Vectors at A-Level
You met vector notation, addition, subtraction and simple vector-geometry proofs at GCSE. A-Level Vectors builds directly on that: a second notation (i-j form) alongside column vectors, the magnitude of a vector via Pythagoras' theorem, position vectors relative to a fixed origin, and harder geometry problems — dividing a line in a given ratio, and proving three points are collinear.
Five skills make up this topic:
- Notation — converting fluently between column-vector form and i-j form
- Vector arithmetic — addition, subtraction and scalar multiplication in i-j form
- Magnitude — finding \(|\mathbf{v}|\) and the unit vector in a given direction
- Position vectors — finding the vector between two points, \(\vec{AB}=\mathbf{b}-\mathbf{a}\)
- Vector geometry problems — dividing a line in a given ratio, and proving collinearity
2D vectors and notation
A 2D vector can be written as a column vector, \(\binom{x}{y}\), or in i-j form, \(x\mathbf{i}+y\mathbf{j}\), where \(\mathbf{i}=\binom{1}{0}\) and \(\mathbf{j}=\binom{0}{1}\) are the standard unit vectors along the \(x\)- and \(y\)-axes. Both forms describe exactly the same vector — the top entry of the column vector is always the i-component, and the bottom entry is always the j-component.
Worked Example 1
Write the vector \(\binom{-3}{5}\) using \(\mathbf{i}\)-\(\mathbf{j}\) notation. Then write \(4\mathbf{i}-2\mathbf{j}\) as a column vector.
1
The top entry is the \(\mathbf{i}\)-component and the bottom entry is the \(\mathbf{j}\)-component: \(\binom{-3}{5} = -3\mathbf{i}+5\mathbf{j}\)
2
Reversing the process, \(4\mathbf{i}-2\mathbf{j}\) has \(\mathbf{i}\)-component \(4\) and \(\mathbf{j}\)-component \(-2\)
Answer\(-3\mathbf{i}+5\mathbf{j}\); \(\binom{4}{-2}\)
Vector arithmetic
Adding, subtracting or scalar-multiplying vectors in i-j form works component-by-component, exactly as it does for column vectors at GCSE.
Worked Example 2
Given \(\mathbf{a}=3\mathbf{i}-2\mathbf{j}\) and \(\mathbf{b}=-\mathbf{i}+5\mathbf{j}\), find \(2\mathbf{a}+3\mathbf{b}\).
1
\(2\mathbf{a}=6\mathbf{i}-4\mathbf{j}\)
2
\(3\mathbf{b}=-3\mathbf{i}+15\mathbf{j}\)
3
Add component-wise: \((6-3)\mathbf{i}+(-4+15)\mathbf{j}\)
Answer\(3\mathbf{i}+11\mathbf{j}\)
Magnitude of a vector
The magnitude of \(\mathbf{v}=x\mathbf{i}+y\mathbf{j}\) is \(|\mathbf{v}|=\sqrt{x^2+y^2}\) — Pythagoras' theorem applied to the vector's components. Dividing a vector by its own magnitude gives the unit vector in the same direction, a vector of length exactly 1.
Worked Example 3
Find the magnitude of \(\mathbf{v}=5\mathbf{i}-12\mathbf{j}\), and hence find the unit vector in the direction of \(\mathbf{v}\).
1
\(|\mathbf{v}|=\sqrt{5^2+(-12)^2}=\sqrt{25+144}\)
2
\(|\mathbf{v}|=\sqrt{169}=13\)
3
Unit vector \(=\dfrac{\mathbf{v}}{|\mathbf{v}|}=\dfrac{5\mathbf{i}-12\mathbf{j}}{13}\)
Answer\(|\mathbf{v}|=13\); unit vector \(=\dfrac{5}{13}\mathbf{i}-\dfrac{12}{13}\mathbf{j}\)
Position vectors
The position vector of a point is the vector from a fixed origin \(O\) to that point. If \(A\) and \(B\) have position vectors \(\mathbf{a}\) and \(\mathbf{b}\), the vector from \(A\) to \(B\) is always "arrival minus start": \(\vec{AB}=\mathbf{b}-\mathbf{a}\).
