A LevelFurther MathsEdexcelFurther Pure 1Revision NotesFurther VectorsFurther VectorsDirection Ratios & Direction Cosines

Direction Ratios & Direction Cosines (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Exam code: 9FM0

Written by: Mark Curtis

Updated on 20 January 2026

Direction ratios & direction cosines

What are direction ratios and direction cosines?

The straight line $r = a + λ b$ goes through the point with position vector $a$ in the direction of the vector $b$
- The direction vector is $b = b_{1} i + b_{2} j + b_{3} k$
- The direction ratio is $b_{1} : b_{2} : b_{3}$
- The direction cosines are $l$ , $m$ and $n$ where
  - $l = \frac{b_{1}}{| b |} = \cos α$
  - $m = \frac{b_{2}}{| b |} = \cos β$
  - $n = \frac{b_{3}}{| b |} = \cos γ$
- where
  - $α$ is the angle between $b$ and the positive $x$ -axis
  - $β$ is the angle between $b$ and the positive $y$ -axis
  - $γ$ is the angle between $b$ and the positive $z$ -axis

Three diagrams show vector b in 3D space with axes x, y, z. The vector b makes an angle of α with the x-axis (with x-component b_1), an angle of β with the y-axis (with y-component b_2) and angle of γ with the z-axis (with z-component b_3).

Examiner Tips and Tricks

You need to learn the formulae for the direction cosines $l$ , $m$ and $n$ as they are not given in the formula booklet.

Worked Example

Given the straight line $r = i + 2 j + 3 k + λ (- i + 5 j - 3 k)$ , find

(a) the direction ratio,

(b) the direction cosine for the angle made with the positive $y$ -axis.

Answer:

(a)

Identify the direction vector, $b$

$b = - i + 5 j - 3 k$

The direction ratio is $b_{1} : b_{2} : b_{3}$

$- 1 : 5 : - 3$

(b)

Identify the correct direction cosine, $m = \frac{b_{2}}{| b |} = \cos β$

$m = \frac{5}{\sqrt{{(- 1)}^{2} + 5^{2} + {(- 3)}^{2}}}$

Simplify

$\begin{array}{rcl} m & = & \frac{5}{\sqrt{35}} \\ = & \frac{5}{\sqrt{35}} \times \frac{\sqrt{35}}{\sqrt{35}} \\ = & \frac{5 \sqrt{35}}{35} \end{array}$

This is the direction cosine $\cos β$

$\frac{\sqrt{35}}{7}$

What is the direction cosine identity?

The direction cosines $l$ , $m$ and $n$ satisfy the identity
- $l^{2} + m^{2} + n^{2} \equiv 1$
This is because
- $\frac{{b_{1}}^{2}}{| b |^{2}} + \frac{{b_{2}}^{2}}{| b |^{2}} + \frac{{b_{3}}^{2}}{| b |^{2}} = \frac{| b |^{2}}{| b |^{2}} = 1$

Examiner Tips and Tricks

You need to learn the identity as it is not given in the formula booklet.

How do I write the Cartesian equation of a line in terms of direction cosines?

The Cartesian equation $\frac{x - a_{1}}{b_{1}} = \frac{y - a_{2}}{b_{2}} = \frac{z - a_{3}}{b_{3}}$ of the straight line $r = a + λ b$ can be written in terms of the direction cosines $l$ , $m$ and $n$
- by multiplying by $| b |$
  - $| b | \times \frac{x - a_{1}}{b_{1}} = | b | \times \frac{y - a_{2}}{b_{2}} = | b | \times \frac{z - a_{3}}{b_{3}}$
- then rearranging
  - $\frac{x - a_{1}}{\frac{b_{1}}{| b |}} = \frac{y - a_{2}}{\frac{b_{2}}{| b |}} = \frac{z - a_{3}}{\frac{b_{3}}{| b |}}$
- giving
  - $\frac{x - a_{1}}{l} = \frac{y - a_{2}}{m} = \frac{z - a_{3}}{n}$

Examiner Tips and Tricks

You need to learn the Cartesian equation in direction cosine form, as it is not given in the formula booklet.

Worked Example

Find the Cartesian equations of the two lines that

pass through the origin
make an angle of 45° with the $y$ -axis
make an angle of 60° with the $z$ -axis

Answer:

Identify which out of $α$ , $β$ and $γ$ are the angles given in the question

$β = 45 °$ and $γ = 60 °$

Find the direction cosines $l$ , $m$ and $n$

$\begin{array}{rcl} l & = & \cos α \\ m & = & \cos 45 = \frac{\sqrt{2}}{2} \\ n & = & \cos 60 = \frac{1}{2} \end{array}$

Find the two possible values of $l$ using the identity $l^{2} + m^{2} + n^{2} \equiv 1$

$\begin{array}{rcl} l^{2} + \frac{1}{2} + \frac{1}{4} & = & 1 \\ l^{2} & = & \frac{1}{4} \\ l & = & \pm \frac{1}{2} \end{array}$

Use that the Cartesian equation of the line through $a$ is $\frac{x - a_{1}}{l} = \frac{y - a_{2}}{m} = \frac{z - a_{3}}{n}$

$\begin{array}{rcl} \frac{x - 0}{\pm \frac{1}{2}} & = & \frac{y - 0}{\frac{\sqrt{2}}{2}} = \frac{z - 0}{\frac{1}{2}} \\ \pm 2 x & = & \frac{2 y}{\sqrt{2}} = 2 z \\ \pm 2 x & = & \sqrt{2} y = 2 z \end{array}$

Write out the two possible answers

$\begin{array}{rcl} 2 x & = & \sqrt{2} y = 2 z \end{array}$ and $\begin{array}{rcl} - 2 x & = & \sqrt{2} y = 2 z \end{array}$

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Direction Ratios & Direction Cosines (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Direction ratios & direction cosines

What are direction ratios and direction cosines?

What is the direction cosine identity?

How do I write the Cartesian equation of a line in terms of direction cosines?

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Mark Curtis