What is a vector?
Many quantities in mathematics and physics have only a size — temperature, mass, speed. These are called scalars. But other quantities have both a size and a direction — displacement, velocity, force. These are called vectors.
A vector can be represented by a directed line segment: an arrow whose length gives the magnitude (size) and whose direction gives the direction. In print, vectors are written in bold: \(\mathbf{a}\), \(\mathbf{b}\). In handwriting, they are underlined: \(\underline{a}\), \(\underline{b}\).
Where do vectors appear?
The three examples below all need vectors to be described precisely.
Navigation A girl walks 2 km east then 3 km south. Her total displacement combines both distance and direction — a scalar distance alone cannot describe where she ends up.
The parallelogram law Two vectors \(\mathbf{a}\) and \(\mathbf{b}\) can be added using the parallelogram rule. The resultant \(\mathbf{a} + \mathbf{b}\) is the diagonal of the parallelogram they form.
Geometry proofs Vectors allow us to prove geometric results algebraically — for example, that medians of a triangle bisect each other, or that the diagonals of a parallelogram bisect each other.
Choose a topic to explore:
Directed line segments
A vector is represented by a directed line segment — an arrow from a start point to an end point. The vector from \(P\) to \(Q\) is written \(\overrightarrow{PQ}\).
Two vectors are equal if they have the same magnitude and direction — they do not need to start from the same point.
Negative vectors and zero vectors
The vector \(-\mathbf{a}\) has the same magnitude as \(\mathbf{a}\) but the opposite direction. In terms of a directed line segment: \(\overrightarrow{BA} = -\overrightarrow{AB}\).
If you travel from \(P\) to \(Q\) and back again, your total displacement is the zero vector \(\mathbf{0}\):
\[\overrightarrow{PQ} + \overrightarrow{QP} = \mathbf{0}\]The triangle law for vector addition
To add vectors \(\mathbf{a}\) and \(\mathbf{b}\), place them nose-to-tail. The resultant goes from the start of \(\mathbf{a}\) to the end of \(\mathbf{b}\):
\[\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}\]Parallel vectors
A vector parallel to \(\mathbf{a}\) can be written as \(\lambda\mathbf{a}\), where \(\lambda\) is a non-zero scalar (real number).
| Expression | Effect |
|---|---|
| \(3\mathbf{a}\) | Same direction as \(\mathbf{a}\), three times as long |
| \(-2\mathbf{a}\) | Opposite direction to \(\mathbf{a}\), twice as long |
| \(\frac{1}{2}\mathbf{a}\) | Same direction as \(\mathbf{a}\), half as long |
Worked Example — Triangle law (bookwork style)
In the diagram, \(\overrightarrow{QP} = \mathbf{a}\), \(\overrightarrow{QR} = \mathbf{b}\), \(\overrightarrow{QS} = \mathbf{c}\) and \(\overrightarrow{RT} = \mathbf{d}\). Express each of the following in terms of \(\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}\):
(a) \(\overrightarrow{PS}\) (b) \(\overrightarrow{RP}\) (c) \(\overrightarrow{PT}\)
Worked Example — Parallelogram
\(ABCD\) is a parallelogram. \(\overrightarrow{AB} = \mathbf{a}\) and \(\overrightarrow{AD} = \mathbf{b}\). Find \(\overrightarrow{AC}\).
Subtracting vectors
Subtracting \(\mathbf{b}\) from \(\mathbf{a}\) is the same as adding \(-\mathbf{b}\):
\[\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b})\]Geometrically, reverse the direction of \(\mathbf{b}\) then add it nose-to-tail after \(\mathbf{a}\).
Scalar multiplication
Multiplying a vector by a positive scalar \(k\) stretches it by factor \(k\) without changing direction. A negative scalar reverses the direction as well.
Midpoints using vectors
If \(M\) is the midpoint of \(AB\), then \(\overrightarrow{AM} = \frac{1}{2}\overrightarrow{AB}\).
