Vectors in IGCSE Maths
Column vectors, position vectors, magnitude, and the geometry proof questions that appear on every Paper 4 series.
0580 Extended · E7.1 - E7.3Column vectors & notation
A vector is a quantity with both magnitude (size) and direction. In IGCSE Maths we represent vectors as column vectors:
\[ \begin{pmatrix} x \\ y \end{pmatrix} \]The top number gives horizontal movement (positive = right, negative = left). The bottom number gives vertical movement (positive = up, negative = down).
\(\vec{AB}\) — the vector from point \(A\) to point \(B\).
\(\mathbf{a}\) — a named vector, printed in bold. When writing by hand, underline it instead: \(\underline{a}\,\).
Reversing direction flips the signs. If \(\vec{AB} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}\) then \(\vec{BA} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}\).
Operations
Work component by component:
\[ \begin{aligned} \begin{pmatrix} a \\ b \end{pmatrix} \pm \begin{pmatrix} c \\ d \end{pmatrix} &= \begin{pmatrix} a \pm c \\ b \pm d \end{pmatrix} \end{aligned} \]Geometrically, adding vectors places them head to tail.
Multiply each component by the scalar:
\[ k\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} ka \\ kb \end{pmatrix} \]If \(\mathbf{b} = k\mathbf{a}\) for some non-zero scalar \(k\), the two vectors are parallel. This underpins most proof questions.
Given \(\mathbf{a} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} -2 \\ 5 \end{pmatrix}\), find \(3\mathbf{a} - 2\mathbf{b}\).
Scale each vector:
\[ \begin{aligned} 3\mathbf{a} &= \begin{pmatrix} 12 \\ -3 \end{pmatrix} \\[6pt] 2\mathbf{b} &= \begin{pmatrix} -4 \\ 10 \end{pmatrix} \end{aligned} \]Subtract:
\[ \begin{aligned} 3\mathbf{a} - 2\mathbf{b} &= \begin{pmatrix} 12 - (-4) \\ -3 - 10 \end{pmatrix} \\[6pt] &= \begin{pmatrix} 16 \\ -13 \end{pmatrix} \end{aligned} \]Magnitude
The magnitude (length) of a vector uses Pythagoras:
\[ \left| \begin{pmatrix} x \\ y \end{pmatrix} \right| = \sqrt{x^2 + y^2} \]On the non-calculator paper, leave your answer as a surd if it doesn't simplify to a whole number.
Find \(|\mathbf{v}|\) where \(\mathbf{v} = \begin{pmatrix} -5 \\ 12 \end{pmatrix}\).
Position vectors
The position vector of a point \(P\) is the vector from the origin \(O\) to \(P\). If \(P = (3, 7)\) then \(\vec{OP} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}\).
For two points \(A\) and \(B\) with position vectors \(\mathbf{a}\) and \(\mathbf{b}\):
\[ \vec{AB} = \mathbf{b} - \mathbf{a} \]Destination minus start.
\(P = (2,\, 5)\) and \(Q = (8,\, -1)\). Find \(\vec{PQ}\).
Apply \(\vec{PQ} = \mathbf{q} - \mathbf{p}\):
\[ \begin{aligned} \vec{PQ} &= \begin{pmatrix} 8 \\ -1 \end{pmatrix} - \begin{pmatrix} 2 \\ 5 \end{pmatrix} \\[6pt] &= \begin{pmatrix} 6 \\ -6 \end{pmatrix} \end{aligned} \]Vector geometry
Paper 4 vector questions give you a shape with labelled vectors and ask you to express other vectors in terms of \(\mathbf{a}\) and \(\mathbf{b}\). The approach is always the same:
Pick any path from start to end through known vectors. Going with the arrow is positive, against the arrow is negative. Add them up.
The midpoint result \(\frac{1}{2}(\mathbf{a} + \mathbf{b})\) appears frequently.
More generally, if a point divides \(AB\) in the ratio \(m : n\), its position vector is \(\dfrac{n\mathbf{a} + m\mathbf{b}}{m + n}\).
In triangle \(OAB\), \(\vec{OA} = \mathbf{a}\) and \(\vec{OB} = \mathbf{b}\). \(M\) is the midpoint of \(AB\). Find \(\vec{OM}\).
Find \(\vec{AB}\):
\[ \vec{AB} = \mathbf{b} - \mathbf{a} \]Find \(\vec{AM}\) (halfway along \(AB\)):
\[ \vec{AM} = \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) \]Route \(O \to A \to M\):
\[ \begin{aligned} \vec{OM} &= \mathbf{a} + \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) \\[6pt] &= \tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b} \\[6pt] &= \boxed{\tfrac{1}{2}(\mathbf{a} + \mathbf{b})} \end{aligned} \]Proving collinearity
To show that three points \(X\), \(Y\), \(Z\) lie on a straight line, find two vectors that share a common point (e.g. \(\vec{XY}\) and \(\vec{XZ}\)). If one is a scalar multiple of the other, they are parallel. Since they share point \(X\), the three points must be collinear. State both facts for full marks.
