A LevelFurther MathsEdexcelCore PureRevision NotesFurther VectorsVector LinesAngle between Lines

Angle between Lines (Edexcel A Level Further Maths: Core Pure): Revision Note

Exam code: 9FM0

Written by: Amber

Reviewed by: Dan Finlay

Updated on 2 July 2026

Scalar product

The scalar product is an important link between the algebra of vectors and the trigonometry of vectors. We shall see that the scalar product is somewhat comparable to the operation of multiplication on real numbers.

What is the scalar (dot) product?

The scalar product between two vectors a and b is represented by $a \cdot b$
- This is also called the dot product because of the symbol used
The scalar product between two vectors $a = a_{1} i + a_{2} j + a_{3} k$ and $b = b_{1} i + b_{2} j + b_{3} k$ is defined as $a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$
The result of taking the scalar product of two vectors is a real number
- i.e. a scalar
For example,

$(3 i - k) \cdot (2 i + 9 j + k) = 3 \times 2 + 0 \times 9 + (- 1) \times 1 = 6 + 0 - 1 = 5$

and

$(\begin{matrix} 2 \\ 7 \end{matrix}) \cdot (\begin{matrix} - 8 \\ 2 \end{matrix}) = 2 \times (- 8) + 7 \times 2 = - 16 + 14 = - 2$

The scalar product has some important properties:
- The order of the vectors doesn’t affect the result:

$a \cdot b = b \cdot a$

In effect we can ‘multiply out’ brackets:

$a \cdot (b + c) = a \cdot b + a \cdot c$

This means that we can do many of the same things with vectors as we can do when operating on real numbers – for example,

$(a - b) \cdot (a - b) = a \cdot a - 2 a \cdot b + b \cdot b$

The scalar product between a vector and itself is equal to the square of its magnitude:

$a \cdot a = {| a |}^{2}$

For example,

$(\begin{matrix} 2 \\ 7 \end{matrix}) \cdot (\begin{matrix} 2 \\ 7 \end{matrix}) = 2^{2} + 7^{2} = 53$ and ${| (\begin{matrix} 2 \\ 7 \end{matrix}) |}^{2} = 2^{2} + 7^{2} = 53$

What is the connection between the scalar product and trigonometry?

There is another important method for finding $a \cdot b$ involving the angle between the two vectors $θ$ :

$a \cdot b = | a | | b | \cos θ$

Here $θ$ is the angle between the vectors when they are placed ‘base to base’
- when the vectors are placed so that they begin at the same point
- This formula can be derived using the cosine rule and expanding $(a - b) \cdot (a - b)$
The scalar product of two vectors gives information about the angle between the two vectors
- If the scalar product is positive then the angle between the two vectors is acute (less than 90°)
- If the scalar product is negative then the angle between the two vectors is obtuse (between 90° and 180°)
- If the scalar product is zero then the angle between the two vectors is 90° (the two vectors are perpendicular)

How do I tell if vectors or lines are perpendicular?

Two (non-zero) vectors $a$ and $b$ are perpendicular if, and only if, $a \cdot b = 0$
- If the a and b are perpendicular then:
  - $θ = 90 ° \Rightarrow \cos θ = 0 \Rightarrow | a | | b | \cos θ = 0 \Rightarrow a \cdot b = 0$
- If $a \cdot b = 0$ then:
  - $| a | | b | \cos θ = 0 \Rightarrow \cos θ = 0 \Rightarrow θ = 90 ° \Rightarrow$ a and b are perpendicular
- For example, the vectors $2 i - 3 j + 5 k$ and $- 4 i - j + k$ are perpendicular since

$(2 i - 3 j + 5 k) \cdot (- 4 i - j + k) = 2 \times (- 4) + (- 3) \times (- 1) + 5 \times 1 = - 8 + 3 + 5 = 0$

Examiner Tips and Tricks

When writing a scalar product, it’s important to write a distinctive dot between the vectors – otherwise your meaning will not be clear.

Worked Example

Find the value of t such that the two vectors $v = (\binom{2}{\begin{matrix} t \\ 5 \end{matrix}})$ and $w = (t - 1) i - j + k$ are perpendicular to each other.