1. “Teach A Level Maths” Vol. 2: A2 Core Modules
51: The Vector Equation of a Line
© Christine Crisp

2. The vector AB and the vector equation of the line through AB are very different things.
The line through A and B. x x
The vector AB
The vector AB has a definite length ( magnitude ).
The line AB is a line passing through the points A and B and has infinite length.

3. Finding the Equation of a Line
In coordinate geometry, the equation of a line is
e.g.
The equation gives the value (coordinate) of y for any point which lies on the line.
The vector equation of a line must give us the position vector of any point on the line.
We start with fixing a line in space.
We can do this by fixing 2 points, A and B. There is only one line passing through these points.

4. A and B are fixed points. a r1
We consider several more points on the line. x a r1
We need an equation for r, the position vector of any point R on the line. x
Starting with R1:

5. A and B are fixed points. a r2
We consider several more points on the line. x a
We need an equation for r, the position vector of any point R on the line. x x r2
Starting with R1:

6. A and B are fixed points. r3 a
We consider several more points on the line. r3 x a
We need an equation for r, the position vector of any point R on the line. x x
Starting with R1:

7. t is called a parameter and can have any real value.x
So for R1, R2 and R3 x x a x x
For any position of R, we have
t is called a parameter and can have any real value.
It is a scalar not a vector.

8. a b We can substitute for Instead of using a here . . .x x
We can substitute for x a b x x
Instead of using a here . . .
we could use b.
The value of t would then be different to get to any particular point.
We say the equation is not unique.

9. p a b We can substitute forx x
We can substitute for x a b x x
Also, instead of we can equally well use any vector
which is parallel to AB. p
If is not the same length as AB, t will have a different value for any particular R.

10. x x a r1 x
e.g. or

11. SUMMARY
The vector equation of the line through 2 fixed points A and B is given by
The vector equation of the line through 1 fixed point A and parallel to the vector is given by
position vector . . .
direction vector . . .
of a known point on the line
of the line

12. e.g. 1 Find the equation of the line passing through the points A and B with position vectors a and b where
and
Solution:
So,

13. Diagram

14. In this example we had
and
giving
We can replace r
with
and/or replace a
with b
e.g.

15. So,
is the same line as
However, the value of t for any particular point has now changed.
ANS:
Which point is given by t = 2 in the 1st version?
What value of t in the 2nd version gives the same point?
ANS: t = 1

16. e.g. 2 Find the equation of the line passing through the point , parallel to the vector
Solution:
So, or
Demo

17. e.g. 3 Show that C with position vector
lies on the line
Solution: If C lies on the line, there is a value of t that makes r = c.
So,

18. 7 2 -5
We have -6 -1 5 3 2 -1
The top row of the vectors gives the x-components, so,

19. 7 2 -5
We have -6 -1 5 3 2 -1
The top row of the vectors gives the x-components, so,
However, for the point to lie on the line, this value of t must also give the y- and z- components.

20. 7 2 -5
We have -6 -1 5 3 2 -1
The top row of the vectors gives the x-components, so,
However, for the point to lie on the line, this value of t must also give the y- and z- components. y:

21. 7 2 -5
We have -6 -1 5 3 2 -1
The top row of the vectors gives the x-components, so,
However, for the point to lie on the line, this value of t must also give the y- and z- components. y:

22. 7 2 -5
We have -6 -1 5 3 2 -1
The top row of the vectors gives the x-components, so,
However, for the point to lie on the line, this value of t must also give the y- and z- components. y: z:
Notice how we write this. We mustn’t start with
as we are trying to show that this is true.

23. 7 2 -5
We have -6 -1 5 3 2 -1
The top row of the vectors gives the x-components, so,
However, for the point to lie on the line, this value of t must also give the y- and z- components. y: z:

24. 7 2 -5
We have -6 -1 5 3 2 -1
The top row of the vectors gives the x-components, so,
However, for the point to lie on the line, this value of t must also give the y- and z- components. y: z:

25. 7 2 -5
We have -6 -1 5 3 2 -1
The top row of the vectors gives the x-components, so,
However, for the point to lie on the line, this value of t must also give the y- and z- components. y: z:
Since all 3 equations are satisfied, C lies on the line.

26. Exercise
1. Find a vector equation for the line AB for each of the following:
(a)
(b)
(c)
and AB is parallel to the vector
2. Does the point lie on the line AB in 1(a)?

27. Solutions
(a)
(b)
(c)
parallel to

28. x: y: z: Solutions 2. Does the point lie on the line AB in 1(a)?
From 1(a), x: y: z:
The point does not lie on AB.

29. Stop here for C4
Carry on for Further Maths

30. The Cartesian Form of the Equation of a Line
The equation is the Cartesian equation of a line but only if it lies in the x-y plane.
The more general form can be easily found from the vector form.
The Cartesian form does not contain a parameter.

31. e.g.
We can extract the 3 components from this equation:
To eliminate the parameter, t, we rearrange to find t :
So,
This 3 part equation is the Cartesian equation.

32. (1) doesn’t contain t so just gives x = -3
e.g. Find the Cartesian equation of the line
Solution:
(1) doesn’t contain t so just gives x = -3
So the line is

33. Exercise
1. Write the following lines in Cartesian form:
(b)
(a)
(a)
Answers:
(b)

35. The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.
For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

36. direction vector
SUMMARY
The vector equation of the line through 2 fixed points A and B is given by
The vector equation of the line through 1 fixed point A and parallel to the vector is given by
position vector
of a known point on the line
of the line

37. e.g. 1 Find the equation of the line passing through the points A and B with position vectors a and b where
and
Solution:
So,

38. We can replace r
with
In this example we had
and
giving
and/or replace a
with b
e.g.

39. However, the value of t for any particular point has now changed.
So,
is the same line as
In the 1st version t = 2 gives
The same point is given by t =1 in the 2nd version.

40. e.g. 2 Find the equation of the line passing through the point , parallel to the vector
Solution:
So, or

41. Solution: If C lies on the line, there is a value of t that makes r = c.
e.g. 3 Show that C with position vector
lies on the line
So,

42. y: z: The top row of the vectors gives the x-components, so,
However, for the point to lie on the line, this value of t must also give the y- and z- components. z: y:
Since all 3 equations are satisfied, C lies on the line.
The top row of the vectors gives the x-components, so,

43. OCR/MEI only The Cartesian Equation of a Line e.g.
We can extract the 3 components from this equation:
To eliminate the parameter, t, we rearrange to find t :
So,
This 3 part equation is the Cartesian equation.
The Cartesian Equation of a Line
OCR/MEI only

44. are the elements of the direction vector.
We can generalise the result by comparing the 2 forms:
and
The denominators of the Cartesian form
The numerators of the Cartesian form are
where the position vector of the point on the line is

45. The Cartesian equation of a line is given by
SUMMARY
where the position vector of the point on the line is
and the direction vector is

46. It is easy to make a mistake when writing the Cartesian equation of a line.
For both these reasons you may prefer to rearrange the vector equation to find the parameter rather then quote the formula.
Also, it isn’t obvious what to do if an element of is zero.

47. (1) doesn’t contain t so just gives x = -3
e.g. Find the Cartesian equation of the line
Solution:
(1) doesn’t contain t so just gives x = -3
So the line is