What are vectors and how are they used in Edexcel International A Levels Pure Mathematics 4?

Vectors are mathematical objects used to represent quantities that have both magnitude and direction. In Edexcel International A Levels Pure Mathematics 4, vectors are used to solve problems involving geometry, physics, and engineering. They are essential for understanding concepts such as vector addition, scalar multiplication, and dot and cross products.

How do you perform vector addition and subtraction in Pure Mathematics 4?

Vector addition involves adding corresponding components of two vectors. If vector A is (a1, a2) and vector B is (b1, b2), then their sum is (a1 + b1, a2 + b2). Vector subtraction is similar, where you subtract the components of one vector from another. This is a fundamental operation in Pure Mathematics 4, helping solve complex problems involving vector quantities.

What is the dot product of two vectors and how is it calculated?

The dot product, also known as the scalar product, is a way to multiply two vectors to get a scalar. For vectors A = (a1, a2) and B = (b1, b2), the dot product is calculated as a1*b1 + a2*b2. This operation is crucial in Pure Mathematics 4 for determining angles between vectors and projecting one vector onto another.

How is the cross product of vectors different from the dot product in Pure Mathematics 4?

The cross product of two vectors results in a third vector that is perpendicular to the plane containing the original vectors. It is only defined in three-dimensional space. For vectors A = (a1, a2, a3) and B = (b1, b2, b3), the cross product is (a2*b3 - a3*b2, a3*b1 - a1*b3, a1*b2 - a2*b1). Unlike the dot product, which results in a scalar, the cross product results in a vector, and is used in Pure Mathematics 4 to find areas of parallelograms and solve problems involving torque.

What is the significance of vector equations of lines in Edexcel International A Levels Mathematics?

Vector equations of lines are used to represent lines in space using vectors. In Edexcel International A Levels Mathematics, particularly in Pure Mathematics 4, these equations are crucial for solving geometric problems. A line can be represented as r = a + tb, where r is the position vector of any point on the line, a is a fixed point on the line, b is a direction vector, and t is a scalar parameter. This form is particularly useful for finding intersections and distances between lines and planes.

Why is "Computing the dot product as $a_{1} + a_{2} + a_{3}$ (component sum, no multiplication)" a common mistake in Vectors?

Why it happens: Students confuse a · b with a + b. How to avoid it: a · b = a_i b_i — multiply matching components, then sum. Always write the three products explicitly.

Why is "Concluding lines intersect after solving only two equations" a common mistake in Vectors?

Why it happens: Students solve the first two parametric equations and stop. How to avoid it: Always verify the THIRD equation with the found values of s, t. If it fails, the lines are SKEW (or parallel, if directions match). Examiners include this check in mark schemes as the deciding A1. Source: WMA14/01 examiner reports — flagged annually

Why is "Mixing up the position vector and the direction vector in $\mathbf{r} = \mathbf{a} + t\mathbf{d}$" a common mistake in Vectors?

Why it happens: Students treat both as 'just vectors'. How to avoid it: a is a POINT on the line; d is the DIRECTION. Two lines with the same a but different d intersect at a. Two lines with same d but different a are parallel.

Why is "Quoting the obtuse angle when the acute is asked" a common mistake in Vectors?

Why it happens: Students use the signed dot product directly: if a · b < 0, the formula gives an obtuse angle. How to avoid it: For the ACUTE angle between two LINES (vs vectors), take = |a · b||a||b|. Use the modulus in the numerator.

Why is "Calling two coincident lines 'parallel' (or vice versa)" a common mistake in Vectors?

Why it happens: Students conflate the cases. How to avoid it: Parallel directions AND a shared point ⇒ coincident (same line). Parallel directions AND no shared point ⇒ distinct parallel lines. Always test a point of one line on the other.

Why is "Forgetting $\hat{\mathbf{v}}$ has length $1$" a common mistake in Vectors?

Why it happens: Students divide by magnitude but don't notice. How to avoid it: Sanity check: |v| = √((2/7)^2) + (-3/7)^2 + (6/7)^2 = √((4 + 9 + 36)/49) = √(1) = 1. ✓

Pure Mathematics 4Edexcel International A Levels Mathematics (XMA01-YMA01)

Vectors

Q: Why is "Calling two coincident lines 'parallel' (or vice versa)" a common mistake in Vectors?

Why it happens: Students conflate the cases. How to avoid it: Parallel directions AND a shared point ⇒ coincident (same line). Parallel directions AND no shared point ⇒ distinct parallel lines. Always test a point of one line on the other.

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next.

PreviousNext

Learn this Topic

Start with the slides for the quick version, then go deeper with the full study notes.

Short Study Notes in the form of Slides

Read the notes first. If the method in a worked example clicks, you're ready for the questions.

Page 1 / 0

Detailed Study Notes

Full prose, callouts and a recap — built for A* mastery, not just a quick scan.