Worked Example 4
Points \(A\) and \(B\) have position vectors \(\mathbf{a}=6\mathbf{i}+2\mathbf{j}\) and \(\mathbf{b}=2\mathbf{i}+5\mathbf{j}\) relative to a fixed origin \(O\). Find \(\vec{AB}\) and its magnitude.
1
\(\vec{AB}=\mathbf{b}-\mathbf{a}\)
2
\(=(2-6)\mathbf{i}+(5-2)\mathbf{j}=-4\mathbf{i}+3\mathbf{j}\)
3
\(|\vec{AB}|=\sqrt{(-4)^2+3^2}=\sqrt{25}\)
Answer\(\vec{AB}=-4\mathbf{i}+3\mathbf{j}\), magnitude \(5\)
Position vectors \(\mathbf{a}=\vec{OA}\) and \(\mathbf{b}=\vec{OB}\) (teal), and \(\vec{AB}=\mathbf{b}-\mathbf{a}\) (amber, dashed) — the vector "arrival minus start" that goes directly from \(A\) to \(B\). Not to scale.
Vector geometry problems
Two problem types come up repeatedly: finding the position vector of a point that divides a line in a given ratio, and proving three points lie on the same straight line. For a ratio problem, convert the ratio directly into a fraction of the connecting vector — never average the endpoints numerically. For a collinearity proof, show two vectors sharing a common point are parallel (one is a scalar multiple of the other); the shared point is what turns "parallel" into "the same line".
Worked Example 5
The point \(M\) lies on \(AB\) such that \(AM:MB=2:1\). Given \(\vec{OA}=\mathbf{a}\) and \(\vec{OB}=\mathbf{b}\), find \(\vec{OM}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
1
\(\vec{AB}=\mathbf{b}-\mathbf{a}\)
2
Since \(AM:MB=2:1\), \(\vec{AM}=\dfrac{2}{3}\vec{AB}=\dfrac{2}{3}(\mathbf{b}-\mathbf{a})\)
3
\(\vec{OM}=\vec{OA}+\vec{AM}=\mathbf{a}+\dfrac{2}{3}(\mathbf{b}-\mathbf{a})\)
4
Simplify: \(=\dfrac{1}{3}\mathbf{a}+\dfrac{2}{3}\mathbf{b}\)
Answer\(\vec{OM}=\dfrac{1}{3}\mathbf{a}+\dfrac{2}{3}\mathbf{b}\)
Worked Example 6
Relative to \(O\), the position vectors of \(P\), \(Q\) and \(R\) are \(\mathbf{p}=2\mathbf{i}+\mathbf{j}\), \(\mathbf{q}=5\mathbf{i}+4\mathbf{j}\) and \(\mathbf{r}=11\mathbf{i}+10\mathbf{j}\). Show that \(P\), \(Q\) and \(R\) are collinear.
1
\(\vec{PQ}=\mathbf{q}-\mathbf{p}=3\mathbf{i}+3\mathbf{j}\)
2
\(\vec{QR}=\mathbf{r}-\mathbf{q}=6\mathbf{i}+6\mathbf{j}\)
3
\(\vec{QR}=2\vec{PQ}\), so \(\vec{PQ}\) and \(\vec{QR}\) are parallel
4
\(Q\) is a point common to both vectors, so \(P\), \(Q\) and \(R\) lie on the same straight line
Answer\(P\), \(Q\), \(R\) are collinear, since \(\vec{QR}=2\vec{PQ}\), as required
Exam tips
💡 Tip 1
Always work "arrival minus start"
The vector from \(A\) to \(B\) is \(\vec{AB}=\mathbf{b}-\mathbf{a}\), never \(\mathbf{a}-\mathbf{b}\) — reversing the order gives a vector of the correct magnitude but pointing the wrong way.
💡 Tip 2
Turn a ratio directly into a fraction of the vector
\(AM:MB=2:1\) means \(\vec{AM}=\dfrac{2}{3}\vec{AB}\) — write the ratio as "part over whole" immediately, rather than trying to average coordinates.
