1. Vectors

2. After completing this chapter you should be able to:
Know the difference between a scalar and a vector quantity
Draw a vector diagram
Perform simple vector arithmetic and know the definition of a unit vector
Use position vectors to describe points in two or three dimensions
Use Cartesian coordinates in three dimensions, including finding the distance between two points
Find a scalar product and know how it can be used to find the angle between two vectors
Write down the equation of a line in vector form
Determine whether or not two given straight lines intersect (if they do you should be able to find the point of intersection)
Find the angle between two intersecting lines

3. 5.1 You need to know the difference between a scalar and a vector, and how to write down vectors and draw vector diagrams.
Scalar quantities can be described by using a single number – the magnitude
A vector quantity has both magnitude and direction
Scalar
The distance from P to Q is 100m
Vector
From P to Q you go 100m north
A ship sails at 12 km h -1
A ship is sailing at 12 km h-1, on a bearing of 067°

4. Vectors that at equal have both the same magnitude and same directionP
𝑃𝑄 = 𝑅𝑆

5. Two vectors are added using the ‘triangle law’Q
𝑃𝑄 + 𝑄𝑅 = 𝑃𝑅 R
𝑃𝑄 + 𝑄𝑃 =0 P b a
a + b

7. the modulus of a vector is another name for it’s magnitude
The modulus of vector a is written as 𝒂
The modulus of the vector 𝑃𝑄 is written as 𝑃𝑄
Example
the vector a is due east and 𝒂 =7. The vector b is due north and 𝒃 =24. Find 𝒂+𝒃 .
𝒂+𝒃 𝟐 = =625
𝒂+𝒃 =25
a + b b a

8. Exercise 5A page 55

9. in general the unit vector is 𝒂 𝒂
5.2 You need to be able to perform simple vector arithmetic, and to know the definition of a unit vector
a unit vector is a vector which has magnitude (or modulus) 1 unit.
if a vector a has magnitude 25 units,
then the unit vector parallel to a is 𝒂 𝟐𝟓 or 𝒂
in general the unit vector is 𝒂 𝒂

10. tips for vector arithmetic
any vector parallel to the vector a may be written as λa, where λ is a non zero scalar a 4a
-2a
Subtracting a vector is equivalent to ‘adding a negative vector’, so a – b is defined to be a + (-b)
If λa + μb = αa + βb, and the non zero vectors a and b are not parallel,
then λ = α and μ = β
You can prove this as follows
λa + μb = αa + βb can be written as (λ – α)a = (β – μ)b
but the two vectors can’t be equal unless they are parallel or zero!
so (λ – α) = 0, and (β – μ) = 0, so λ = α and β = μ

11. Equating coefficients: 5λ = μ and (1 – λ) = (μ – 3) so 1 – λ = 5λ – 3
Example
Given that 5λa + (1 – λ)b = μa + ( μ – 3 )b, where a and b are non-parallel, non-zero vectors, find the value of λ and the value of μ.
Equating coefficients:
5λ = μ and (1 – λ) = (μ – 3)
so 1 – λ = 5λ – 3
6λ = 4
λ = 2 3
μ = 10 3

12. Exercise 5B page 58

13. 5.3 You need to be able to use vectors to describe the position of a point in two or three dimensions. a O A
𝑂𝐴 =𝒂

14. EXAMPLE
For the quadrilateral OABC, the position vectors of A, B, C referred to O are a, b and c respectively.
The point M is the mid-point of AB and the point N divides BC in the ratio 2:1.
Find an expression for 𝑀𝑁 I n terms of a, b and c. M B A b
𝑀𝑁 = 𝑀𝐵 + 𝐵𝑁 a N O c C
𝑀𝐵 = 1 2 𝐴𝐵= ( 𝒃 −𝒂)
𝑀𝑁 = 1 2 𝒃 − 1 2 𝒂 𝒄 − 2 3 𝒃
𝐵𝑁 = 𝐵𝐶 = 2 3 (𝒄 −𝒃)
𝑀𝑁 =− 1 2 𝒂 − 1 6 𝒃 𝒄

