The scalar product, or dot product, of two vectors is a single number: a · b = a₁b₁ + a₂b₂ + a₃b₃, which also equals |a||b| cos θ, where θ is the angle between them. It gives the angle between two vectors or two lines, and it tests for perpendicular vectors: a · b = 0 exactly when the vectors are at right angles.
The scalar product
- a · b = a₁b₁ + a₂b₂ + a₃b₃
- a · b = |a||b| cos θ
- cos θ = (a · b) / (|a||b|)
- Perpendicular (non-zero) vectors: a · b = 0
- Acute angle between lines: cos θ = |d₁ · d₂| / (|d₁||d₂|)
Calculating the scalar product
Multiply the matching components and add the results. The answer is a scalar, a number, which may be positive, negative or zero. The scalar product is commutative, so a · b = b · a, and a vector dotted with itself gives its magnitude squared: a · a = |a|².
The angle between two vectors
Rearranging a · b = |a||b| cos θ gives cos θ = (a · b) / (|a||b|). The sign of the scalar product tells you the type of angle: positive means acute, zero means a right angle, and negative means obtuse.
Perpendicular vectors
Two non-zero vectors are perpendicular exactly when their scalar product is 0, because cos 90° = 0. This is a quick test for right angles, and setting a scalar product equal to zero lets you find an unknown component that makes two vectors perpendicular.
The angle between two lines
The angle between two lines is the angle between their direction vectors. Because a direction can point either way, use the modulus of the scalar product to get the acute angle: cos θ = |d₁ · d₂| / (|d₁||d₂|). Read the direction vectors from the λ and μ parts of the line equations, not the position vectors.
Angles inside shapes
To find the angle at a vertex B of a triangle ABC, use the two vectors that start at B: BA = a − b and BC = c − b. Using AB instead of BA reverses one vector and gives the supplementary angle by mistake. Angles in a triangle can be obtuse, so here you do not take the modulus.
Worked examples
Example 1. Find a · b for a = 2i + 3j − k and b = 4i − j + 2k.
- (2)(4) + (3)(−1) + (−1)(2).
- = 8 − 3 − 2.
Answer: a · b = 3.
Marks: M1 for multiplying matching components, A1 for 3.
Example 2. Find the angle between a = 2i + 2j + k and b = 6i + 3j + 2k.
- a · b = 12 + 6 + 2 = 20.
- |a| = √9 = 3 and |b| = √49 = 7.
- cos θ = 20/21.
Answer: θ = 17.8° (1 d.p.).
Marks: M1 for a · b, M1 for cos θ, A1 for 17.8°.
Example 3. Show that a = 2i − j + 4k and b = 3i + 2j − k are perpendicular.
- a · b = (2)(3) + (−1)(2) + (4)(−1) = 6 − 2 − 4 = 0.
Answer: a · b = 0, so the vectors are perpendicular.
Marks: M1 for the scalar product, A1 for 0 and the conclusion.
Example 4. Find the acute angle between lines with directions d₁ = i + 2j + 2k and d₂ = 2i − j + 2k.
- d₁ · d₂ = 2 − 2 + 4 = 4.
- |d₁| = 3 and |d₂| = 3, so cos θ = 4/9.
Answer: θ = 63.6° (1 d.p.).
Marks: M1 for the scalar product, M1 for cos θ, A1 for 63.6°.
Example 5. Find angle ABC for A(3, 4, 1), B(1, 1, 1) and C(2, −1, 3).
- BA = a − b = 2i + 3j and BC = c − b = i − 2j + 2k.
- BA · BC = 2 − 6 + 0 = −4.
- |BA| = √13 and |BC| = 3, so cos ABC = −4/(3√13).
Answer: Angle ABC = 111.7° (1 d.p.).
Marks: M1 for the vectors from B, M1 for cos, A1 for 111.7°.
Example 6. Find p so that the vector AB is perpendicular to the vector BC, for A(1, 2, 3), B(3, 1, 4) and C(2, 4, p).
- AB = 2i − j + k and BC = −i + 3j + (p − 4)k.
- AB · BC = −2 − 3 + p − 4 = 0.
Answer: p = 9.
Marks: M1 for the vectors, M1 for setting the scalar product to 0, A1 for p = 9.
Common mistakes
- Adding instead of multiplying components. a · b multiplies matching components, then adds.
- Using position vectors for the angle between lines. Use the direction vectors only.
- Using vectors that do not start at the vertex. For the angle at B, use BA and BC.
- Taking the modulus for an angle in a shape. Triangle angles can be obtuse; only take |d₁ · d₂| for the angle between lines.
- Forgetting to square negative components in |a|. Every component is squared before adding.
How the scalar product is examined
The scalar product is Pure 4 content (WMA14) in Edexcel International A-Level; in Edexcel A-Level it is taught in Further Maths rather than single Maths (9MA0). Questions ask you to:
- calculate a scalar product and the angle between two vectors
- find an unknown that makes two vectors perpendicular
- find the acute angle between two lines
- find an angle at a vertex of a triangle
Practise these, with mark schemes, in the Pure 4 vectors exam questions. The Pure 4 vectors worksheet and the A-Level vectors revision engine give further practice.
The scalar product is used in geometric problems with vectors and to find angles between intersecting lines.
Frequently asked questions
What is the scalar product?
The number a₁b₁ + a₂b₂ + a₃b₃ found by multiplying matching components of two vectors and adding. It also equals |a||b| cos θ.
How do you show two vectors are perpendicular?
Show that their scalar product is 0.
How do you find the angle between two vectors?
Use cos θ = (a · b) / (|a||b|), then take the inverse cosine.