The Scalar Product (Dot Product)

This lesson covers the scalar product (also called the dot product) of two vectors — as required by the Edexcel A-Level Mathematics specification (9MA0). You need to know both the geometric and component forms, use it to find angles between vectors, and apply the condition for perpendicularity.

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Definition — Geometric Form

The scalar product of two vectors a and b is defined as:

a · b = |a| |b| cos θ

where θ is the angle between the vectors (when placed tail-to-tail), and 0° ≤ θ ≤ 180°.

Key Properties of the Geometric Form

• The result is a scalar (a number), not a vector.
• If θ = 90° (vectors are perpendicular), then a · b = 0 (since cos 90° = 0).
• If θ = 0° (vectors are parallel, same direction), then a · b = |a||b|.
• If θ = 180° (vectors are anti-parallel), then a · b = -|a||b|.

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Definition — Component Form

For vectors a = (a₁, a₂, a₃) and b = (b₁, b₂, b₃):

a · b = a₁b₁ + a₂b₂ + a₃b₃

This is the sum of the products of corresponding components.

Example 1: (2, 3, -1) · (4, -2, 5) = (2)(4) + (3)(-2) + (-1)(5) = 8 - 6 - 5 = -3

Example 2: (1, 0, 3) · (2, 5, -1) = 2 + 0 - 3 = -1

Example 3: (3, 4) · (4, -3) (2D) = 12 - 12 = 0

Since the dot product is zero, these 2D vectors are perpendicular.

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Finding the Angle Between Two Vectors

Combining both forms:

a · b = |a||b| cos θ = a₁b₁ + a₂b₂ + a₃b₃

Rearranging for the angle:

cos θ = (a₁b₁ + a₂b₂ + a₃b₃) / (|a||b|)

Worked Example 1

Find the angle between a = (1, 2, 3) and b = (4, -1, 2).

Step 1: Calculate a · b = 4 - 2 + 6 = 8

Step 2: Calculate magnitudes. |a| = √(1 + 4 + 9) = √14 |b| = √(16 + 1 + 4) = √21

Step 3: Find cos θ. cos θ = 8 / (√14 × √21) = 8 / √294 = 8 / (7√6) ≈ 0.4665

Step 4: θ = arccos(0.4665) ≈ 62.2°

Worked Example 2

Find the angle between p = (3, 1) and q = (-1, 3) (2D vectors).

p · q = -3 + 3 = 0

Since the dot product is zero, the angle is 90° — the vectors are perpendicular.

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Perpendicularity Condition

Two vectors are perpendicular if and only if their scalar product is zero:

a · b = 0 ⟺ a ⊥ b (provided neither is the zero vector)

This is an extremely useful test and is frequently examined.

Example

Show that a = (2, -1, 3) and b = (1, 5, 1) are perpendicular.

a · b = (2)(1) + (-1)(5) + (3)(1) = 2 - 5 + 3 = 0

Since a · b = 0, the vectors are perpendicular. ✓

Finding a Perpendicular Vector

Example: Find a value of p such that (p, 3, 2) is perpendicular to (1, -2, 4).

(p, 3, 2) · (1, -2, 4) = p - 6 + 8 = p + 2

For perpendicularity: p + 2 = 0 → p = -2

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Properties of the Scalar Product

1. Commutative: a · b = b · a
2. Distributive: a · (b + c) = a · b + a · c
3. Scalar factor: (ka) · b = k(a · b)
4. Self dot product: a · a = |a|²

Property 4 is particularly useful: if you need |a|², you can compute a · a.

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Scalar Product with Unit Vectors

The scalar product of a with a unit vector n̂ gives the component of a in the direction of n̂:

Component of a in direction of n̂ = a · n̂

Example: Find the component of a = (3, 4, 0) in the direction of b = (1, 0, 0).

Since b is already a unit vector (|b| = 1): Component = a · b = 3

The component of a in the x-direction is 3 (as expected).

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Angles in Geometric Figures

Angle of a Triangle

To find the angle at vertex B in triangle ABC:

1. Find BA and BC (vectors FROM B to the other vertices).
2. Calculate cos(angle B) = (BA · BC) / (|BA| × |BC|).

Example: Find angle ABC where A(1, 2, 0), B(3, 1, -1), C(2, 3, 1).