💡 Tip 3
A collinearity proof needs both parts stated
Showing two vectors are parallel is not enough on its own — always state the shared point explicitly ("since \(Q\) is common to both...") to conclude the points are collinear.
💡 Tip 4
Keep i-j and column-vector notation separate within a solution
Convert everything to one notation before adding or subtracting — mixing \(\binom{x}{y}\) and \(x\mathbf{i}+y\mathbf{j}\) mid-calculation is where sign and component errors creep in.
💡 Tip 5
Leave magnitudes exact unless told otherwise
If \(x^2+y^2\) isn't a perfect square, give \(|\mathbf{v}|\) as a simplified surd rather than rounding early — rounding before the final answer is a common source of lost accuracy marks.
Common mistakes
Common Mistake 1
Reversing the subtraction order
Writing \(\vec{AB}=\mathbf{a}-\mathbf{b}\) instead of \(\mathbf{b}-\mathbf{a}\) gives a vector pointing in exactly the opposite direction — always start from the destination point's position vector.
Common Mistake 2
Using the whole vector instead of the required fraction
In a ratio problem, adding the full \(\vec{AB}\) instead of the fraction that corresponds to the given ratio (e.g. \(\frac{2}{3}\vec{AB}\) for \(AM:MB=2:1\)) is the single most common error in this topic.
Common Mistake 3
Forgetting to state the common point in a collinearity proof
\(\vec{PQ}\) parallel to \(\vec{QR}\) alone only proves the two vectors point in the same (or opposite) direction — the mark scheme also requires noting that \(Q\) is shared by both, which is what makes the three points lie on one line.
Common Mistake 4
Sign errors when finding a unit vector
Dividing only the first component by the magnitude and leaving the second unchanged — both components must be divided by \(|\mathbf{v}|\).
Common Mistake 5
Mixing up column-vector rows and i-j components
Swapping the top and bottom entries when converting \(\binom{x}{y}\) to \(x\mathbf{i}+y\mathbf{j}\) — the top entry is always the \(\mathbf{i}\)-component (horizontal), the bottom always the \(\mathbf{j}\)-component (vertical).
Practice questions
Work through each question before checking the worked solution.
Core Skills
Q1Write \(\binom{4}{-7}\) using \(\mathbf{i}\)-\(\mathbf{j}\) notation.Core Skills
Q2Write \(-2\mathbf{i}+6\mathbf{j}\) as a column vector.Core Skills
Q3Given \(\mathbf{a}=4\mathbf{i}+3\mathbf{j}\), find \(|\mathbf{a}|\).Core Skills
Q4Given \(\mathbf{p}=2\mathbf{i}-\mathbf{j}\) and \(\mathbf{q}=3\mathbf{i}+4\mathbf{j}\), find \(\mathbf{p}+\mathbf{q}\).Core Skills
Q5Given \(\mathbf{a}=\mathbf{i}-3\mathbf{j}\), find \(2\mathbf{a}\).Core Skills
Exam-Style
Q6Given \(\mathbf{a}=5\mathbf{i}-\mathbf{j}\) and \(\mathbf{b}=-2\mathbf{i}+3\mathbf{j}\), find \(3\mathbf{a}-2\mathbf{b}\).Exam-Style