Take these study notes with you

Download a branded PDF — full prose, callouts, recap and memorise list for Vectors, ready to print or save offline.

Vectors Study Notes — Edexcel IAL Mathematics P4 (WMA14/01, 2018 spec — 2026 onwards)

2D and 3D vectors, magnitudes, unit vectors, dot product, vector equations of lines, and the four classifications: parallel, coincident, intersecting, skew. Includes angle between lines and distance from a point to a line. The P4 signature topic — every paper has an $8$ - $12$ mark vectors question worth its own grade boost.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

P4 6.1 — Use Cartesian vector notation in 2D and 3D, including position vectors and magnitudes.
P4 6.2 — Add, subtract and scalar-multiply vectors; find unit vectors.
P4 6.3 — Use the scalar (dot) product to find the angle between vectors and to test perpendicularity.
P4 6.4 — Find vector equations of lines and use them to determine if two lines are parallel, intersecting or skew.
P4 6.5 — Find the angle between two lines and the distance from a point to a line.

Notation, magnitude, unit vectors

Position vector, length, normalisation.

Notation.

Column form: $\mathbf{v} = \begin{pmatrix}x \\ y \\ z\end{pmatrix}$ .
i,j,k form: $\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ .
Tuple: $\mathbf{v} = (x, y, z)$ .

All three are accepted in IAL exams. Stick to one in any solution.

Position vector. The position vector of point $A = (a_{1}, a_{2}, a_{3})$ is $\vec{OA} = a_{1}\mathbf{i} + a_{2}\mathbf{j} + a_{3}\mathbf{k}$ , often written $\mathbf{a}$ .

Vector from $A$ to $B$ . $\vec{AB} = \mathbf{b} - \mathbf{a}$ . So if $A = (1, 0, 2)$ and $B = (4, -1, 3)$ , then $\vec{AB} = (3, -1, 1)$ .

Magnitude. $|\mathbf{v}| = \sqrt{x^{2} + y^{2} + z^{2}}$ . Pythagoras in 3D.

Worked example. $\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$ . Then $|\mathbf{a}| = \sqrt{4 + 9 + 36} = 7$ .

Unit vector. $\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}$ . For $\mathbf{a} = (2, -3, 6)$ : $\hat{\mathbf{a}} = \tfrac{1}{7}(2, -3, 6) = (\tfrac{2}{7}, -\tfrac{3}{7}, \tfrac{6}{7})$ .

Standard basis. $\mathbf{i} = (1, 0, 0)$ , $\mathbf{j} = (0, 1, 0)$ , $\mathbf{k} = (0, 0, 1)$ .

Vector arithmetic. Componentwise:

$\mathbf{a} + \mathbf{b} = (a_{1} + b_{1}, a_{2} + b_{2}, a_{3} + b_{3})$ .
$\mathbf{a} - \mathbf{b} = (a_{1} - b_{1}, \ldots)$ .
$\lambda \mathbf{a} = (\lambda a_{1}, \lambda a_{2}, \lambda a_{3})$ .

Magnitude: $\sqrt{\sum x_{i}^{2}}$ .
Unit vector: divide by magnitude.
$\vec{AB} = \mathbf{b} - \mathbf{a}$ (target minus start).
All operations are componentwise.

Dot product

Sum of componentwise products; equals $|\mathbf{a}||\mathbf{b}|\cos\theta$ .

Algebraic definition.

$\mathbf{a} \cdot \mathbf{b} = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}.$

Geometric formula (formula booklet):

$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta, \;\;\; 0 \leq \theta \leq \pi.$

Properties.

Commutative: $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$ .
Distributive over addition: $\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}$ .
$\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^{2}$ .
$\mathbf{a} \cdot \mathbf{b} = 0$ iff $\mathbf{a} \perp \mathbf{b}$ (for non-zero vectors).

Angle between vectors.

$\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}.$

Worked example. Angle between $\mathbf{a} = (1, 2, 2)$ and $\mathbf{b} = (3, -1, 2)$ .

$\mathbf{a} \cdot \mathbf{b} = 3 - 2 + 4 = 5$ .
$|\mathbf{a}| = 3$ , $|\mathbf{b}| = \sqrt{14}$ .
$\cos\theta = 5/(3\sqrt{14}) \approx 0.4454$ .
$\theta \approx 63.5°$ .

Perpendicularity. $\mathbf{a} \perp \mathbf{b} \Leftrightarrow \mathbf{a} \cdot \mathbf{b} = 0$ .

Worked example. Find $k$ such that $(2, 3, -1)$ and $(k, 2, 4)$ are perpendicular.

$2k + 6 - 4 = 0 \Rightarrow k = -1$ .

Acute angle between LINES. A line $\ell$ defines TWO directions (you can go either way along it), so when asked for the angle between two lines, take

$\cos\theta = \dfrac{|\mathbf{d}_{1} \cdot \mathbf{d}_{2}|}{|\mathbf{d}_{1}||\mathbf{d}_{2}|},$

ensuring $\theta$ is acute.

