Straight lines

*Please note: you may not see animations, interactions or images that are potentially on this page because you have not allowed Flash to run on S-cool. To do this, click here.*

Straight lines

The cartesian equation for a straight line is y = mx + c, where m represents the gradient of the line, and c is the point where the line crosses the y-axis.

A vector equation for a line similarly needs 2 pieces of information:


  1. A point on the line.
  2. The direction of the line
  3. .


To illustrate - Leeds lies on the M1, to get to Leeds you firstly need to get on the M1 and then travel along it until you arrive.

So to get from the origin to a point on the line you firstly need to get to the line, and then you need to move along the line.

This means that if A is a point on the line with position vector [u]a[/u], and [u]b[/u] is a vector parallel to the line, them the vector equation of the line is:

[u]r[/u] = [u]a[/u] + λ[u]b[/b]

Where [u]r[/u] represents any point on the line, and λ is any number.

Vector Equation of a Straight Line



(These are the y-intercept and gradient.)

Therefore the vector equation of the line is:



Find the vector equation of the line that passes through (1, 4, 0) and (5, 2, 1).


Therefore, the vector equation of the line is:


It is possible to rewrite this in cartesian form (i.e. in terms of x, y, and z). To do this we use the fact that:

x = 1 + 4λ, y = 4 - 2λ, and z = λ

Rewriting each expression we get:


Note: Each has been written in the same style.

As each expression is the same we get the cartesian equation,


(The denominators give us the vector parallel to the line; the numerators give us minus the vector to the line.)

In 2-D two lines are guaranteed to cross unless they are parallel. At the point of intersection the coordinates for both lines are the same, and are usually found by solving simultaneous equations.

To make the simultaneous equations we match the x-values and then the y-values.


Find where the lines



Matching the x-values gives: 2 + λ = 3 - μ

Matching the y-values gives: 3 - 2λ = -4 + 3μ

Rearrange these to get:

λ + μ = 1

2λ + 3μ = 7

These solve to give: λ = -4, μ = 5

Substituting these values into our line equations we get the point of intersection as being:

x = -2, y = 11.

Therefore the lines meet at (-2, 11).

In 3-D two lines do not have to meet. If they do not meet then the lines are called 'skew'.

Finding where two lines meet is the same as for 2-D, except there will be three simultaneous equations (in two variables) to solve. (A solution that does not work for all three equations means the lines do not meet and are therefore skew.)

To find the angle between two lines find the angle between the two direction vectors. This means that it is also possible to find the angle between skew lines.


Find whether the lines


'intersect', and find the angle between them.

Matching the x-values gives: 2 - 2λ = 3 + μ

Matching the y-values gives: 4λ = -5 - 3μ

Matching the z-values gives: -1 - λ = -5 - μ

Solving the first two simultaneous equations gives: λ = 1, μ = -3.

These values do work in the third equation therefore the lines meet. Substituting λ = 1, μ = -3 into the equation of lines gives the point of intersection as being:

x = 0, y = 4, z = -2.

Therefore the lines meet at (0, 4, -2)

The angle between the lines is the angle between the direction vectors, so using the scalar product we get,


This gives us θ = 148.8o, but you must remember that when the two vectors intersect, two angles are formed (see diagram below):


Therefore the angle normally quoted is θ = 31.2o (to 1dp).

S-cool Exclusive Offers