Vectors — A-Level Mathematics Pure 3 Study Guide

For: A-Level Mathematics candidates sitting Paper 3 (Pure Mathematics 3).

Covers: Position vectors, displacement calculations, vector magnitude and unit vectors, dot product and angle between vectors, vector equation of a line, skew/parallel/intersecting line analysis, plane equations, and perpendicular distance from a point to a plane.

You should already know: A-Level Mathematics Pure 1 (functions, calculus, trigonometry, basic 2D vector operations).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Vectors?

Vectors are mathematical quantities that have both magnitude (size) and direction, unlike scalars which only have magnitude. In A-Level Mathematics Pure 3, we work almost exclusively with 3D vectors applied to geometry problems involving lines and planes, with notation using either column vectors $\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$ or unit base vectors $\mathbf{i},\mathbf{j},\mathbf{k}$ aligned to the x, y, and z axes respectively. This topic accounts for 10-15% of marks on Paper 3, with exam questions almost always combining multiple subtopics in a single multi-part problem.

2. Position vectors and displacement; $\overrightarrow{AB}=\vec{b}-\vec{a}$

A position vector describes the location of a point relative to the fixed origin $O$ of the coordinate system. The position vector of point $A$ is written as $\vec{a}$, so if $A$ has coordinates $(x_1,y_1,z_1)$, $\vec{a}=x_1\mathbf{i}+y_1\mathbf{j}+z_1\mathbf{k}$.

A displacement vector describes the movement from one point to another, independent of the origin. The displacement vector from point $A$ to point $B$ ($\overrightarrow{AB}$) is calculated as the position vector of the end point minus the position vector of the start point: $$\overrightarrow{AB}=\vec{b}-\vec{a}$$ This comes from the vector addition rule: to get from $A$ to $B$, you travel from $A$ to $O$ ($-\vec{a}$) then $O$ to $B$ ($\vec{b}$).

Worked Example

Point $A$ has coordinates $(2,-1,4)$ and point $B$ has coordinates $(5,3,-2)$. Calculate $\overrightarrow{AB}$:

1. Write position vectors: $\vec{a}=2\mathbf{i}-\mathbf{j}+4\mathbf{k}$, $\vec{b}=5\mathbf{i}+3\mathbf{j}-2\mathbf{k}$
2. Subtract: $\overrightarrow{AB}=(5-2)\mathbf{i}+(3-(-1))\mathbf{j}+(-2-4)\mathbf{k}=3\mathbf{i}+4\mathbf{j}-6\mathbf{k}$

Exam tip: Examiners regularly penalize students for reversing the order of this calculation, so always confirm you are using end point minus start point for displacement vectors.

3. Magnitude $|\vec{a}|$ and unit vectors

The magnitude of a vector is its length, calculated using a 3D extension of Pythagoras' theorem. For a vector $\vec{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$: $$|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$$

A unit vector is a vector with magnitude 1 that points in the same direction as the original vector. It is denoted $\hat{a}$ and calculated by dividing the original vector by its magnitude: $$\hat{a}=\frac{\vec{a}}{|\vec{a}|}$$

Worked Example

Calculate the magnitude of $\overrightarrow{AB}$ from the previous example, then find the unit vector in the direction of $\overrightarrow{AB}$:

1. Magnitude calculation: $|\overrightarrow{AB}|=\sqrt{3^2+4^2+(-6{)}^2}=\sqrt{9+16+36}=\sqrt{61}$
2. Unit vector: $\widehat{AB}=\frac{1}{\sqrt{61}}(3\mathbf{i}+4\mathbf{j}-6\mathbf{k})$

Exam note: You do not need to rationalize the denominator of unit vectors unless explicitly asked to do so, but simplified fractions are always preferred.