Q7Find the unit vector in the direction of \(\mathbf{v}=8\mathbf{i}-6\mathbf{j}\).Exam-Style
Q8Points \(A\) and \(B\) have position vectors \(\mathbf{a}=\mathbf{i}+4\mathbf{j}\) and \(\mathbf{b}=7\mathbf{i}-4\mathbf{j}\). Find \(\vec{AB}\) and its magnitude.Exam-Style
Q9The point \(N\) lies on \(AB\) such that \(AN:NB=1:2\). Given \(\vec{OA}=\mathbf{a}\) and \(\vec{OB}=\mathbf{b}\), find \(\vec{ON}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).Exam-Style
Q10Find the positive value of \(k\) for which \(k\mathbf{i}+12\mathbf{j}\) has magnitude \(13\).Exam-Style
A* Challenge
Q11Relative to \(O\), points \(P\), \(Q\), \(R\) have position vectors \(\mathbf{p}=3\mathbf{i}+2\mathbf{j}\), \(\mathbf{q}=7\mathbf{i}+5\mathbf{j}\), \(\mathbf{r}=15\mathbf{i}+11\mathbf{j}\). Show that \(P\), \(Q\), \(R\) are collinear.A* Challenge
Q12\(OABC\) is a parallelogram, with \(\vec{OA}=\mathbf{a}\) and \(\vec{OC}=\mathbf{c}\). Find, in terms of \(\mathbf{a}\) and \(\mathbf{c}\): (a) \(\vec{OB}\), (b) the position vector of the midpoint of \(AC\).A* Challenge
Q13Given \(\mathbf{a}=2\mathbf{i}-\mathbf{j}\) and \(\mathbf{b}=\mathbf{i}+3\mathbf{j}\), find the value of \(\lambda\) for which \(\mathbf{a}+\lambda\mathbf{b}\) is parallel to \(3\mathbf{i}+5\mathbf{j}\).A* Challenge
Q14Points \(A\), \(B\), \(C\) have position vectors \(\mathbf{a}\), \(\mathbf{b}\), \(\mathbf{c}\). \(M\) is the midpoint of \(BC\). Find \(\vec{AM}\) in terms of \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\).A* Challenge
Q15Relative to \(O\), \(\vec{OA}=4\mathbf{i}+3\mathbf{j}\) and \(\vec{OB}=-2\mathbf{i}+11\mathbf{j}\). The point \(C\) lies on \(AB\) extended beyond \(B\), such that \(AC=2AB\). Find the position vector of \(C\).A* Challenge
Answers — full worked solutions
Core Skills (Q1–Q5)
Q1 — \(4\mathbf{i}-7\mathbf{j}\)
\(4\mathbf{i}-7\mathbf{j}\)
Q2 — \(\binom{-2}{6}\)
\(\binom{-2}{6}\)
Q3 — \(5\)
1
\(|\mathbf{a}|=\sqrt{4^2+3^2}=\sqrt{16+9}\)
\(=\sqrt{25}=5\)
Q4 — \(5\mathbf{i}+3\mathbf{j}\)
\((2+3)\mathbf{i}+(-1+4)\mathbf{j}=5\mathbf{i}+3\mathbf{j}\)
Q5 — \(2\mathbf{i}-6\mathbf{j}\)
\(2\mathbf{i}-6\mathbf{j}\)
Exam-Style (Q6–Q10)
Q6 — \(19\mathbf{i}-9\mathbf{j}\)
1
\(3\mathbf{a}=15\mathbf{i}-3\mathbf{j}\); \(2\mathbf{b}=-4\mathbf{i}+6\mathbf{j}\)
\(3\mathbf{a}-2\mathbf{b}=(15-(-4))\mathbf{i}+(-3-6)\mathbf{j}=19\mathbf{i}-9\mathbf{j}\)
Q7 — \(\dfrac{4}{5}\mathbf{i}-\dfrac{3}{5}\mathbf{j}\)
1
\(|\mathbf{v}|=\sqrt{8^2+(-6)^2}=\sqrt{100}=10\)
\(\dfrac{8\mathbf{i}-6\mathbf{j}}{10}=\dfrac{4}{5}\mathbf{i}-\dfrac{3}{5}\mathbf{j}\)
Q8 — \(\vec{AB}=6\mathbf{i}-8\mathbf{j}\), magnitude \(10\)
1
\(\vec{AB}=\mathbf{b}-\mathbf{a}=(7-1)\mathbf{i}+(-4-4)\mathbf{j}=6\mathbf{i}-8\mathbf{j}\)
\(|\vec{AB}|=\sqrt{6^2+(-8)^2}=\sqrt{100}=10\)
