CAIE A-Level · Mathematics 9709 · Vectors

The Scalar Product — Pure Mathematics 3 (9709)

11 min readSyllabus 3.7PreviewBy Uzair Khan

Syllabus objective

Use formulae to calculate the scalar product of two vectors, and use scalar products in problems involving lines and points, e.g. finding the angle between two lines, and finding the foot of the perpendicular from a point to a line; questions may involve 3D objects such as cuboids and tetrahedra. Knowledge of the vector product is not required.

Introduction

The scalar product (also called the dot product) is an operation that combines two vectors to produce a single scalar (a real number). For the 9709 exam, it is the essential tool for two key tasks: finding the angle between two lines, and finding the foot of the perpendicular from a point to a line. It also underpins questions set on 3D objects such as cuboids and tetrahedra. Mastering this topic unlocks the final layer of the Vectors unit, building directly on your knowledge of vector equations of lines.


Core Concept

Given two vectors a\mathbf{a} and b\mathbf{b}, the scalar product measures "how much" one vector points in the direction of the other. Geometrically, if θ\theta is the angle between the vectors when placed tail-to-tail (0°θ180°0° \leq \theta \leq 180°), then:

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

Key geometric consequences:

  • If ab=0\mathbf{a} \cdot \mathbf{b} = 0 and neither vector is zero, then cosθ=0\cos\theta = 0, so θ=90°\theta = 90°: the vectors are perpendicular.
  • If ab>0\mathbf{a} \cdot \mathbf{b} > 0 then θ\theta is acute; if ab<0\mathbf{a} \cdot \mathbf{b} < 0 then θ\theta is obtuse.

Angle between two lines: Lines have directions, not orientations — so the angle between two lines is always taken as the acute (or right) angle, i.e. between 0° and 90°90°. If the formula gives an obtuse angle θ\theta, use 180°θ180° - \theta instead.

Foot of the perpendicular: Given a point AA not on line \ell, the foot of the perpendicular FF is the unique point on \ell such that AF\overrightarrow{AF} \perp \ell. This is found by expressing FF in terms of the parameter λ\lambda, then imposing AFd=0\overrightarrow{AF} \cdot \mathbf{d} = 0 (where d\mathbf{d} is the direction vector of \ell), and solving for λ\lambda.


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