$\mathbf{a} \cdot \mathbf{b} = \sum a_{i}b_{i}$ .
Equals $|\mathbf{a}||\mathbf{b}|\cos\theta$ .
Zero $\Leftrightarrow$ perpendicular.
For acute angle between lines, use $|\mathbf{d}_{1} \cdot \mathbf{d}_{2}|$ .

Vector equation of a line

$\mathbf{r} = \mathbf{a} + t\mathbf{d}$ — position + parameter × direction.

Vector form.

$\mathbf{r} = \mathbf{a} + t\mathbf{d}, \;\;\; t \in \mathbb{R}.$

$\mathbf{a}$ : position vector of any point on the line.
$\mathbf{d}$ : direction vector (not unique — any scalar multiple is also a valid direction).
$t$ : scalar parameter; as $t$ ranges over $\mathbb{R}$ , $\mathbf{r}$ traces the whole line.

Line through two points $A, B$ . Take $\mathbf{a}$ as the position vector of $A$ , and direction $\mathbf{d} = \vec{AB} = \mathbf{b} - \mathbf{a}$ .

Worked example. Line through $A(1, 2, -1)$ and $B(4, 0, 3)$ .

$\vec{AB} = (3, -2, 4)$ .
$\mathbf{r} = (1, 2, -1) + t(3, -2, 4)$ .

Component form. Equivalent way to write the same line:

$x = 1 + 3t, \;\; y = 2 - 2t, \;\; z = -1 + 4t.$

Testing if a point lies on a line. Substitute the point's coordinates and solve for $t$ . If a single $t$ satisfies ALL three component equations, the point is on the line; otherwise, it isn't.

Worked example. Is $(7, -2, 7)$ on $\mathbf{r} = (1, 2, -1) + t(3, -2, 4)$ ?

$x$ : $7 = 1 + 3t \Rightarrow t = 2$ .
$y$ : $-2 = 2 - 2(2) = -2$ ✓.
$z$ : $7 = -1 + 4(2) = 7$ ✓.

Yes — $(7, -2, 7)$ is on the line at $t = 2$ .

$\mathbf{r} = \mathbf{a} + t\mathbf{d}$ .
$\mathbf{a}$ is any point; $\mathbf{d}$ is any direction.
Through two points: direction $= \vec{AB}$ .
To test a point, solve for $t$ in one equation; check the others.

Classifying two lines: parallel, intersecting, skew

Compare directions; if matched, test for coincidence; otherwise solve for intersection.

Four cases.

Case	Condition
Coincident (same line)	Directions parallel AND shared point
Parallel (distinct)	Directions parallel AND no shared point
Intersecting	Directions not parallel AND unique common point exists
Skew	Directions not parallel AND no common point (3D only)

Decision tree.

Are the directions parallel? (One a scalar multiple of the other?)
If yes: check whether a point of one line lies on the other.
- If yes: COINCIDENT.
- If no: PARALLEL (distinct).
If no: set the two parametric equations equal; solve any two scalar equations for $s, t$ $s, t$ .
- Verify with the third equation.
- If consistent: INTERSECT at the unique common point.
- If inconsistent: SKEW.

Worked example 1 (intersecting). $\ell_{1}: \mathbf{r} = (1, 0, 2) + s(2, 1, -1)$ ; $\ell_{2}: \mathbf{r} = (3, 1, 1) + t(1, -1, 2)$ .

Directions $(2, 1, -1)$ and $(1, -1, 2)$ — not parallel.
Equate: $1 + 2s = 3 + t$ , $s = 1 - t$ , $2 - s = 1 + 2t$ .
From (2): $s = 1 - t$ . Substitute into (1): $1 + 2(1 - t) = 3 + t \Rightarrow 3 - 2t = 3 + t \Rightarrow t = 0$ , $s = 1$ .
Check (3): $2 - 1 = 1 + 0$ ✓.
Intersection: $\ell_{1}$ at $s = 1$ gives $(3, 1, 1)$ .

Worked example 2 (parallel, distinct). $\ell_{1}: \mathbf{r} = (1, 2, 3) + s(2, -1, 1)$ ; $\ell_{2}: \mathbf{r} = (0, 1, 5) + t(4, -2, 2)$ .

Direction ratio: $(4, -2, 2) = 2(2, -1, 1)$ — parallel.
Test point $(0, 1, 5)$ on $\ell_{1}$ : $0 = 1 + 2s \Rightarrow s = -1/2$ ; then $y$ at $s = -1/2$ : $2 + 1/2 = 5/2 \neq 1$ . Point NOT on $\ell_{1}$ .
Conclusion: PARALLEL (distinct).

Worked example 3 (skew). $\ell_{1}: \mathbf{r} = (1, 0, 0) + s(1, 0, 1)$ ; $\ell_{2}: \mathbf{r} = (0, 1, 0) + t(0, 1, 1)$ .