Q9 — \(\vec{ON}=\dfrac{2}{3}\mathbf{a}+\dfrac{1}{3}\mathbf{b}\)
1
\(AN:NB=1:2\), so \(\vec{AN}=\dfrac{1}{3}\vec{AB}=\dfrac{1}{3}(\mathbf{b}-\mathbf{a})\)
2
\(\vec{ON}=\mathbf{a}+\dfrac{1}{3}(\mathbf{b}-\mathbf{a})\)
\(=\dfrac{2}{3}\mathbf{a}+\dfrac{1}{3}\mathbf{b}\)
Q10 — \(k=5\)
1
\(\sqrt{k^2+12^2}=13 \implies k^2+144=169\)
\(k^2=25 \implies k=5\)
A* Challenge (Q11–Q15)
Q11 — Collinear, since \(\vec{QR}=2\vec{PQ}\)
1
\(\vec{PQ}=\mathbf{q}-\mathbf{p}=4\mathbf{i}+3\mathbf{j}\)
2
\(\vec{QR}=\mathbf{r}-\mathbf{q}=8\mathbf{i}+6\mathbf{j}=2(4\mathbf{i}+3\mathbf{j})\)
\(\vec{QR}=2\vec{PQ}\), and \(Q\) is common to both, so \(P\), \(Q\), \(R\) are collinear
Q12 — (a) \(\mathbf{a}+\mathbf{c}\), (b) \(\dfrac{1}{2}(\mathbf{a}+\mathbf{c})\)
1
(a) In parallelogram \(OABC\), \(\vec{AB}=\vec{OC}=\mathbf{c}\), so \(\vec{OB}=\vec{OA}+\vec{AB}=\mathbf{a}+\mathbf{c}\)
(b) Midpoint of \(AC=\dfrac{1}{2}(\mathbf{a}+\mathbf{c})\)
Q13 — \(\lambda=\dfrac{13}{4}\)
1
\(\mathbf{a}+\lambda\mathbf{b}=(2+\lambda)\mathbf{i}+(-1+3\lambda)\mathbf{j}\)
2
Parallel to \(3\mathbf{i}+5\mathbf{j}\) requires \(\dfrac{2+\lambda}{3}=\dfrac{-1+3\lambda}{5}\)
3
\(5(2+\lambda)=3(-1+3\lambda) \implies 10+5\lambda=-3+9\lambda\)
\(13=4\lambda \implies \lambda=\dfrac{13}{4}\)
Q14 — \(\vec{AM}=\dfrac{1}{2}(\mathbf{b}+\mathbf{c})-\mathbf{a}\)
1
\(M\) is the midpoint of \(BC\): \(\vec{OM}=\dfrac{1}{2}(\mathbf{b}+\mathbf{c})\)
\(\vec{AM}=\vec{OM}-\vec{OA}=\dfrac{1}{2}(\mathbf{b}+\mathbf{c})-\mathbf{a}\)
Q15 — \(\vec{OC}=-8\mathbf{i}+19\mathbf{j}\)
1
\(\vec{AB}=\mathbf{b}-\mathbf{a}=(-2-4)\mathbf{i}+(11-3)\mathbf{j}=-6\mathbf{i}+8\mathbf{j}\)
2
\(AC=2AB\): \(\vec{AC}=2(-6\mathbf{i}+8\mathbf{j})=-12\mathbf{i}+16\mathbf{j}\)
\(\vec{OC}=\vec{OA}+\vec{AC}=(4-12)\mathbf{i}+(3+16)\mathbf{j}=-8\mathbf{i}+19\mathbf{j}\)
Summary
- A 2D vector can be written as a column vector \(\binom{x}{y}\) or in \(\mathbf{i}\)-\(\mathbf{j}\) form \(x\mathbf{i}+y\mathbf{j}\) — the same vector, two notations.
- Vector arithmetic (addition, subtraction, scalar multiplication) works component-by-component in either notation.
- The magnitude \(|\mathbf{v}|=\sqrt{x^2+y^2}\) is Pythagoras' theorem applied to a vector's components; dividing by \(|\mathbf{v}|\) gives the unit vector.
- For position vectors, always work "arrival minus start": \(\vec{AB}=\mathbf{b}-\mathbf{a}\).
- Turn a given ratio directly into a fraction of the connecting vector, and always state the shared point explicitly when proving three points are collinear.
This topic builds on GCSE Vectors and GCSE Pythagoras' Theorem, and reuses the proof techniques from Algebraic Methods and the coordinate-geometry skills from Straight Line Graphs. Next in Pure Year 1: Differentiation.
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