4. Dot product and the angle between two vectors

The dot product (or scalar product) is an operation that takes two vectors and returns a scalar value. It has two equivalent definitions, which you will use interchangeably to solve for unknown angles or vector components:

1. Component form: For $\vec{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$ and $\vec{b}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}$: $$\vec{a}\cdot \vec{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$$
2. Geometric form: Where $\theta$ is the angle between the two vectors when placed tail to tail: $$\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta$$

To find the angle between two vectors, equate the two definitions: $$\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$$

Key special cases:

• If vectors are perpendicular, $\vec{a}\cdot \vec{b}=0$ (since $\cos 90^{\circ }=0$)
• If vectors are parallel, $\vec{a}\cdot \vec{b}=\pm |\vec{a}||\vec{b}|$ (since $\cos 0^{\circ }=1$, $\cos 180^{\circ }=-1$)

Worked Example

Find the acute angle between $\vec{a}=2\mathbf{i}-\mathbf{j}+3\mathbf{k}$ and $\vec{b}=\mathbf{i}+4\mathbf{j}-2\mathbf{k}$:

1. Calculate dot product: $\vec{a}\cdot \vec{b}=(2)(1)+(-1)(4)+(3)(-2)=2-4-6=-8$
2. Calculate magnitudes: $|\vec{a}|=\sqrt{4+1+9}=\sqrt{14}$, $|\vec{b}|=\sqrt{1+16+4}=\sqrt{21}$
3. Calculate cosine: $\cos \theta =\frac{-8}{\sqrt{14}\sqrt{21}}\approx -0.466$
4. Obtuse angle from calculation is $\approx 117.8^{\circ }$, so acute angle is $180^{\circ }-117.8^{\circ }=62.2^{\circ }$

Exam tip: Always check if the question asks for the acute or obtuse angle, as 70% of students forget to adjust for negative cosine values when asked for the acute angle.

5. Equation of a line in vector form $\vec{r}=\vec{a}+t\vec{d}$

A straight line in 3D space is uniquely defined by two pieces of information: a fixed point that lies on the line, and a direction vector that is parallel to the line. The general vector equation of a line is: $$\vec{r}=\vec{a}+t\vec{d}$$ Where:

• $\vec{r}$ = position vector of any general point on the line
• $\vec{a}$ = position vector of a known fixed point on the line
• $\vec{d}$ = direction vector parallel to the line
• $t$ = scalar parameter that can take any real value to generate all points on the line

You can also rewrite this in two other common forms:

1. Parametric form: $x=a_1+t{d}_1$, $y=a_2+t{d}_2$, $z=a_3+t{d}_3$
2. Cartesian form: $\frac{x-a_1}{d_1}=\frac{y-a_2}{d_2}=\frac{z-a_3}{d_3}=t$

Worked Example

Find the vector and Cartesian equations of the line passing through point $A(1,-3,2)$ parallel to $\vec{d}=2\mathbf{i}+\mathbf{j}-4\mathbf{k}$:

1. Vector equation: $\vec{r}=\left(\begin{array}{c}1\\ -3\\ 2\end{array}\right)+t\left(\begin{array}{c}2\\ 1\\ -4\end{array}\right)$, $t\in \mathbb{R}$
2. Cartesian form: $\frac{x-1}{2}=\frac{y+3}{1}=\frac{z-2}{-4}$

Exam note: Direction vectors can be any scalar multiple of the original vector, so using $4\mathbf{i}+2\mathbf{j}-8\mathbf{k}$ instead of $2\mathbf{i}+\mathbf{j}-4\mathbf{k}$ is still correct, but simplified direction vectors make later calculations much easier.

6. Skew, parallel, intersecting lines — and where they meet

Unlike 2D space, two lines in 3D space can have three possible relationships:

1. Parallel: The direction vectors of the two lines are scalar multiples of each other, so $\overrightarrow{d_1}=k\overrightarrow{d_2}$ for some constant $k$. Parallel lines never meet unless they are coincident (identical lines).
2. Intersecting: The lines are not parallel, and there exist unique values of the parameters $t$ and $s$ such that the position vectors of points on the two lines are equal: $\overrightarrow{a_1}+t\overrightarrow{d_1}=\overrightarrow{a_2}+s\overrightarrow{d_2}$.
3. Skew: The lines are not parallel, and there is no solution for $t$ and $s$ that makes the position vectors equal, so the lines never meet and are not parallel. Skew lines only exist in 3D space.

