A vector has both size and direction. This note shows how to write vectors as columns, add and subtract them, multiply them by a number, find their magnitude, and use them to prove geometric facts such as parallel lines and points lying on a straight line. Vectors are used for forces, velocities and movement in computer graphics.
Key ideas
A scalar has size only (for example, a mass of 5 kg). A vector has size and direction (for example, a velocity of 5 m/s north). On a diagram a vector is an arrow. The vector from point \(A\) to point \(B\) is written \(\overrightarrow{AB}\). A single vector can also be named with a bold or underlined small letter, such as \(\mathbf{a}\).
Column vectors. The vector \(\begin{pmatrix} x \\ y \end{pmatrix}\) means move \(x\) units right (left if negative) and \(y\) units up (down if negative). Two vectors are equal if they have the same size and direction, wherever they are drawn.
Adding and subtracting. Add or subtract the matching components:
\(\begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a + c \\ b + d \end{pmatrix}\). On a diagram, put the vectors nose to tail, and the sum is the single arrow from the start to the end. Reversing a vector changes its sign: \(\overrightarrow{BA} = -\overrightarrow{AB}\).
Multiplying by a scalar. \(k\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} ka \\ kb \end{pmatrix}\). If \(k\) is positive the direction is the same, and if \(k\) is negative the direction is reversed. The vector \(2\mathbf{a}\) is parallel to \(\mathbf{a}\) and twice as long.
Magnitude. The magnitude (length) of \(\begin{pmatrix} x \\ y \end{pmatrix}\) is written \(\left|\begin{pmatrix} x \\ y \end{pmatrix}\right|\) and found with Pythagoras: \(\sqrt{x^2 + y^2}\). Magnitude is a scalar and is never negative.
Position vectors. The position vector of a point \(P\) is \(\overrightarrow{OP}\), the vector from the origin \(O\) to \(P\). If \(P\) has coordinates \((x, y)\), then \(\overrightarrow{OP} = \begin{pmatrix} x \\ y \end{pmatrix}\). The vector from \(A\) to \(B\) is found by going back to the origin and then out to \(B\):
\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = \overrightarrow{OB} - \overrightarrow{OA}\).
Vector geometry. To solve a geometry problem, write the unknown vector as a journey along known vectors, using several routes if you need to. Useful facts:
- If one vector is a scalar multiple of another, they are parallel.
- If two vectors are parallel and share a point, the three points lie on a straight line (collinear).
- If \(M\) is the midpoint of \(AB\), then \(\overrightarrow{OM} = \dfrac12(\overrightarrow{OA} + \overrightarrow{OB})\).
- If \(P\) divides \(AB\) in the ratio \(m:n\), then \(\overrightarrow{AP} = \dfrac{m}{m + n}\overrightarrow{AB}\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\overrightarrow{AB}\) | The vector from \(A\) to \(B\). |
| \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\) | Find the vector between two points from their position vectors. |
| \(\overrightarrow{BA} = -\overrightarrow{AB}\) | Reversing a journey reverses the sign. |
| \(k\begin{pmatrix} a \\ b \end{pmatrix}\) | Scalar multiple: multiply both components by \(k\). |
| \(\left|\begin{pmatrix} x \\ y \end{pmatrix}\right| = \sqrt{x^2 + y^2}\) | Magnitude (length) of a vector. |
| Parallel vectors | One is a scalar multiple of the other. |
| Collinear points | Points on one straight line. |
Worked examples
Example 1 Given \(\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} -1 \\ 5 \end{pmatrix}\), find (a) \(2\mathbf{a} - 3\mathbf{b}\) and (b) the magnitude of \(\mathbf{a} + \mathbf{b}\).
(a) \(2\mathbf{a} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}\) and \(3\mathbf{b} = \begin{pmatrix} -3 \\ 15 \end{pmatrix}\).
\(2\mathbf{a} - 3\mathbf{b} = \begin{pmatrix} 6 - (-3) \\ -4 - 15 \end{pmatrix} = \begin{pmatrix} 9 \\ -19 \end{pmatrix}\).
(b) \(\mathbf{a} + \mathbf{b} = \begin{pmatrix} 3 + (-1) \\ -2 + 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\).
Magnitude \(= \sqrt{2^2 + 3^2} = \sqrt{13} = 3.605\ldots\)
Answer: (a) \(\begin{pmatrix} 9 \\ -19 \end{pmatrix}\). (b) \(\sqrt{13} = 3.61\).
