Vectors

A complete guide to A-Level vectors: position vectors, magnitude, unit vectors, the scalar (dot) product, and geometric proofs using vectors.

Vectors are a fundamental mathematical tool used to represent quantities that have both magnitude (size) and direction. At A-Level, vectors appear in both pure mathematics and mechanics, making them one of the most versatile topics in the specification.

At AS Level, AQA, Edexcel, and OCR all require you to understand vector notation, perform arithmetic with vectors, calculate magnitudes, find unit vectors, use the scalar (dot) product, and apply vectors to geometric problems. This topic builds directly on the basic vector work done at GCSE and extends it significantly with the introduction of the scalar product and formal geometric reasoning.

Vectors provide an elegant alternative to coordinate geometry for proving geometric results. They are also essential in mechanics for resolving forces and describing motion. Mastering vectors at this stage will serve you well throughout the rest of your A-Level studies and beyond.

Core Concepts

Vector Notation and Representation

A vector can be represented in several ways:

  • Column vector: a=(a1a2)\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} in 2D, or a=(a1a2a3)\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} in 3D.
  • Component form: a=a1i+a2j\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} in 2D, or a=a1i+a2j+a3k\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} in 3D, where i\mathbf{i}, j\mathbf{j}, k\mathbf{k} are unit vectors along the xx-, yy-, and zz-axes respectively.
  • Directed line segment: AB\overrightarrow{AB} represents the vector from point AA to point BB.

Vectors are typically printed in bold (a\mathbf{a}) or with an underline (a\underline{a}) in handwriting. Scalars (ordinary numbers) are not bold.

Position Vectors

The position vector of a point PP is the vector OP\overrightarrow{OP} from the origin OO to the point PP. If P=(3,1,4)P = (3, -1, 4), then the position vector of PP is:

OP=(314)=3ij+4k\overrightarrow{OP} = \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} = 3\mathbf{i} - \mathbf{j} + 4\mathbf{k}

The vector from point AA (with position vector a\mathbf{a}) to point BB (with position vector b\mathbf{b}) is:

AB=ba\overrightarrow{AB} = \mathbf{b} - \mathbf{a}

This is one of the most frequently used results in vector problems.

Vector Arithmetic

Addition: Vectors are added component by component. (a1a2)+(b1b2)=(a1+b1a2+b2)\begin{pmatrix} a_1 \\ a_2 \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \end{pmatrix}

Scalar multiplication: Each component is multiplied by the scalar. λ(a1a2)=(λa1λa2)\lambda\begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \begin{pmatrix} \lambda a_1 \\ \lambda a_2 \end{pmatrix}

Subtraction: ab=a+(b)\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b}).

Parallel vectors: Two vectors are parallel if one is a scalar multiple of the other: b=λa\mathbf{b} = \lambda\mathbf{a} for some scalar λ\lambda.

Magnitude of a Vector

The magnitude (or modulus) of a vector a=(a1a2a3)\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} is:

a=a12+a22+a32|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}

In 2D, this simplifies to a=a12+a22|\mathbf{a}| = \sqrt{a_1^2 + a_2^2}.

The magnitude represents the length of the vector. The distance between points AA and BB is AB|\overrightarrow{AB}|.

Unit Vectors

A unit vector has magnitude 1. The unit vector in the direction of a\mathbf{a} is:

a^=aa\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}

Unit vectors are used to specify direction without regard to magnitude. The standard unit vectors i\mathbf{i}, j\mathbf{j}, k\mathbf{k} are unit vectors along the coordinate axes.

The Scalar (Dot) Product

The scalar product (or dot product) of two vectors a\mathbf{a} and b\mathbf{b} is defined in two equivalent ways:

Algebraic definition: ab=a1b1+a2b2+a3b3\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3

Geometric definition: ab=abcosθ\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta

where θ\theta is the angle between the two vectors.

From the geometric definition, the angle between two vectors can be found:

cosθ=abab\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}

Key properties:

  • ab=ba\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} (commutative)
  • aa=a2\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2
  • If ab=0\mathbf{a} \cdot \mathbf{b} = 0 and neither vector is the zero vector, then a\mathbf{a} and b\mathbf{b} are perpendicular.

Midpoints and Section Formulae

The midpoint MM of points AA and BB with position vectors a\mathbf{a} and b\mathbf{b} has position vector:

OM=a+b2\overrightarrow{OM} = \frac{\mathbf{a} + \mathbf{b}}{2}

More generally, the point dividing ABAB in the ratio m:nm:n has position vector:

na+mbm+n\frac{n\mathbf{a} + m\mathbf{b}}{m + n}

Strategy Tips

Tip 1: Draw a Diagram

Always start vector problems with a clear diagram showing the points and vectors involved. Label position vectors and direction vectors. This helps you set up the correct vector equations.