More generally, if \(P\) divides \(AB\) in the ratio \(\lambda : \mu\), then:
\[\overrightarrow{AP} = \frac{\lambda}{\lambda + \mu}\,\overrightarrow{AB}\]Worked Example — Midpoints in a triangle
In triangle \(ABC\), \(\overrightarrow{AB} = \mathbf{a}\) and \(\overrightarrow{AC} = \mathbf{b}\). \(P\) is the midpoint of \(AB\) and \(Q\) divides \(AC\) in the ratio \(3:2\). Find, in terms of \(\mathbf{a}\) and \(\mathbf{b}\): (a) \(\overrightarrow{BC}\) (b) \(\overrightarrow{AP}\) (c) \(\overrightarrow{AQ}\) (d) \(\overrightarrow{PQ}\)
Column vector notation
A vector can be described by its horizontal and vertical components. We write it as a column vector:
\[\mathbf{a} = \begin{pmatrix} x \\ y \end{pmatrix}\]where \(x\) is the change in the horizontal direction and \(y\) is the change in the vertical direction.
Operations in column vector form
To add or subtract column vectors, operate on each component separately:
\[\begin{pmatrix} p \\ q \end{pmatrix} + \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p+r \\ q+s \end{pmatrix}, \qquad \lambda\begin{pmatrix} p \\ q \end{pmatrix} = \begin{pmatrix} \lambda p \\ \lambda q \end{pmatrix}\]Unit vectors i and j
The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) point along the positive \(x\)- and \(y\)-axes respectively:
\[\mathbf{i} = \begin{pmatrix}1\\0\end{pmatrix}, \qquad \mathbf{j} = \begin{pmatrix}0\\1\end{pmatrix}\]Any 2D vector can be written as \(p\mathbf{i} + q\mathbf{j}\), which is equivalent to the column vector \(\begin{pmatrix}p\\q\end{pmatrix}\).
Worked Example — Column vector arithmetic
Given \(\mathbf{a} = 3\mathbf{i} - 4\mathbf{j}\) and \(\mathbf{b} = 2\mathbf{i} + 7\mathbf{j}\), find: (a) \(\frac{1}{2}\mathbf{a}\) (b) \(\mathbf{a} + \mathbf{b}\) (c) \(3\mathbf{a} - 2\mathbf{b}\)
Worked Example — Parallel vectors
Show that the vectors \(6\mathbf{a} + 8\mathbf{b}\) and \(9\mathbf{a} + 12\mathbf{b}\) are parallel.
Magnitude of a vector
The magnitude (or modulus) of a vector is its length. For the vector \(\mathbf{a} = x\mathbf{i} + y\mathbf{j} = \begin{pmatrix}x\\y\end{pmatrix}\), Pythagoras' theorem gives:
\[|\mathbf{a}| = \sqrt{x^2 + y^2}\]The magnitude is always a non-negative scalar.
Unit vectors
A unit vector has magnitude 1. To find a unit vector in the direction of \(\mathbf{a}\), divide by its magnitude:
\[\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}\]The notation \(\hat{\mathbf{a}}\) (read "a hat") denotes a unit vector in the direction of \(\mathbf{a}\).
Direction of a vector
The direction is given by the angle the vector makes with a reference axis (usually the positive \(x\)-axis or, in navigation problems, North). For \(\mathbf{a} = x\mathbf{i} + y\mathbf{j}\):
\[\tan\theta = \frac{y}{x}\]Always draw a sketch to identify which angle you need — the formula gives the angle in the right-angled triangle, which may need adjusting by 90°, 180°, etc.
Worked Example — Magnitude, unit vector and angle
Given \(\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}\) and \(\mathbf{b} = -2\mathbf{i} - 4\mathbf{j}\): (a) Find \(|\mathbf{a}|\) (b) Find a unit vector in the direction of \(\mathbf{a}\) (c) Find the exact value of \(|2\mathbf{a} + \mathbf{b}|\)
Worked Example — Angle with the positive x-axis
Find the angle the vector \(-5\mathbf{i} + 2\mathbf{j}\) makes with the positive \(x\)-axis, to 1 d.p.