Worked Example

Determine if lines $L_1:\vec{r}=\left(\begin{array}{c}2\\ 0\\ 1\end{array}\right)+t\left(\begin{array}{c}1\\ -1\\ 3\end{array}\right)$ and $L_2:\vec{r}=\left(\begin{array}{c}1\\ 2\\ -2\end{array}\right)+s\left(\begin{array}{c}2\\ 1\\ 1\end{array}\right)$ are parallel, intersecting, or skew:

1. Check parallel: Direction vectors (1,-1,3) and (2,1,1) are not scalar multiples, so not parallel.
2. Set equal to solve for t and s:

• $2+t=1+2s$ → $t-2s=-1$
• $0-t=2+s$ → $-t-s=2$
• $1+3t=-2+s$ → $3t-s=-3$

3. Solve first two equations: Add them to eliminate t: $-3s=1$ → $s=-\frac{1}{3}$, then $t=-\frac{5}{3}$
4. Check third equation: Left = $3(-\frac{5}{3})-(-\frac{1}{3})=-5+\frac{1}{3}=-\frac{14}{3}\ne -3$. No consistent solution, so lines are skew.

7. Equation of a plane and distance from a point

A plane is a flat 2D surface in 3D space, uniquely defined by a point on the plane and a normal vector that is perpendicular to all lines lying on the plane. The most commonly used forms of the plane equation are:

1. Scalar product form: $\vec{r}\cdot \mathbf{n}=d$, where $\mathbf{n}$ is the normal vector to the plane, and $d=\vec{a}\cdot \mathbf{n}$ for a point $\vec{a}$ on the plane.
2. Cartesian form: Expanding the scalar product gives $n_{1}x+n_{2}y+n_{3}z=d$, where $\mathbf{n}=n_1\mathbf{i}+n_2\mathbf{j}+n_3\mathbf{k}$.

The perpendicular distance from a point $P$ with position vector $\vec{p}$ to the plane $\vec{r}\cdot \mathbf{n}=d$ is given by: $$\text{Distance}=\frac{|\vec{p}\cdot \mathbf{n}-d|}{|\mathbf{n}|}$$ The absolute value ensures the distance is always positive.

Worked Example

A plane has normal vector $3\mathbf{i}-\mathbf{j}+2\mathbf{k}$ and passes through point $A(2,1,-4)$. (a) Find the Cartesian equation of the plane. (b) Calculate the distance from point $P(1,-2,3)$ to the plane.

1. Calculate d: $d=\vec{a}\cdot \mathbf{n}=(2)(3)+(1)(-1)+(-4)(2)=6-1-8=-3$
2. Cartesian equation: $3x-y+2z=-3$
3. Distance calculation:

• Numerator: $|(1)(3)+(-2)(-1)+(3)(2)-(-3)|=|3+2+6+3|=14$
• $|\mathbf{n}|=\sqrt{9+1+4}=\sqrt{14}$
• Distance = $\frac{14}{\sqrt{14}}=\sqrt{14}$

8. Common Pitfalls (and how to avoid them)

• Mistake: Calculating $\overrightarrow{AB}$ as $\vec{a}-\vec{b}$ instead of $\vec{b}-\vec{a}$ Why it happens: Students associate the first letter with the first vector. Fix: Remember displacement is always end point minus start point, and label vectors on a quick sketch if you get confused.
• Mistake: Assuming lines intersect after solving only the first two simultaneous equations Why it happens: Students carry over 2D line assumptions to 3D, forgetting skew lines exist. Fix: Always substitute your calculated t and s values into the third component equation to confirm consistency before stating an intersection point.
• Mistake: Using a direction vector of a line on the plane as the normal vector for the plane equation Why it happens: Confusion between line and plane requirements. Fix: The normal vector is perpendicular to all lines on the plane, so calculate it via cross product of two non-parallel vectors lying on the plane if it is not given.
• Mistake: Omitting the absolute value sign in the point-to-plane distance formula, leading to negative distance values Why it happens: Students forget distance is a scalar magnitude, not a directed vector. Fix: Always apply the absolute value to the numerator to return a positive distance.
• Mistake: Reporting the obtuse angle between vectors when the question explicitly asks for the acute angle Why it happens: Students use the direct output of arccos without checking the sign of the dot product. Fix: If your calculated $\cos \theta$ is negative, subtract the resulting angle from $180^{\circ }$ to get the acute angle.

