overrightarrow{AB}$.
Parallel, with common point $B$, so collinear.
$A(-2, 6)$ and $B(6, -2)$. $P$ lies on $AB$ with $AP : PB = 1 : 3$. Find $P$.
▶ Show solution
$\overrightarrow{AB} = 8\mathbf{i} - 8\mathbf{j}$; $P$ is $\tfrac14$ of the way along.
$\overrightarrow{OP} = (-2\mathbf{i} + 6\mathbf{j}) + (2\mathbf{i} - 2\mathbf{j}) = 4\mathbf{j}$, so $P(0, 4)$.
Forces $(3\mathbf{i} - 4\mathbf{j})$ N and $(-\mathbf{i} + 7\mathbf{j})$ N act on a particle. Find the force $\mathbf{F}$ which must be added for equilibrium, and its magnitude.
▶ Show solution
Resultant so far: $2\mathbf{i} + 3\mathbf{j}$. So $\mathbf{F} = (-2\mathbf{i} - 3\mathbf{j})$ N.
$|\mathbf{F}| = \sqrt{4 + 9} = \sqrt{13} \approx 3.61$ N.
$OABC$ is a parallelogram with $\overrightarrow{OA} = 4\mathbf{i} + \mathbf{j}$ and $\overrightarrow{OC} = \mathbf{i} + 3\mathbf{j}$.
(a) Find $\overrightarrow{OB}$ and $\overrightarrow{AC}$. (b) Find the exact lengths of the two diagonals. (c) Show that the midpoint of $OB$ is also the midpoint of $AC$. (d) Show that $OABC$ is not a rectangle.
▶ Show solution
(a) $\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = (4\mathbf{i} + \mathbf{j}) + (\mathbf{i} + 3\mathbf{j}) = 5\mathbf{i} + 4\mathbf{j}$.
$\overrightarrow{AC} = \mathbf{c} - \mathbf{a} = -3\mathbf{i} + 2\mathbf{j}$.
(b) $OB = \sqrt{25 + 16} = \sqrt{41}$; $AC = \sqrt{9 + 4} = \sqrt{13}$.
(c) Midpoint of $OB$: $\tfrac12(5\mathbf{i} + 4\mathbf{j}) = 2.5\mathbf{i} + 2\mathbf{j}$.
Midpoint of $AC$: $\mathbf{a} + \tfrac12\overrightarrow{AC} = (4\mathbf{i} + \mathbf{j}) + (-1.5\mathbf{i} + \mathbf{j}) = 2.5\mathbf{i} + 2\mathbf{j}$ ✓
(d) A rectangle's diagonals are equal in length, but $\sqrt{41} \neq \sqrt{13}$, so $OABC$ is not a rectangle.