15. Exercise 5C page 60

16. 5.4 You need to know how to write down and use the Cartesian components of a vector in two dimensions
the vectors I and j are unit vectors parallel to the x-axis and the y-axis, and in the direction of x increasing y increasing, respectively
Example
The points A and B have coordinates (3, 4) and (11,2) respectively. Find in terms of i and j:
the position vector of A
the position vector of B
the vector 𝐴𝐵
A (3, 4)
b - a a
B (11, 2)
a = 𝑂𝐴 = 3i + 4j
b = 𝑂𝐵 = 11i + 2j
𝐴𝐵 =𝒃 −𝒂
= (11i + 2j) – (3i + 4j)
= 8i – 2j b O

17. You can write a vector with Cartesian components as a column matrix
xi+ yj = 𝑥 𝑦
The modulus (or magnitude) of xi+ yj is 𝑥 2 + 𝑦 2

18. 5.5 You need to know how to use Cartesian coordinates in three dimensions
the distance from the origin to a point (x, y, z) is 𝑥 2 + 𝑦 2 + 𝑧 2
the distance between two points (x1, y1, z1) and (x2, y2, z2) is
𝑥 1 − 𝑥 𝑦 1 − 𝑦 𝑧 1 − 𝑧 2 2
Exercise 5E page 66

19. 5.6 You can extend the two-dimensional vector results to three dimensions, using k as the unit vector parallel to the z-axis, in the direction of z increasing.
three dimensions – 3 unit vectors i, j, k (x. y, z respectively)
the vector xi + yj + zk may be written as a column vector
the modulus (or magnitude) of xi + yj + zk is 𝑥 2 + 𝑦 2 + 𝑧 2

20. Example
The points A and B have position vectors (5i – 2j + 3k) and (8i + 8j +tk) respectively, and 𝐴𝐵 =5 5 .
Find the possible values of t.
𝐴𝐵 = 𝑡 − 5 −2 3 = 𝑡 −3
𝐴𝐵 = 𝑡 −3 2
= 𝑡 2 −6𝑡+9
= 𝑡 2 −6𝑡+118
so 𝑡 2 −6𝑡 =5 5
t² - 6t = 125
t² - 6t – 7 = 0
(t – 7)(t – 1) = t = 7 or t = -1

21. 5.7 You need to know the definition of the scalar product of two vectors (in either two or three dimensions), and how it can be used to find the angle between two vectors. a X Ө b
To find the angle between two vectors , use the angle when both vector a and b are directed away from X a X b

22. where Ө is the angle between a and b
The scalar product of two vectors a and b is written as a.b ( we say ‘ a dot b’), and defined by
a.b = 𝒂 𝒃 𝒄𝒐𝒔 𝜽
where Ө is the angle between a and b A a Ө B O b
If a and b are the position vectors of the points A and B then cos AOB = 𝒂.𝒃 𝒂 𝒃
If two vectors a and b are perpendicular, a.b = 0 because cos 90° = 0
If two vectors a and b are parallel, a.b = 𝒂 𝒃
a.a = 𝑎 2

23. If a = a1i + a2j + a3k and b = b1i + b2j + b3k
then a.b = 𝑎 1 𝑎 2 𝑎 𝑏 1 𝑏 2 𝑏 3 = 𝑎 1 𝑏 1 + 𝑎 2 𝑏 2 + 𝑎 3 𝑏 3
Example
Given that a = 7i – 4j + 2k and b = 2i + 8j + tk
find the value of t for which a and b are perpendicular
find, to the nearest degree, the angle between a and b when t = 11
𝒂.𝒃= 7 − 𝑡 =14 −32+2𝑡=2𝑡 −18
If a and b are perpendicular , 2t – 18 = 0
t = 9

24. cosӨ = 𝒂.𝒃 𝒂 𝒃 = 4 69 189 when t = 11 a.b = 2t – 18 = 4
𝒂 = − = 69
𝒃 = = 189
cosӨ = 𝒂.𝒃 𝒂 𝒃 =
angle between a and b is 88° (nearest degree)
Exercise 5G page 74

25. 5.8 You need to know how to write the equation of a straight line in vector form.
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.
The direction of the line.
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.