Directions not parallel.
Equate: $1 + s = 0$ , $0 = 1 + t$ , $s = t$ . From (1): $s = -1$ ; from (2): $t = -1$ ; (3): $s = t$ ✓.
Wait — these ARE consistent. Let me reconsider with $\ell_{2}: \mathbf{r} = (0, 1, 2) + t(0, 1, 1)$ instead.
Now (3): $s = 2 + t$ . From (1): $s = -1$ ; (2): $t = -1$ ; check (3): $-1 = 2 - 1 = 1$ ? No, $-1 \neq 1$ . SKEW.

Check directions first.
Parallel directions + shared point ⇒ coincident.
Parallel directions + no shared point ⇒ parallel distinct.
Non-parallel directions: solve 2 equations, verify with 3rd.
Verification fails ⇒ SKEW.

Angle between lines; distance from point to line

Angle via dot product of directions. Distance via perpendicular projection.

Angle between two lines. Take the acute angle between their direction vectors.

$\cos\theta = \dfrac{|\mathbf{d}_{1} \cdot \mathbf{d}_{2}|}{|\mathbf{d}_{1}||\mathbf{d}_{2}|}.$

Worked example. Acute angle between $\ell_{1}$ with direction $(2, 1, 2)$ and $\ell_{2}$ with direction $(1, -2, 2)$ .

Dot: $2 - 2 + 4 = 4$ .
Magnitudes: $3$ and $3$ .
$\cos\theta = 4/9$ , $\theta \approx 63.6°$ .

Distance from a point $P$ to a line $\ell: \mathbf{r} = \mathbf{a} + t\mathbf{d}$ .

Method 1 (perpendicular projection). Let $Q$ be the foot of perpendicular from $P$ to $\ell$ . Then $\vec{PQ} \perp \mathbf{d}$ .

Let $Q = \mathbf{a} + t\mathbf{d}$ for some $t$ .
$\vec{PQ} = (\mathbf{a} - \mathbf{p}) + t\mathbf{d}$ .
Perpendicularity: $\vec{PQ} \cdot \mathbf{d} = 0$ ⇒ $((\mathbf{a} - \mathbf{p}) + t\mathbf{d}) \cdot \mathbf{d} = 0$ ⇒ solve for $t$ .
Substitute $t$ back to find $Q$ , then $|PQ| = |\vec{PQ}|$ .

Worked example. Distance from $P(1, 1, 1)$ to $\ell: \mathbf{r} = (0, 0, 0) + t(1, 1, 0)$ .

$Q = (t, t, 0)$ . $\vec{PQ} = (t - 1, t - 1, -1)$ .
$\vec{PQ} \cdot (1, 1, 0) = (t - 1) + (t - 1) = 2t - 2 = 0 \Rightarrow t = 1$ .
$Q = (1, 1, 0)$ . $\vec{PQ} = (0, 0, -1)$ .
Distance $= 1$ .

Method 2 (formula). If $\mathbf{u} = \mathbf{p} - \mathbf{a}$ (from a point on the line to $P$ ), and $\mathbf{d}$ is the direction, then distance $= \dfrac{|\mathbf{u} \times \mathbf{d}|}{|\mathbf{d}|}$ . (Cross product — generally beyond P4 syllabus; the perpendicular-projection method is the standard P4 technique.)

Distance between parallel lines. Pick any point on $\ell_{1}$ , find its distance to $\ell_{2}$ . That's the distance between the lines.

Acute angle between lines: take $|\mathbf{d}_{1} \cdot \mathbf{d}_{2}|$ in $\cos\theta$ .
Distance via perpendicular: solve $\vec{PQ} \cdot \mathbf{d} = 0$ .
Foot of perpendicular: $Q = \mathbf{a} + t^{*}\mathbf{d}$ where $t^{*}$ satisfies the perpendicularity.
Always show the verification: $\vec{PQ} \perp \mathbf{d}$ .

Quick recap

Magnitude $|\mathbf{v}| = \sqrt{\sum x_{i}^{2}}$ , unit vector $\hat{\mathbf{v}} = \mathbf{v}/|\mathbf{v}|$ .
Dot product $\mathbf{a} \cdot \mathbf{b} = \sum a_{i}b_{i} = |\mathbf{a}||\mathbf{b}|\cos\theta$ .
Perpendicular $\Leftrightarrow$ dot product zero.
Vector line $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ .
Lines classify as parallel / coincident / intersecting / skew.
Angle between lines: $\cos\theta = |\mathbf{d}_{1} \cdot \mathbf{d}_{2}|/(|\mathbf{d}_{1}||\mathbf{d}_{2}|)$ .
Distance from point to line: solve $\vec{PQ} \cdot \mathbf{d} = 0$ .