BA = a - b = (1 - 3, 2 - 1, 0 - (-1)) = (-2, 1, 1) BC = c - b = (2 - 3, 3 - 1, 1 - (-1)) = (-1, 2, 2)

BA · BC = 2 + 2 + 2 = 6 |BA| = √(4 + 1 + 1) = √6 |BC| = √(1 + 4 + 4) = 3

cos(ABC) = 6 / (√6 × 3) = 6 / (3√6) = 2/√6 = √6/3

Angle ABC = arccos(√6/3) ≈ 35.3°

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Common Mistakes

1. Forgetting the dot product gives a scalar, not a vector — If your answer for a · b is a vector, something is wrong.
2. Using the wrong angle — The vectors must be placed tail-to-tail to measure the angle between them.
3. Not simplifying — When showing perpendicularity, you only need to show the dot product is zero; no need to find magnitudes.
4. Sign errors — Be careful with negative components.

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Summary

• Component form: a · b = a₁b₁ + a₂b₂ + a₃b₃.
• Geometric form: a · b = |a||b| cos θ.
• Finding angles: cos θ = (a · b) / (|a||b|).
• Perpendicularity: a · b = 0 if and only if a ⊥ b.
• The dot product is commutative and distributive.
• a · a = |a|².

A-Level Deep Dive: The Scalar Product (Dot Product)

Spec mapping

Edexcel 9MA0-02 specification section 12 — Vectors (Year 2 Pure) covers vectors in three dimensions; calculate the magnitude of a 3D vector; understand and use position vectors; calculate the distance between two points represented by position vectors; use vectors to solve problems in pure mathematics and in context (including forces); understand and use the scalar product (refer to the official specification document for exact wording). The scalar product is a Year 2 only topic — it does not appear on AS Paper 1 or 9MA0-01. It is examined on 9MA0-02 Paper 2 — Pure Mathematics, and reappears synoptically on 9MA0-03 Paper 3 — Statistics and Mechanics (section 7, Forces and motion) when work done is computed as $W = \vec{F} \cdot \vec{d}$W=F⋅d. The Edexcel formula booklet does list $\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\vec{a}||\vec{b}|\cos\theta$a⋅b=a1​b1​+a2​b2​+a3​b3​=∣a∣∣b∣cosθ, but does not list the projection formula or the perpendicularity test — those must be derived from the listed identity.

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Worked example with full mark scheme

Question (8 marks):

The points $A$A, $B$B and $C$C have position vectors $\vec{a} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}$a=2i−j+3k, $\vec{b} = 4\mathbf{i} + 2\mathbf{j} - \mathbf{k}$b=4i+2j−k and $\vec{c} = \mathbf{i} + 5\mathbf{j} + 2\mathbf{k}$c=i+5j+2k.

(a) Find the angle $BAC$BAC, giving your answer in degrees to 1 d.p. (5)

(b) Determine whether $\overrightarrow{AB}$AB is perpendicular to the vector $\vec{n} = 3\mathbf{i} - 5\mathbf{j} - 4\mathbf{k}$n=3i−5j−4k. (3)

Solution with mark scheme:

(a) Step 1 — find the displacement vectors from $A$A.

$\overrightarrow{AB} = \vec{b} - \vec{a} = (4-2)\mathbf{i} + (2-(-1))\mathbf{j} + (-1-3)\mathbf{k} = 2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}$AB=b−a=(4−2)i+(2−(−1))j+(−1−3)k=2i+3j−4k

$\overrightarrow{AC} = \vec{c} - \vec{a} = (1-2)\mathbf{i} + (5-(-1))\mathbf{j} + (2-3)\mathbf{k} = -\mathbf{i} + 6\mathbf{j} - \mathbf{k}$AC=c−a=(1−2)i+(5−(−1))j+(2−3)k=−i+6j−k

M1 — subtracting position vectors from $A$A (not toward $A$A). Common error: candidates compute $\vec{a} - \vec{b}$a−b instead of $\vec{b} - \vec{a}$b−a. The angle is the same numerically because $\cos\theta$cosθ is even, but the convention $\overrightarrow{AB} = \vec{b} - \vec{a}$AB=b−a is what examiners expect to see.