\(\vec{OD} = \mathbf{a} + 2\mathbf{b}\), \(\vec{OE} = 2\mathbf{a}\), \(\vec{OC} = 3\mathbf{a} - 2\mathbf{b}\). Show \(D\), \(E\), \(C\) are collinear.
Find \(\vec{DE}\):
\[ \vec{DE} = 2\mathbf{a} - (\mathbf{a} + 2\mathbf{b}) = \mathbf{a} - 2\mathbf{b} \]Find \(\vec{DC}\):
\[ \begin{aligned} \vec{DC} &= (3\mathbf{a} - 2\mathbf{b}) - (\mathbf{a} + 2\mathbf{b}) \\[6pt] &= 2\mathbf{a} - 4\mathbf{b} \end{aligned} \]Compare:
\[ \vec{DC} = 2(\mathbf{a} - 2\mathbf{b}) = 2\,\vec{DE} \]\(\vec{DC}\) is a scalar multiple of \(\vec{DE}\) (parallel), and they share point \(D\), so \(D\), \(E\), \(C\) are collinear.
Exam-style question
\(OABC\) is a parallelogram.
\[ \begin{aligned} \vec{OA} &= \begin{pmatrix} 1 \\ 2 \end{pmatrix} \\[6pt] \vec{OC} &= \begin{pmatrix} 4 \\ 0 \end{pmatrix} \end{aligned} \]\(X\) is the point on \(OB\) such that \(OX = kOB\), where \(0 < k < 1\).
(a) Find the vector \(\vec{OB}\) as a column vector. [1]
(b) Find, in terms of \(k\): (i) \(\vec{OX}\) (ii) \(\vec{AX}\) (iii) \(\vec{XC}\) [3]
(c) Find the value of \(k\) for which \(\vec{AX} = \vec{XC}\). [2]
(d) Use your answer to part (c) to show that the diagonals of parallelogram \(OABC\) bisect one another. [2]
Part (a)
In a parallelogram, \(\vec{AB} = \vec{OC}\). So:
\[ \begin{aligned} \vec{OB} &= \vec{OA} + \vec{AB} \\[6pt] &= \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \begin{pmatrix} 4 \\ 0 \end{pmatrix} \\[6pt] &= \begin{pmatrix} 5 \\ 2 \end{pmatrix} \end{aligned} \]Part (b)(i)
\(X\) lies on \(OB\) with \(OX = k \cdot OB\):
\[ \begin{aligned} \vec{OX} &= k\begin{pmatrix} 5 \\ 2 \end{pmatrix} \\[6pt] &= \begin{pmatrix} 5k \\ 2k \end{pmatrix} \end{aligned} \]Part (b)(ii)
Route \(A \to O \to X\):
\[ \begin{aligned} \vec{AX} &= -\vec{OA} + \vec{OX} \\[6pt] &= \begin{pmatrix} -1 \\ -2 \end{pmatrix} + \begin{pmatrix} 5k \\ 2k \end{pmatrix} \\[6pt] &= \begin{pmatrix} 5k - 1 \\ 2k - 2 \end{pmatrix} \end{aligned} \]Part (b)(iii)
Route \(X \to O \to C\):
\[ \begin{aligned} \vec{XC} &= -\vec{OX} + \vec{OC} \\[6pt] &= \begin{pmatrix} -5k \\ -2k \end{pmatrix} + \begin{pmatrix} 4 \\ 0 \end{pmatrix} \\[6pt] &= \begin{pmatrix} 4 - 5k \\ -2k \end{pmatrix} \end{aligned} \]Part (c)
Set \(\vec{AX} = \vec{XC}\) and equate components:
\[ \begin{aligned} 5k - 1 &= 4 - 5k \\[6pt] 10k &= 5 \\[6pt] k &= \tfrac{1}{2} \end{aligned} \]Check with bottom components: \(2(\tfrac{1}{2}) - 2 = -1\) and \(-2(\tfrac{1}{2}) = -1\). ✓ \(\boxed{k = \tfrac{1}{2}}\)
Part (d)
When \(k = \frac{1}{2}\), we have \(OX = \frac{1}{2}OB\), so \(X\) is the midpoint of diagonal \(OB\).
At this same point, \(\vec{AX} = \vec{XC}\), so \(X\) is also the midpoint of diagonal \(AC\).
Since both diagonals pass through the same midpoint \(X\), the diagonals of \(OABC\) bisect one another.
This is a typical 8-mark Paper 4 question. Each part builds on the one before, so getting part (a) right is essential. The proof in part (d) is where most students go quiet — practise writing that final explanation clearly.
Common mistakes
\(\vec{AB} = \mathbf{b} - \mathbf{a}\), not \(\mathbf{a} - \mathbf{b}\). The letter you're going to comes first in the subtraction.
Parallel vectors could sit on two separate lines. For collinearity you need to state they also share a point.
\(\vec{AM} = \frac{1}{2}\vec{AB}\), not \(\frac{1}{2}\mathbf{a}\). The half applies to the vector along the side, not to the position vector.
If \(\vec{OA} = \mathbf{a}\) then \(\vec{AO} = -\mathbf{a}\). Forgetting that negative is the most common route-finding error.
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