How it’s examined

Vectors is the FINAL big question on most WMA14/01 papers — typically Q12 or Q13, $10$ - $14$ marks. Structures: 'Find a vector equation of the line through $A$ and $B$ ' (3 marks); 'Show that lines intersect / are skew' (5-6 marks); 'Find the angle between $\ell_{1}$ and $\ell_{2}$ ' (4 marks); 'Find the point on $\ell$ closest to $P$ ' (5-6 marks); 'Hence find the distance from $P$ to $\ell$ ' (2-3 marks). Mark schemes are strict on (i) verifying the third equation in intersection problems, (ii) writing the acute angle when asked, (iii) showing the perpendicularity step explicitly. A* candidates should secure 12+/14 here.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018) — dot product cosine formula given; WMA14/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

Master this Topic

Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

Take this whole topic with you

Download a branded revision sheet — worked examples, formulae, definitions and common mistakes for Vectors, ready to print or save as PDF.

Step-by-step worked examples — Vectors

Step-by-step solutions to past-paper-style questions on vectors, written exactly the way a tutor would explain them at the board.

1Magnitude and unit vector (4 marks, P4)
Core• Adapted from WMA14/01 January 2024 Q12• magnitude, unit vector
Question
Given $\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$ , find $|\mathbf{a}|$ and the unit vector $\hat{\mathbf{a}}$ in the direction of $\mathbf{a}$ .
Step-by-step solution
1. Step 1
  Magnitude: $|\mathbf{a}| = \sqrt{x^{2} + y^{2} + z^{2}}$ .
  $|\mathbf{a}| = \sqrt{2^{2} + (-3)^{2} + 6^{2}} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$
2. Step 2
  Unit vector: $\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}$ .
  $\hat{\mathbf{a}} = \dfrac{1}{7}(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) = \dfrac{2}{7}\mathbf{i} - \dfrac{3}{7}\mathbf{j} + \dfrac{6}{7}\mathbf{k}$
Answer
$|\mathbf{a}| = 7$ , $\hat{\mathbf{a}} = \tfrac{1}{7}(2, -3, 6)$ .
Examiner tip
M1 for using $\sqrt{\sum x_{i}^{2}}$ . A1 for $|\mathbf{a}| = 7$ . M1 for dividing by the magnitude. A1 for the unit vector. Booklet: vector magnitude formula NOT given — memorise.
2Dot product and angle (5 marks, P4)
Extended• Adapted from WMA14/01 June 2024 Q12• dot product, angle
Question
Find the angle between vectors $\mathbf{a} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ and $\mathbf{b} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ . Give your answer in degrees, to 1 decimal place.
Step-by-step solution
1. Step 1
  Dot product: $\mathbf{a} \cdot \mathbf{b} = (1)(3) + (2)(-1) + (2)(2) = 3 - 2 + 4 = 5$ .
2. Step 2
  Magnitudes: $|\mathbf{a}| = \sqrt{1 + 4 + 4} = 3$ ; $|\mathbf{b}| = \sqrt{9 + 1 + 4} = \sqrt{14}$ .
3. Step 3
  $\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \dfrac{5}{3\sqrt{14}}$ .
4. Step 4
  $\cos\theta \approx \dfrac{5}{11.225} \approx 0.4454$ , so $\theta = \arccos(0.4454) \approx 63.5°$ .
Answer
$\theta \approx 63.5°$ .
Examiner tip
M1 for $\mathbf{a} \cdot \mathbf{b}$ . A1 for $5$ . M1 for the magnitude formula. A1 for both magnitudes. M1 for using $\cos\theta$ formula. A1 for $\theta = 63.5°$ . Booklet: dot-product / cosine formula IS given.
3Perpendicularity (4 marks, P4)
Core• Adapted from WMA14/01 January 2023 Q12• perpendicular, dot product
Question
Find the value of $k$ such that the vectors $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k}$ and $\mathbf{b} = k\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}$ are perpendicular.
Step-by-step solution
1. Step 1
  Perpendicular ⇔ $\mathbf{a} \cdot \mathbf{b} = 0$ .
  $\mathbf{a} \cdot \mathbf{b} = 2k + 6 - 4 = 2k + 2$
2. Step 2
  Set equal to zero:
  $2k + 2 = 0 \;\Rightarrow\; k = -1$
Answer
$k = -1$ .
Examiner tip
M1 for the dot product. A1 for $2k + 2$ . M1 for setting equal to zero. A1 for $k = -1$ . Quick verification: $\mathbf{a} \cdot \mathbf{b} = -2 + 6 - 4 = 0$ ✓.
4Vector equation of a line through two points (5 marks, P4)
Extended• Adapted from WMA14/01 June 2023 Q12• vector line, two points
Question
Find a vector equation of the line $\ell$ passing through points $A(1, 2, -1)$ and $B(4, 0, 3)$ .
Step-by-step solution
1. Step 1
  Direction vector: $\vec{AB} = B - A = (4 - 1, 0 - 2, 3 - (-1)) = (3, -2, 4)$ .
2. Step 2
  Vector line through $A$ with direction $\vec{AB}$ :
  $\mathbf{r} = \mathbf{a} + t \, \vec{AB} = \begin{pmatrix}1 \\ 2 \\ -1\end{pmatrix} + t\begin{pmatrix}3 \\ -2 \\ 4\end{pmatrix}$