27. Where t is a scalar parameter
Now suppose a straight line passes through two give points C and D, with position vectors c and d respectively. Only one such line is possible D
You can use CD as a direction vector for the line
𝐶𝐷 =𝒅 −𝒄 C d c
Then you can use one of the two given points and the direction vector to form an equation for the straight line O
A vector equation of a straight line passing through the points C and D, with position vectors c and d respectively, is
r = c + t(d-c)
Where t is a scalar parameter

29. Example
The points A and B have position vectors (5i + 8j – 4k) and (8i + 2j + 5k) respectively.
Find the vector equation for the line l which passes through A and B.
Given that the point with coordinates (p, 4p, q) lies on l, find the value of p and the value of q.
𝒂= −4 𝒃=
Find the direction vector for the line
b – a = − −4 = 3 −6 9
equation of l: r = −4 +𝑡 3 −6 9
or r = −4 +𝑡 1 −2 3
common factor of 3 removed

30. or 𝑝 4𝑝 𝑞 = 5 8 −4 +𝑡 1 −2 3 𝑝 4𝑝 𝑞 = 5 8 −4 +𝑡 3 −6 9 So p = 5 + t ①
𝑝 4𝑝 𝑞 = −4 +𝑡 3 −6 9
or 𝑝 4𝑝 𝑞 = −4 +𝑡 1 −2 3
So p = 5 + t ①
4p = 8 – 6t ②
q = t ③
sub p from ① into ②
4(5 + t) = 8 – 6t
10t = -12
t = -1.2
sub in ① gives p = 3.8
sub in ③ gives q = -14.8
so p = 5 + t ①
4p = 8 -2t ②
q = t ③
sub ① into ②
4(5 + t) = 8 - 2t
6t = -12
t = -2
sub in ① gives p = 3
sub in ③ gives q = -10

31. 5.9 You need to be able to determine whether two given straight lines intersect.
we will be looking at two lines so we use a different parameter for each line.
s and t are commonly used, as are λ and μ
It is easier to show this with an example

32. The lines l1 and l2 have vector equations
r = 𝑡 3 − and r = 5 −1 9 +𝑠 2 − respectively.
show that l1 and l2 intersect
find the coordinates of their point of intersection
At an intersection point l1 = l2
5+3𝑡 −4𝑡 4+2𝑡 = 𝑠 −1 −3𝑠 9+3𝑠
Equate the x and y components then solve these two simultaneous equations
5 + 3t = 5 + 2s ① ⇒ t = 2𝑠 3
-4t = -1 – 3s ②
sub ① into ②

33. Equate the x and y components then solve these two simultaneous equations
sub ① into ②
-8s = -3 – 9s
s = -3
⇒ t = -2
if the lines intersect these values will be true for the z components
4 + 2t = 9 + 3s
substituting in the values for s and t gives 0 = 0
so lines l1 and l2 do intersect

34. 𝒓= 5+3𝑡 −4𝑡 4+2𝑡 = −1 8 0 so coordinates are (-1, 8, 0)
and the coordinates of their point of intersection are:
𝒓= 5+3𝑡 −4𝑡 4+2𝑡 = −1 8 0
so coordinates are (-1, 8, 0)
Exercise 5.l page 80

35. 5.10 You need to be able to calculate the angle between two straight lines.
The acute angle Ө between two straight lines is given by
𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 𝒃
where a and b are direction vectors of the lines

36. 𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 𝒃 = 𝟐𝟒 𝟐𝟗 𝟐𝟐 angle between l1 and l2 is 18.2° (1 d.p.)
The lines l1 and l2 have vector equations
r = 𝑡 3 − and r = 5 −1 9 +𝑠 2 − respectively.
find the acute angle between l1 and l2, giving your answer in degrees to 1 decimal place
using the direction vectors
𝒂= 3 −4 2
𝒃= 2 −3 3
𝒂.𝒃= 3 − −3 3 =6+12+6=24
𝒂 = − = 29
𝒃 = − = 22
𝑐𝑜𝑠𝜃= 𝒂.𝒃 𝒂 𝒃 = 𝟐𝟒 𝟐𝟗 𝟐𝟐
angle between l1 and l2 is 18.2° (1 d.p.)