Tip 2: Use AB=ba\overrightarrow{AB} = \mathbf{b} - \mathbf{a}

The vector from AA to BB is the position vector of BB minus the position vector of AA. This is the single most important formula in vector geometry. Getting the order wrong (subtracting the wrong way) is a very common error.

Tip 3: Check Perpendicularity with the Dot Product

To show two vectors are perpendicular, compute their dot product. If ab=0\mathbf{a} \cdot \mathbf{b} = 0, they are perpendicular. This is cleaner and more efficient than using gradients.

Tip 4: Use Parallel Vectors for Collinearity

To show that three points AA, BB, CC are collinear (lie on the same straight line), show that AB=λAC\overrightarrow{AB} = \lambda\overrightarrow{AC} for some scalar λ\lambda. This means the vectors are parallel and share a common point.

Tip 5: Be Precise with Notation

Use bold or underlined letters for vectors in your written work. Clearly distinguish between vectors and scalars. Write a|\mathbf{a}| for the magnitude, not a\mathbf{|a|}. Correct notation demonstrates understanding and avoids ambiguity.

Worked Example: Example 1

Problem

Points AA and BB have position vectors a=2i+3jk\mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k} and b=5ij+2k\mathbf{b} = 5\mathbf{i} - \mathbf{j} + 2\mathbf{k}. Find AB\overrightarrow{AB} and AB|\overrightarrow{AB}|.

Solution

AB=ba=(52)i+(13)j+(2(1))k=3i4j+3k\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (5 - 2)\mathbf{i} + (-1 - 3)\mathbf{j} + (2 - (-1))\mathbf{k} = 3\mathbf{i} - 4\mathbf{j} + 3\mathbf{k}

AB=32+(4)2+32=9+16+9=34|\overrightarrow{AB}| = \sqrt{3^2 + (-4)^2 + 3^2} = \sqrt{9 + 16 + 9} = \sqrt{34}

Worked Example: Example 2

Problem

Find the unit vector in the direction of v=(623)\mathbf{v} = \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix}.

Solution

v=62+(2)2+32=36+4+9=49=7|\mathbf{v}| = \sqrt{6^2 + (-2)^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7

v^=17(623)=(6/72/73/7)\hat{\mathbf{v}} = \frac{1}{7}\begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix} = \begin{pmatrix} 6/7 \\ -2/7 \\ 3/7 \end{pmatrix}

Worked Example: Example 3

Problem

Find the angle between a=(121)\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} and b=(312)\mathbf{b} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}.

Solution

ab=(1)(3)+(2)(1)+(1)(2)=322=1\mathbf{a} \cdot \mathbf{b} = (1)(3) + (2)(-1) + (-1)(2) = 3 - 2 - 2 = -1

a=1+4+1=6|\mathbf{a}| = \sqrt{1 + 4 + 1} = \sqrt{6}

b=9+1+4=14|\mathbf{b}| = \sqrt{9 + 1 + 4} = \sqrt{14}

cosθ=1614=184=1221\cos\theta = \frac{-1}{\sqrt{6}\sqrt{14}} = \frac{-1}{\sqrt{84}} = \frac{-1}{2\sqrt{21}}

θ=arccos(1221)96.3°\theta = \arccos\left(\frac{-1}{2\sqrt{21}}\right) \approx 96.3°

Worked Example: Example 4

Problem

Show that the vectors p=2i+3j\mathbf{p} = 2\mathbf{i} + 3\mathbf{j} and q=6i4j\mathbf{q} = 6\mathbf{i} - 4\mathbf{j} are perpendicular.

Solution

pq=(2)(6)+(3)(4)=1212=0\mathbf{p} \cdot \mathbf{q} = (2)(6) + (3)(-4) = 12 - 12 = 0

Since pq=0\mathbf{p} \cdot \mathbf{q} = 0 and neither p\mathbf{p} nor q\mathbf{q} is the zero vector, the vectors are perpendicular. \square

Worked Example: Example 5

Problem

Points AA, BB, and CC have position vectors a=(13)\mathbf{a} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}, b=(49)\mathbf{b} = \begin{pmatrix} 4 \\ 9 \end{pmatrix}, and c=(613)\mathbf{c} = \begin{pmatrix} 6 \\ 13 \end{pmatrix}. Prove that AA, BB, CC are collinear.

Solution

AB=ba=(36)\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 3 \\ 6 \end{pmatrix}

AC=ca=(510)\overrightarrow{AC} = \mathbf{c} - \mathbf{a} = \begin{pmatrix} 5 \\ 10 \end{pmatrix}

Now AC=53AB\overrightarrow{AC} = \frac{5}{3}\overrightarrow{AB}, so AB\overrightarrow{AB} and AC\overrightarrow{AC} are parallel.