Worked Example — Navigation (bearings)
A girl walks 2 km due east from a fixed point \(O\) to \(A\), then 3 km due south from \(A\) to \(B\). Find: (a) total distance walked (b) \(\overrightarrow{OB}\) (c) \(|\overrightarrow{OB}|\) (d) bearing of \(B\) from \(O\).
(In mechanics/navigation, \(\mathbf{i}\) = East, \(\mathbf{j}\) = North.)
What is a position vector?
A position vector gives the location of a point relative to a fixed origin \(O\). The position vector of point \(A\) is \(\overrightarrow{OA}\).
If \(\overrightarrow{OA} = a\mathbf{i} + b\mathbf{j}\), then \(A\) has coordinates \((a, b)\) and position vector \(\begin{pmatrix}a\\b\end{pmatrix}\).
Dividing a line segment in a ratio
If point \(P\) divides \(AB\) in the ratio \(\lambda : \mu\), then the position vector of \(P\) is:
\[\overrightarrow{OP} = \overrightarrow{OA} + \frac{\lambda}{\lambda+\mu}\overrightarrow{AB} = \frac{\mu\,\overrightarrow{OA} + \lambda\,\overrightarrow{OB}}{\lambda + \mu}\]Special case — midpoint (\(\lambda = \mu = 1\)):
\[\overrightarrow{OM} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2}\]Worked Example — Position vectors and distance
\(\overrightarrow{OA} = 5\mathbf{i} - 2\mathbf{j}\) and \(\overrightarrow{AB} = 3\mathbf{i} + 4\mathbf{j}\). Find: (a) the position vector of \(B\) (b) \(|\overrightarrow{OB}|\) in simplified surd form.
Worked Example — Point dividing a segment
\(OABC\) is a parallelogram. \(\overrightarrow{OA} = 2\mathbf{i} + 4\mathbf{j}\), \(\overrightarrow{OC} = 7\mathbf{i}\). \(P\) divides \(AC\) in the ratio \(3:2\). Find, in \(\mathbf{i},\mathbf{j}\) format and column vector format: (a) \(\overrightarrow{AC}\) (b) \(\overrightarrow{OP}\)
Proving geometric results with vectors
Vectors are a powerful tool for proving geometric results. The key technique is to express unknown vectors in terms of known ones, using the triangle law and the property of parallel vectors.
| Geometric fact | Vector statement |
|---|---|
| \(AB \parallel CD\) | \(\overrightarrow{CD} = \lambda\,\overrightarrow{AB}\) for some scalar \(\lambda\) |
| \(A\), \(B\), \(C\) are collinear | \(\overrightarrow{AB} = \lambda\,\overrightarrow{AC}\) for some scalar \(\lambda\) |
| \(M\) is midpoint of \(AB\) | \(\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OB})\) |
| \(AB = CD\) (equal length & direction) | \(\overrightarrow{AB} = \overrightarrow{CD}\) |
Worked Example — Diagonals of a parallelogram bisect each other
\(OABC\) is a parallelogram. The vectors \(\mathbf{a}\) and \(\mathbf{c}\) are equal to \(\overrightarrow{OA}\) and \(\overrightarrow{OC}\) respectively. \(P\) is the point where diagonals \(OB\) and \(AC\) intersect. Prove that the diagonals bisect each other.
Worked Example — Collinearity
\(OAB\) is a triangle. \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). The point \(M\) divides \(OA\) in the ratio \(3:2\) and \(MN\) is parallel to \(OB\). Show that \(A\), \(N\) and \(B\) are collinear and find \(AN:NB\).
Exam-style Problem — Parallelogram with a trapezium
\(ABCD\) is a trapezium with \(AB \parallel DC\) and \(DC = 4AB\). \(M\) divides \(DC\) such that \(DM:MC = 3:2\). \(\overrightarrow{AB} = \mathbf{a}\) and \(\overrightarrow{BC} = \mathbf{b}\). Find \(\overrightarrow{AM}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).