Maths revision: Vector equation of a line and the intersection of two lines

It’s everybody’s favourite: AS and A level maths revision of vectors, YAY! Here we’re looking at the vector equation of a line, converting between vector form and cartesian form and the intersection of two lines.

This guide assumes you know about and have revised vectors already!

As always, let’s start with the basics:

The vector equation of a line consists of two vectors – a position vector, a vector from the origin to any point on the line – and a direction vector – which describes the gradient of the line. The vector equation of a line is given by:

Dynamic image 0

Where Dynamic image 1 is the position vector, Dynamic image 2 is the direction vector and Dynamic image 3 is a parameter which can take any value. Each point on the line has a different value of Dynamic image 3

Due to the nature of the vector equation of the line, the same line can have multiple equations. The position vector can be the position vector of any point on the line. The direction vector too can be written in many different ways. For example Dynamic image 5 means the line goes 4 units up for every 2 across. This is the same as Dynamic image 6 or even Dynamic image 7

General vector form of the equation of a line

In general, the vector form of the equation of the line joining two points, A and B, with position vectors a and b is given by:

Dynamic image 8

Converting to the cartesian form of a line

You may want to (actually, you probably won’t, but the examiner might want you to!) convert from the vector equation of a line to the cartesian form – involving x and y values. To do this firstly write r as Dynamic image 9, for example:

Dynamic image 10

This enables you to see equations for x and for y:

Dynamic image 11
Dynamic image 12

From here, rearrange both to make Dynamic image 3 the subject:

Dynamic image 14
Dynamic image 15

You can now eliminate Dynamic image 3 by equating the two expressions:

Dynamic image 17

This is the correct answer, however you may wish to tidy it up to make it look nicer! If you’re asked to simplify then make sure you do for those extra marks.

General form (2 and 3 dimensions)

In general the vector form of an equation:

Dynamic image 18

Can be written in cartesian form as:

Dynamic image 19

For 3 dimensions:

Dynamic image 20

Can be written in cartesian form as:

Dynamic image 21

Converting from cartesian form to vector form

But what if the examiner is an awkward sod and wants you to convert the other way around? To convert from cartesian form to vector form you first want to find a position vector, to do this find any point on the line, the easiest way is to substitute x = 0 into the equation and find a value for y. You then need to calculate then convert the gradient of the line into vector form to give you your direction vector.


Example question: Write y = 2x + 3 in vector form

Worked solution: Firstly find a point on the line, in this case substituting x = 0 gives y = 3, this point (0, 3) has the position vector Dynamic image 22 and gives the first part of the vector form of the equation. The gradient in this example is 2 as the equation is in the form y = mx + c. This needs to be converted into vector form, a gradient of 2 means the line goes up 2 units for every 1 across, thus the direction vector is Dynamic image 6. Hence the final equation is:

Dynamic image 24

Intersection of two lines

Finding the points of intersection of two lines in vector form is similar to that involving lines in cartesian form. For example, take these two lines in vector form

Dynamic image 25

Dynamic image 26

Dynamic image 27 is just a parameter like Dynamic image 3.

Start by equating the two equations together:

Dynamic image 29

You can now create two equations involving Dynamic image 3 and Dynamic image 27

Dynamic image 32

Dynamic image 33

You now have to two simultaneous equations (lucky you!) which can be solved to give value(s) of Dynamic image 27 and Dynamic image 3 which satisfy both equations.

This is example is pretty easy as the second equation gives us the value of Dynamic image 27 right away. Substituting this into the top equation gives Dynamic image 37. You’re not finished yet though, to find the position vector of the this point you need to substitute either Dynamic image 38 or Dynamic image 39 into the original equations.

Dynamic image 40


Dynamic image 41

Hence the position vector of the point of intersection is Dynamic image 42

If you have the time, substitute it into both equations to make sure you get the same answer: If you don’t, you’ve gone wrong!