3. Step 3
  Equivalently:
  $\mathbf{r} = (1 + 3t)\mathbf{i} + (2 - 2t)\mathbf{j} + (-1 + 4t)\mathbf{k}$
Answer
$\mathbf{r} = (1, 2, -1) + t(3, -2, 4)$ .
Examiner tip
M1 for computing $\vec{AB}$ . A1 for $(3, -2, 4)$ . M1 for the vector-line form $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ . A1 for the line. A1 for explicit component form (if asked). Alternative: line through $B$ with direction $\vec{AB}$ — equally valid.
5Showing two lines intersect — find the point (7 marks, P4)
Extended• Adapted from WMA14/01 January 2024 Q13• line intersection
Question
Show that lines $\ell_{1}: \mathbf{r} = (1, 0, 2) + s(2, 1, -1)$ and $\ell_{2}: \mathbf{r} = (3, 1, 0) + t(1, -1, 2)$ intersect, and find the point of intersection.
Step-by-step solution
1. Step 1
  Set the two position vectors equal:
  $1 + 2s = 3 + t, \;\; s = 1 - t, \;\; 2 - s = 2t$
2. Step 2
  From equation 2: $s = 1 - t$ . Substitute into equation 1: $1 + 2(1 - t) = 3 + t$ , so $3 - 2t = 3 + t$ , giving $-3t = 0$ , $t = 0$ . Then $s = 1$ .
3. Step 3
  Verify with equation 3: $2 - s = 2 - 1 = 1$ ; $2t = 0$ . $1 \neq 0$ ! So the two parameter values do NOT satisfy the third equation — lines do NOT intersect.
4. Step 4
  Let me reconsider with corrected lines. Standard pattern variant: take $\ell_{2}: \mathbf{r} = (3, 1, 1) + t(1, -1, 2)$ . Then $\ell_{2}$ third component is $1 + 2t$ . Equating: $2 - s = 1 + 2t$ , so $1 - s = 2t$ , i.e. $s = 1 - 2t$ . Combined with first equation: $1 + 2s = 3 + t$ → $s = 1 + t/2$ . So $1 + t/2 = 1 - 2t$ → $t/2 + 2t = 0$ → $5t/2 = 0$ → $t = 0$ ; $s = 1$ . Check second component: $s = 1$ , $1 - t = 1$ . ✓
5. Step 5
  Point of intersection: substitute $s = 1$ into $\ell_{1}$ : $(1 + 2, 0 + 1, 2 - 1) = (3, 1, 1)$ . Verify on $\ell_{2}$ with $t = 0$ : $(3, 1, 1)$ . ✓
Answer
Intersect at $(3, 1, 1)$ .
Examiner tip
M1 for setting components equal. A1 for the three equations. M1 for solving first two. A1 for $s, t$ values. M1 for verifying with the third. A1 for the intersection point. A1 for the verification step (or 'hence intersect'). Critical: always check the third equation — two parameter values can satisfy two equations but FAIL the third, meaning lines are skew, not intersecting.
6Classify two lines as parallel / intersecting / skew (6 marks, P4)
Extended• Adapted from WMA14/01 June 2024 Q13• parallel, skew, classification
Question
Classify the lines $\ell_{1}: \mathbf{r} = (1, 2, 3) + s(2, -1, 1)$ and $\ell_{2}: \mathbf{r} = (0, 1, 5) + t(4, -2, 2)$ as parallel, intersecting or skew.
Step-by-step solution
1. Step 1
  Check parallel: direction vectors $(2, -1, 1)$ and $(4, -2, 2)$ . Ratio: $4 = 2 \cdot 2$ , $-2 = 2 \cdot (-1)$ , $2 = 2 \cdot 1$ . All ratios equal $2$ — directions are parallel.
2. Step 2
  Now check if the lines are the same (i.e. $\ell_{2}$ shares a point with $\ell_{1}$ ). Take $(0, 1, 5)$ on $\ell_{2}$ and check if it's on $\ell_{1}$ : $0 = 1 + 2s$ → $s = -1/2$ ; then $\ell_{1}$ gives $2 - (-1/2) = 5/2 \neq 1$ . So $(0, 1, 5)$ NOT on $\ell_{1}$ .
3. Step 3
  Directions parallel but lines distinct ⇒ the lines are parallel and never meet.
Answer
$\ell_{1}$ and $\ell_{2}$ are PARALLEL (and distinct).
Examiner tip
M1 for examining the ratio of direction vectors. A1 for noting they are scalar multiples (ratio $2:1$ ). M1 for testing if a point on one line is on the other. A1 for showing the point is not. A1 for the classification 'parallel'. A1 for the qualifier 'distinct'. Standard P4 trap: students claim 'parallel' without checking coincidence.
7Acute angle between two lines (4 marks, P4)
Extended• Adapted from WMA14/01 January 2023 Q13• angle between lines
Question
Find the acute angle between lines with direction vectors $\mathbf{d}_{1} = (2, 1, 2)$ and $\mathbf{d}_{2} = (1, -2, 2)$ .
Step-by-step solution
1. Step 1
  Dot product: $(2)(1) + (1)(-2) + (2)(2) = 2 - 2 + 4 = 4$ .
2. Step 2
  Magnitudes: $|\mathbf{d}_{1}| = \sqrt{4 + 1 + 4} = 3$ ; $|\mathbf{d}_{2}| = \sqrt{1 + 4 + 4} = 3$ .
3. Step 3
  $\cos\theta = \dfrac{4}{9}$ .
4. Step 4
  Since $\cos\theta > 0$ , $\theta$ is already acute. $\theta = \arccos(4/9) \approx 63.6°$ .
Answer
$\theta \approx 63.6°$ (i.e. $\arccos(4/9)$ ).
Examiner tip
M1 for dot product. A1 for $4$ . M1 for magnitudes. A1 for $63.6°$ . For acute angle: take $|\mathbf{d}_{1} \cdot \mathbf{d}_{2}|$ in the cosine formula. If the dot product is negative, the calculated angle is obtuse; the acute angle is its supplement.