Since they share the common point AA, the points AA, BB, CC are collinear. \square

Practice Problems

  1. Problem 1

    Given a=3i2j+k\mathbf{a} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k} and b=i+4j3k\mathbf{b} = \mathbf{i} + 4\mathbf{j} - 3\mathbf{k}, find ab\mathbf{a} \cdot \mathbf{b} and the angle between them. [Answer: ab=8\mathbf{a} \cdot \mathbf{b} = -8, θ120.5°\theta \approx 120.5°]

    Problem 2

    Find the value of tt such that (2t1)\begin{pmatrix} 2 \\ t \\ -1 \end{pmatrix} and (426)\begin{pmatrix} 4 \\ -2 \\ 6 \end{pmatrix} are perpendicular. [Answer: t=1t = 1]

    Problem 3

    The position vectors of PP and QQ are 2i+5j2\mathbf{i} + 5\mathbf{j} and 8i3j8\mathbf{i} - 3\mathbf{j} respectively. Find the position vector of the midpoint of PQPQ and the distance PQPQ. [Answer: midpoint =5i+j= 5\mathbf{i} + \mathbf{j}, PQ=10|PQ| = 10]

    Problem 4

    Find a unit vector perpendicular to both (100)\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} and (010)\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}. [Answer: (001)\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} or (001)\begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix}]

    Problem 5

    Points A(1,2,3)A(1, 2, 3), B(3,0,5)B(3, 0, 5), and C(k,6,1)C(k, 6, -1) are such that angle ABC=90°ABC = 90°. Find kk. [Answer: k=7k = 7]

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Common Mistakes

  • Subtracting position vectors in the wrong order. AB=ba\overrightarrow{AB} = \mathbf{b} - \mathbf{a}, not ab\mathbf{a} - \mathbf{b}. Think: "destination minus start".

  • Confusing the scalar product with the cross product. At A-Level, you primarily use the scalar (dot) product, which gives a number. The cross product (which gives a vector) is not on most A-Level syllabuses.

  • Forgetting to square root when finding magnitude. a=a12+a22+a32|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}, not a12+a22+a32a_1^2 + a_2^2 + a_3^2.

  • Assuming perpendicular means the dot product is 11. Perpendicular vectors have a dot product of 00, not 11. A dot product of 11 has no special geometric significance in general.

  • Not stating conclusions in geometric proofs. After computing AB=λAC\overrightarrow{AB} = \lambda\overrightarrow{AC}, you must explicitly state that the vectors are parallel and share a common point, therefore the points are collinear.

  • Mixing up 2D and 3D. When working in 3D, ensure you include all three components. Forgetting the k\mathbf{k} component leads to incorrect magnitudes and dot products.

Frequently Asked Questions

What is the difference between a vector and a position vector?

A vector represents a displacement — it has magnitude and direction but no fixed position. A position vector is specifically the vector from the origin to a given point. So OP\overrightarrow{OP} is a position vector, while AB\overrightarrow{AB} is a general vector.

Can the scalar product be negative?

Yes. The scalar product ab=abcosθ\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta is negative when the angle between the vectors is obtuse (90°<θ<180°90° < \theta < 180°), because cosθ<0\cos\theta < 0 in that range.

How do I show that a triangle is right-angled using vectors?

Find the vectors representing two sides of the triangle that meet at the suspected right angle. Compute their dot product. If it equals zero, the angle between them is 90°90°, confirming the right angle.

Do I need to know the cross product for A-Level?

No. The cross (vector) product is not part of the A-Level Maths specification for AQA, Edexcel, or OCR. It may appear in Further Maths. At A-Level, the scalar (dot) product is sufficient.

When should I use column vectors vs $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ notation?

Both notations are equivalent and equally acceptable. Use whichever the question uses or whichever you find clearer. Column vectors can be neater for calculations; i\mathbf{i}, j\mathbf{j}, k\mathbf{k} notation is sometimes clearer for writing equations.

Key Takeaways

  • Vectors have magnitude and direction. Unlike scalars, vectors encode directional information. This makes them ideal for geometric reasoning and physical applications.

  • AB=ba\overrightarrow{AB} = \mathbf{b} - \mathbf{a} is essential. This formula converts between points and vectors. Master it — it appears in virtually every vector problem.

  • The scalar product has two forms. The algebraic form (a1b1+a2b2+a3b3a_1b_1 + a_2b_2 + a_3b_3) is used for computation. The geometric form (abcosθ|\mathbf{a}||\mathbf{b}|\cos\theta) is used for finding angles and understanding the result.

  • Zero dot product means perpendicular. This is the cleanest test for perpendicularity and is more elegant than using coordinate geometry methods.

  • Parallel vectors are scalar multiples. If b=λa\mathbf{b} = \lambda\mathbf{a}, the vectors point in the same (or opposite) direction. Use this to prove collinearity or parallelism.

  • Unit vectors normalise direction. Dividing a vector by its magnitude gives a unit vector — a pure direction with magnitude 11. This is useful for specifying directions and in mechanics problems.

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