A1 — both displacement vectors correct.

Step 2 — compute the scalar product.

$\overrightarrow{AB} \cdot \overrightarrow{AC} = (2)(-1) + (3)(6) + (-4)(-1) = -2 + 18 + 4 = 20$AB⋅AC=(2)(−1)+(3)(6)+(−4)(−1)=−2+18+4=20

M1 — correct application of the component formula $a_1 b_1 + a_2 b_2 + a_3 b_3$a1​b1​+a2​b2​+a3​b3​. Sign errors on the $(-4)(-1) = +4$(−4)(−1)=+4 term are extremely common.

Step 3 — compute the magnitudes.

$|\overrightarrow{AB}| = \sqrt{2^2 + 3^2 + (-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$∣AB∣=22+32+(−4)2​=4+9+16​=29​

$|\overrightarrow{AC}| = \sqrt{(-1)^2 + 6^2 + (-1)^2} = \sqrt{1 + 36 + 1} = \sqrt{38}$∣AC∣=(−1)2+62+(−1)2​=1+36+1​=38​

M1 — correct magnitudes (both must be right; either wrong loses this mark).

Step 4 — solve for $\theta$θ.

$\cos\theta = \dfrac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|} = \dfrac{20}{\sqrt{29}\sqrt{38}} = \dfrac{20}{\sqrt{1102}}$cosθ=∣AB∣∣AC∣AB⋅AC​=29​38​20​=1102​20​

Numerically $\sqrt{1102} \approx 33.196$1102​≈33.196, so $\cos\theta \approx 0.6025$cosθ≈0.6025, giving $\theta \approx 52.95°$θ≈52.95°.

A1 — final answer $\theta \approx 53.0°$θ≈53.0° (to 1 d.p.).

(b) Step 1 — compute the scalar product $\overrightarrow{AB} \cdot \vec{n}$AB⋅n.

$\overrightarrow{AB} \cdot \vec{n} = (2)(3) + (3)(-5) + (-4)(-4) = 6 - 15 + 16 = 7$AB⋅n=(2)(3)+(3)(−5)+(−4)(−4)=6−15+16=7

M1 — correct component-wise multiplication and sum.

A1 — value $7$7 correctly evaluated.

Step 2 — apply the perpendicularity criterion.

Since $\overrightarrow{AB} \cdot \vec{n} = 7 \neq 0$AB⋅n=7=0, the vectors are not perpendicular.

A1 (AO2.4) — explicit conclusion stated, with reference to the criterion $\vec{a} \cdot \vec{b} = 0 \iff \vec{a} \perp \vec{b}$a⋅b=0⟺a⊥b (for non-zero vectors).

Total: 8 marks (M3 A5).

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): The line $\ell_1$ℓ1​ has equation $\vec{r} = (\mathbf{i} + 2\mathbf{j} - \mathbf{k}) + \lambda(2\mathbf{i} - \mathbf{j} + 2\mathbf{k})$r=(i+2j−k)+λ(2i−j+2k). The line $\ell_2$ℓ2​ has direction vector $\vec{d}_2 = 3\mathbf{i} + \mathbf{j} - \mathbf{k}$d2​=3i+j−k.

(a) Find the acute angle between $\ell_1$ℓ1​ and $\ell_2$ℓ2​, giving your answer in degrees to 1 d.p. (4)

(b) Find the value of the constant $k$k such that $\vec{d}_1 + k\vec{d}_2$d1​+kd2​ is perpendicular to $\vec{d}_1$d1​, where $\vec{d}_1$d1​ is the direction of $\ell_1$ℓ1​. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — identifying the direction vector $\vec{d}_1 = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}$d1​=2i−j+2k and computing $\vec{d}_1 \cdot \vec{d}_2 = 6 - 1 - 2 = 3$d1​⋅d2​=6−1−2=3.
• M1 (AO1.1b) — computing magnitudes: $|\vec{d}_1| = \sqrt{4 + 1 + 4} = 3$∣d1​∣=4+1+4​=3 and $|\vec{d}_2| = \sqrt{9 + 1 + 1} = \sqrt{11}$∣d2​∣=9+1+1​=11​.
• A1 (AO1.1b) — $\cos\theta = \dfrac{3}{3\sqrt{11}} = \dfrac{1}{\sqrt{11}}$cosθ=311​3​=11​1​.
• A1 (AO2.5) — $\theta = \arccos(1/\sqrt{11}) \approx 72.5°$θ=arccos(1/11​)≈72.5°. Since this is already acute (less than 90°), no adjustment is needed.