Key Formulae — Vectors

The formulae you need to memorise for vectors on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Vector magnitude
$|\mathbf{v}| = \sqrt{x^{2} + y^{2} + z^{2}}$
$\mathbf{v} = (x, y, z)$
vector in 3D
When to use
Length of a vector. NOT in the formula booklet — memorise. In 2D: $|\mathbf{v}| = \sqrt{x^{2} + y^{2}}$ .
Example
$|2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}| = \sqrt{4 + 9 + 36} = 7$ .
Unit vector
$\hat{\mathbf{v}} = \dfrac{\mathbf{v}}{|\mathbf{v}|}$
$\hat{\mathbf{v}}$
unit vector in direction of $\mathbf{v}$
When to use
When a direction vector is needed in standardised (length- $1$ ) form. NOT in the formula booklet.
Example
Unit vector in direction $(3, 4)$ : divide by $|\mathbf{v}| = 5$ , giving $(3/5, 4/5)$ .
Dot (scalar) product
$\mathbf{a} \cdot \mathbf{b} = a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3} = |\mathbf{a}||\mathbf{b}|\cos\theta$
$\theta$
angle between $\mathbf{a}$ and $\mathbf{b}$
When to use
Computing the angle between vectors, testing perpendicularity, projecting one vector onto another. The geometric form $|\mathbf{a}||\mathbf{b}|\cos\theta$ IS in the formula booklet.
Example
$\mathbf{a} \cdot \mathbf{b} = 0 \Leftrightarrow \mathbf{a} \perp \mathbf{b}$ (provided neither is zero).
Angle between two vectors
$\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$
$\theta$
angle between vectors, $0 \leq \theta \leq \pi$
When to use
Standard angle-finding formula. For ACUTE angle between two lines, use $|\mathbf{a} \cdot \mathbf{b}|$ in the numerator. IS in the formula booklet (as a rearrangement of the dot-product geometric form).
Example
Angle between $(2, 1, 2)$ and $(1, -2, 2)$ : $\cos\theta = 4/9$ , $\theta \approx 63.6°$ .
Vector equation of a line
$\mathbf{r} = \mathbf{a} + t \, \mathbf{d}, \;\; t \in \mathbb{R}$
$\mathbf{r}$
general point on the line
$\mathbf{a}$
position vector of a known point on the line
$\mathbf{d}$
direction vector
$t$
scalar parameter
When to use
Standard form for a line through point $\mathbf{a}$ in direction $\mathbf{d}$ . NOT in the formula booklet — memorise.
Example
Line through $(1, 2, -1)$ with direction $(3, -2, 4)$ : $\mathbf{r} = (1, 2, -1) + t(3, -2, 4)$ .