9. Practice Questions (A-Level Mathematics Paper 3 Style)

Question 1

Points $A$ and $B$ have coordinates $(4,-2,1)$ and $(-1,3,5)$ respectively. (a) Find the displacement vector $\overrightarrow{AB}$ and its magnitude. [3 marks] (b) Find the unit vector in the direction of $\overrightarrow{BA}$. [2 marks]

Solution

(a) $\overrightarrow{AB}=\vec{b}-\vec{a}=(-1-4)\mathbf{i}+(3-(-2))\mathbf{j}+(5-1)\mathbf{k}=-5\mathbf{i}+5\mathbf{j}+4\mathbf{k}$ Magnitude: $|\overrightarrow{AB}|=\sqrt{(-5{)}^2+5^2+4^2}=\sqrt{25+25+16}=\sqrt{66}$ (b) $\overrightarrow{BA}=-\overrightarrow{AB}=5\mathbf{i}-5\mathbf{j}-4\mathbf{k}$, same magnitude as $\overrightarrow{AB}$. Unit vector = $\frac{1}{\sqrt{66}}(5\mathbf{i}-5\mathbf{j}-4\mathbf{k})$

Question 2

Two lines are given: $L_1:\vec{r}=\left(\begin{array}{c}1\\ -1\\ 3\end{array}\right)+t\left(\begin{array}{c}2\\ 1\\ -1\end{array}\right)$, $t\in \mathbb{R}$ $L_2:\vec{r}=\left(\begin{array}{c}4\\ -2\\ k\end{array}\right)+s\left(\begin{array}{c}1\\ -3\\ 2\end{array}\right)$, $s\in \mathbb{R}$, where $k$ is a constant. (a) Show that $L_1$ and $L_2$ are not parallel. [2 marks] (b) Given that $L_1$ and $L_2$ intersect, find the value of $k$ and the coordinates of the intersection point. [5 marks]

Solution

(a) Direction vectors of $L_1$ and $L_2$ are (2,1,-1) and (1,-3,2). For parallel lines, there must exist $k$ such that $2=k(1)$, $1=k(-3)$, $-1=k(2)$. $k=2$ from first equation, but $1\ne -6$, so no such $k$ exists, lines not parallel. (b) Set equal: $1+2t=4+s$ → $2t-s=3$ $-1+t=-2-3s$ → $t+3s=-1$ $3-t=k+2s$ Solve first two equations: Multiply first by 3: $6t-3s=9$, add to second: $7t=8$ → $t=\frac{8}{7}$, then $s=2(\frac{8}{7})-3=\frac{16}{7}-\frac{21}{7}=-\frac{5}{7}$ Substitute into third equation: $3-\frac{8}{7}=k+2(-\frac{5}{7})$ → $\frac{13}{7}=k-\frac{10}{7}$ → $k=\frac{23}{7}$ Intersection point: Plug $t=\frac{8}{7}$ into $L_1$: $(\frac{23}{7},\frac{1}{7},\frac{13}{7})$

Question 3

A plane passes through points $P(2,0,1)$, $Q(0,1,3)$ and $R(1,-1,2)$. (a) Find the Cartesian equation of the plane. [5 marks] (b) Calculate the perpendicular distance from the origin to the plane. [2 marks]

Solution

(a) Vectors in plane: $\overrightarrow{PQ}=(-2,1,2)$, $\overrightarrow{PR}=(-1,-1,1)$ Normal vector $\mathbf{n}=\overrightarrow{PQ}\times \overrightarrow{PR}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -2& 1& 2\\ -1& -1& 1\end{array}\right|=\mathbf{i}(1+2)-\mathbf{j}(-2+2)+\mathbf{k}(2+1)=3\mathbf{i}+0\mathbf{j}+3\mathbf{k}$ Simplify normal to $\mathbf{i}+\mathbf{k}$. Plane equation: $x+z=d$, plug in $P(2,0,1)$: $2+1=d=3$ Cartesian equation: $x+z=3$ (b) Distance from origin (0,0,0): $\frac{|0+0-3|}{\sqrt{1+0+1}}=\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}$

10. Quick Reference Cheatsheet

11. What's Next

Mastery of 3D vectors is critical for success in the remaining Pure 3 topics, as you will build on this content to solve line-plane intersection problems, calculate angles between lines and planes, and find angles between two planes in subsequent sections. Vectors also form the backbone of A-Level Mathematics Mechanics units, where you will use displacement, magnitude, and dot product operations to solve kinematics, force, and projectile problems.

If you struggled with any of the worked examples, formula applications, or exam-specific tricks covered in this guide, you can get personalized support from Ollie, our AI tutor, who can walk you through step-by-step explanations, generate additional practice questions, or create custom quizzes targeted to your weak spots. Head to the homepage to get started with your A-Level Mathematics Paper 3 revision today.

Aligned with the Cambridge International AS & A Level Mathematics 9709 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.