(b)

• M1 (AO1.1b) — applying the perpendicularity criterion: $(\vec{d}_1 + k\vec{d}_2) \cdot \vec{d}_1 = 0$(d1​+kd2​)⋅d1​=0, expanding to $|\vec{d}_1|^2 + k(\vec{d}_1 \cdot \vec{d}_2) = 0$∣d1​∣2+k(d1​⋅d2​)=0, i.e. $9 + 3k = 0$9+3k=0.
• A1 (AO1.1b) — $k = -3$k=−3.

Total: 6 marks split AO1 = 5, AO2 = 1. This is an AO1-dominated question; the AO2 mark in (a) rewards the explicit handling of "acute" — candidates who simply quote $\arccos$arccos without checking the angle range can lose this mark when the dot product is negative and the obtuse angle would be returned instead.

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Synoptic links

Connects to:

1. Vector lines (section 11) — angle between two lines: the angle between two lines with direction vectors $\vec{d}_1$d1​, $\vec{d}_2$d2​ is computed using the dot product of the directions only. Position vectors of points on the lines are irrelevant for the angle — a fact 9MA0 Paper 2 examiners exploit by giving full line equations and watching whether candidates extract the directions cleanly. If the dot product is negative, take $|\vec{d}_1 \cdot \vec{d}_2|$∣d1​⋅d2​∣ to get the acute angle.

2. Shortest distance from a point to a line: the perpendicular distance from a point $P$P to a line through $A$A with direction $\vec{d}$d is found by writing the foot of perpendicular as $A + \lambda\vec{d}$A+λd and imposing $(\overrightarrow{PF}) \cdot \vec{d} = 0$(PF)⋅d=0. This single dot-product equation determines $\lambda$λ, after which $|\overrightarrow{PF}|$∣PF∣ is the required distance — a routinely-tested 6–8 mark question.

3. Mechanics — work done by a force (9MA0-03 section 7): when a constant force $\vec{F}$F acts through displacement $\vec{d}$d, the work done is $W = \vec{F} \cdot \vec{d} = |\vec{F}||\vec{d}|\cos\theta$W=F⋅d=∣F∣∣d∣cosθ. Negative work means the force opposes the motion ($\theta > 90°$θ>90°); zero work means the force is perpendicular to motion (e.g. normal reaction on a horizontally moving block).

4. Coordinate geometry (section 3) — perpendicularity in the plane: two 2D vectors $(a_1, a_2)$(a1​,a2​) and $(b_1, b_2)$(b1​,b2​) are perpendicular iff $a_1 b_1 + a_2 b_2 = 0$a1​b1​+a2​b2​=0. Specialising to direction vectors of lines, gradients $m_1$m1​ and $m_2$m2​ satisfy $m_1 m_2 = -1$m1​m2​=−1 — exactly the GCSE result, now derived from the dot product rather than memorised.

5. Physics A-Level — flux and projections: the dot product computes the component of one vector along another. Magnetic flux $\Phi = \vec{B} \cdot \vec{A}$Φ=B⋅A, electric work $W = q\vec{E} \cdot \vec{d}$W=qE⋅d, and power $P = \vec{F} \cdot \vec{v}$P=F⋅v all use the same operation. Confidence here pays compound dividends in synoptic A-Level Physics questions.

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Mark-scheme literacy

Scalar product questions on 9MA0-02 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Correct component-wise dot product, magnitude calculation, application of $\cos\theta$cosθ formula, perpendicularity test $\vec{a} \cdot \vec{b} = 0$a⋅b=0
AO2 (reasoning / interpretation)	20–30%	Stating "acute angle" explicitly, justifying perpendicularity conclusion, recognising when to use directions vs displacements, handling sign of dot product
AO3 (problem-solving)	10–20%	Multi-step problems combining angle-between-lines with foot-of-perpendicular or shortest-distance reasoning