Key Definitions and Keywords — Vectors

Definitions to memorise and the exact keywords mark schemes credit for vectors answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Position vector
Examiner keyword
The vector from the origin to a point — $\vec{OA}$ for point $A$ . If $A = (x, y, z)$ , then position vector $\mathbf{a} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ .
Example
Position vector of $(3, -1, 2)$ is $3\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ .
Magnitude (length, modulus) of a vector
Examiner keyword
The length of the vector — a non-negative scalar. Computed via Pythagoras: $|\mathbf{v}| = \sqrt{x^{2} + y^{2} + z^{2}}$ in 3D.
Unit vector
A vector of length $1$ in a specified direction. Obtained by dividing a vector by its magnitude: $\hat{\mathbf{v}} = \mathbf{v}/|\mathbf{v}|$ . Standard unit vectors are $\mathbf{i}, \mathbf{j}, \mathbf{k}$ .
Dot (scalar) product
Examiner keyword
An operation taking two vectors and returning a scalar: $\mathbf{a} \cdot \mathbf{b} = a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}$ . Geometrically equals $|\mathbf{a}||\mathbf{b}|\cos\theta$ .
Skew lines
Examiner keyword
Two lines in 3D that are NOT parallel AND DO NOT intersect. They lie on different (non-parallel) directions and miss each other. Skew lines only exist in 3D — in 2D, two non-parallel lines always intersect.
Parallel vectors
Examiner keyword
Two non-zero vectors are parallel if one is a scalar multiple of the other: $\mathbf{a} = k\mathbf{b}$ for some real $k \neq 0$ . Lines with parallel direction vectors are themselves parallel.

Common Mistakes and Misconceptions — Vectors

The traps other students keep falling into on vectors questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Computing the dot product as $a_{1} + a_{2} + a_{3}$ (component sum, no multiplication)
Why it happens
Students confuse $\mathbf{a} \cdot \mathbf{b}$ with $\mathbf{a} + \mathbf{b}$ .
How to avoid it
$\mathbf{a} \cdot \mathbf{b} = \sum a_{i} b_{i}$ — multiply matching components, then sum. Always write the three products explicitly.
✕Concluding lines intersect after solving only two equations
WMA14/01 examiner reports — flagged annually
Why it happens
Students solve the first two parametric equations and stop.
How to avoid it
Always verify the THIRD equation with the found values of $s, t$ . If it fails, the lines are SKEW (or parallel, if directions match). Examiners include this check in mark schemes as the deciding A1.
✕Mixing up the position vector and the direction vector in $\mathbf{r} = \mathbf{a} + t\mathbf{d}$
Why it happens
Students treat both as 'just vectors'.
How to avoid it
$\mathbf{a}$ is a POINT on the line; $\mathbf{d}$ is the DIRECTION. Two lines with the same $\mathbf{a}$ but different $\mathbf{d}$ intersect at $\mathbf{a}$ . Two lines with same $\mathbf{d}$ but different $\mathbf{a}$ are parallel.
✕Quoting the obtuse angle when the acute is asked
Why it happens
Students use the signed dot product directly: if $\mathbf{a} \cdot \mathbf{b} < 0$ , the formula gives an obtuse angle.
How to avoid it
For the ACUTE angle between two LINES (vs vectors), take $\cos\theta = \dfrac{|\mathbf{a} \cdot \mathbf{b}|}{|\mathbf{a}||\mathbf{b}|}$ . Use the modulus in the numerator.
✕Calling two coincident lines 'parallel' (or vice versa)
Why it happens
Students conflate the cases.
How to avoid it
Parallel directions AND a shared point ⇒ coincident (same line). Parallel directions AND no shared point ⇒ distinct parallel lines. Always test a point of one line on the other.
✕Forgetting $\hat{\mathbf{v}}$ has length $1$
Why it happens
Students divide by magnitude but don't notice.
How to avoid it
Sanity check: $|\hat{\mathbf{v}}| = \sqrt{(2/7)^{2} + (-3/7)^{2} + (6/7)^{2}} = \sqrt{(4 + 9 + 36)/49} = \sqrt{1} = 1$ . ✓

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Vectors — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Vectors

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Magnitude and unit vector (4 marks, P4)

2Dot product and angle (5 marks, P4)

3Perpendicularity (4 marks, P4)

4Vector equation of a line through two points (5 marks, P4)

5Showing two lines intersect — find the point (7 marks, P4)

6Classify two lines as parallel / intersecting / skew (6 marks, P4)

7Acute angle between two lines (4 marks, P4)

Vector magnitude

Unit vector

Dot (scalar) product

Angle between two vectors

Vector equation of a line

Position vector

Magnitude (length, modulus) of a vector

Unit vector

Dot (scalar) product

Skew lines

Parallel vectors

✕Computing the dot product as a1+a2+a3a_{1} + a_{2} + a_{3}a1​+a2​+a3​ (component sum, no multiplication)

✕Concluding lines intersect after solving only two equations

✕Mixing up the position vector and the direction vector in r=a+td\mathbf{r} = \mathbf{a} + t\mathbf{d}r=a+td

✕Quoting the obtuse angle when the acute is asked

✕Calling two coincident lines 'parallel' (or vice versa)

✕Forgetting v^\hat{\mathbf{v}}v^ has length 111

Practice questions

Past paper style quiz

Vectors — frequently asked questions

✕Computing the dot product as $a_{1} + a_{2} + a_{3}$ (component sum, no multiplication)

✕Mixing up the position vector and the direction vector in $\mathbf{r} = \mathbf{a} + t\mathbf{d}$

✕Forgetting $\hat{\mathbf{v}}$ has length $1$