Example 2 A drone flies from point \(A(2, 1)\) to point \(B(8, 5)\), where the coordinates are in hundreds of metres. Find \(\overrightarrow{AB}\) and the distance flown, in metres.
\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 8 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}\).
Length \(= \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 7.211\ldots\) hundred metres.
Distance \(= 7.211 \times 100 = 721.1\ldots\) m.
Answer: \(\overrightarrow{AB} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}\), distance 721 m.
Example 3 In triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). The point \(P\) lies on \(AB\) so that \(AP : PB = 1 : 2\). The point \(Q\) is such that \(\overrightarrow{OQ} = 2\mathbf{a} + \mathbf{b}\). Show that \(O\), \(P\) and \(Q\) lie on a straight line.
First find \(\overrightarrow{AB}\): \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}\).
\(P\) is one third of the way from \(A\) to \(B\), since the ratio is \(1:2\) and \(1 + 2 = 3\). So \(\overrightarrow{AP} = \dfrac13(\mathbf{b} - \mathbf{a})\).
Now go from \(O\) to \(P\) via \(A\): \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \mathbf{a} + \dfrac13(\mathbf{b} - \mathbf{a}) = \dfrac23\mathbf{a} + \dfrac13\mathbf{b} = \dfrac13(2\mathbf{a} + \mathbf{b})\).
Since \(\overrightarrow{OQ} = 2\mathbf{a} + \mathbf{b}\), we have \(\overrightarrow{OQ} = 3\overrightarrow{OP}\).
So \(\overrightarrow{OQ}\) is a scalar multiple of \(\overrightarrow{OP}\), which means they are parallel. Both vectors start at \(O\), so \(O\), \(P\) and \(Q\) are on one straight line. Also \(OQ\) is three times as long as \(OP\).
Answer: shown, because \(\overrightarrow{OQ} = 3\overrightarrow{OP}\) and both start at \(O\).
Common errors
- Common error: writing \(\overrightarrow{AB} = \overrightarrow{OA} - \overrightarrow{OB}\). It must be end minus start: \(\overrightarrow{OB} - \overrightarrow{OA}\).
- Common error: forgetting that subtracting a negative component adds, for example \(6 - (-3) = 9\).
- Common error: adding magnitudes instead of vectors. In general \(|\mathbf{a} + \mathbf{b}|\) is not \(|\mathbf{a}| + |\mathbf{b}|\).
- Common error: showing two vectors are parallel but not saying they share a point, then claiming the points are collinear.
- Common error: giving a negative magnitude, or not square-rooting at the end.
Exam tip: In a proof, write each journey step by step, such as \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP}\), and simplify at the end. State your conclusion in words, such as "parallel and share the point \(O\), so collinear".
Check yourself
- Find \(k\) so that \(\begin{pmatrix} k \\ 6 \end{pmatrix}\) is parallel to \(\begin{pmatrix} 2 \\ 3 \end{pmatrix}\).
Answer
The second component is multiplied by 2 (\(3 \times 2 = 6\)), so the first must be too: \(k = 2 \times 2 = 4\). - Find the magnitude of \(\begin{pmatrix} -8 \\ 6 \end{pmatrix}\).
Answer
\(\sqrt{(-8)^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\). - \(A\) has position vector \(\begin{pmatrix} 1 \\ 7 \end{pmatrix}\) and \(B\) has position vector \(\begin{pmatrix} 5 \\ -1 \end{pmatrix}\). Find \(\overrightarrow{AB}\) and the position vector of the midpoint of \(AB\).
Answer
\(\overrightarrow{AB} = \begin{pmatrix} 5 - 1 \\ -1 - 7 \end{pmatrix} = \begin{pmatrix} 4 \\ -8 \end{pmatrix}\). Midpoint \(= \tfrac12\left(\begin{pmatrix} 1 \\ 7 \end{pmatrix} + \begin{pmatrix} 5 \\ -1 \end{pmatrix}\right) = \tfrac12\begin{pmatrix} 6 \\ 6 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}\). - In triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(M\) is the midpoint of \(AB\). Write \(\overrightarrow{OM}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
Answer
\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\), so \(\overrightarrow{AM} = \tfrac12(\mathbf{b} - \mathbf{a})\). Then \(\overrightarrow{OM} = \mathbf{a} + \tfrac12(\mathbf{b} - \mathbf{a}) = \tfrac12(\mathbf{a} + \mathbf{b})\).