1. 3D Coordinates
In the real world points in space can be located using a 3D coordinate system.
For example, air traffic controllers find the location a plane by its height and grid reference. z
(x, y, z) y O x 1

2. Write down the coordinates for the verticesx y z A C D E F G H B 1 2 6
(0, 1, 2)
(6, 1, 2)
(6, 0, 2)
(0, 0, 2)
(0, 1, 0 )
(6, 1, 0)
(0,0, 0)
(6, 0, 0) 2

3. 3D vectors are defined by 3 components.
For example, the velocity of an aircraft taking off can be illustrated by the vector v. z
(7, 3, 2) 2 v y 2 O 3 x 7 3

4. Any vector can be represented in terms of the i , j and k
Where i, j and k are unit vectors (one unit long)
in the x, y and z directions. z y O k j x i 4

5. Any vector can be represented in terms of the i , j and k
Where i, j and k are unit vectors
in the x, y and z directions. z
(7, 3, 2) v y 2
v = ( 7i+ 3j + 2k ) O 3 x 7 5

6. Magnitude of a Vector
A vector’s magnitude (length) is represented by |v|
A 3D vector’s magnitude is calculated using Pythagoras Theorem in 3D
(3 , 2 , 1) z v y 1 O 2 x 3

7. Find the unit vector in the direction of u

8. Addition of vectors

9. Addition of Vectors
For vectors u and v

10. Negative vector
For any vector u

11. Subtraction of vectors

12. Subtraction of Vectors
For vectors u and v

13. Multiplication by a scalar ( a number)
Hence if u = kv then u is parallel to v
Conversely if u is parallel to v then u = kv

14. CombiningVectors

15. If z = kw then z is parallel to w
Show that the two vectors are parallel.
If z = kw then z is parallel to w

16. The position vector of the point A(3 , 2 , 1) is OA written as a
Position Vectors
A (3,2,1) z a y 1 O 2 x 3
The position vector of the point A(3 , 2 , 1) is OA written as a

17. Position Vectors If R is ( 2 , -5 , 1) and S is (4 , 1 , -3) 4 = 1 -3
= 1 -3 2
– -5 1 2
= 6 -4
Then RS is s – r

18. y component of RS is -5 to 1 = 6 z component of RS is 1 to -3 = -4
Position Vectors as a journey
If R is ( 2 , -5 , 1) and S is (4 , 1 , -3)
x component of RS is 2 to 4 = 2
y component of RS is -5 to 1 = 6
z component of RS is 1 to -3 = -4 2
RS = 6 -4

19. Express VT in terms of f, g and h
Vectors as a journey
Express VT in terms of f, g and h
– h
VT =
VR +
RS + ST
VT =
– h
– f
+ g
– f
+ g

20. Collinear Points BC = 2AB
A is (0 , -3 , 5), B is (7 , -6 , 9) and C is (21 , -12 , 17). Show that A , B and C are collinear stating the ratio AB:BC.
BC = 2AB
AB and BC are parallel (multiples of each other) through the common point B, and so must be collinear A 1 B 2 C
AB : BC = 1 : 2

21. Collinear Points
Given that the points S(–4, 5, 1), T(–16, –4, 16) and U(–24, –10, 26) are collinear, calculate the ratio in which T divides SU. S 3 T 2 U
ST : TU = 3 : 2

22. The journey from Q to R is the same as the journey from P to S
Using Vectors as journeys
PQRS is a parallelogram with P(3 , 4 , 0), Q(7 , 6 , -3) and R(8 , 5 , 2). Find the coordinates of S.
P(3 , 4 , 0)
Q(7 , 6 , -3)
Draw a sketch
The journey from Q to R is the same as the journey from P to S S
R(8 , 5 , 2)
From P to S
S(3 + 1 , 4 – 1 , 0 + 5)
S(4 , 3 , 5)

23. Find the co-ordinates of Q.
The point Q divides the line joining P(–1, –1, 0) to R(5, 2 –3) in the ratio 2:1.
Find the co-ordinates of Q.
P(-1 , -1 , 0)
R(5 , 2 , -3)
PQ = 2/3 PR Q 1 2 6
PQ = 2/ -3 4
= 2 -2
The journey from P to Q
Q(-1 + 4, , 0 – 2)
Q(3 , 1 , – 2)

24. P divides the line joining S(1,0,2) and T(5,4,10) in the ratio 1:3
P divides the line joining S(1,0,2) and T(5,4,10) in the ratio 1:3. Find the coordinates of P.
SP = 1/4 ST
S(1 , 0 , 2)
T(5 , 4 , 10) 4
SP = 1/ 8 1
= 1 2
The journey from S to P P 3 1
P(1 + 1, 0 + 1, 2 + 2)
P(2 , 1 , 4)

25. a and b must be divergent, ie joined tail to tail
The scalar product
The scalar product is defined as being:
a . b = |a| |b| cos θ
0 ≤ θ ≤ 180 a θ
a and b must be divergent, ie joined tail to tail b

26. Find the scalar product for a and b when
|a|= 4 , |b|= 5 when (a) θ = 45o (b) θ = 90o
a . b = |a| |b| cos θ
 a . b = 4 × 5 cos 45o
 a . b = 20 × 1/√2
× √2/√2
 a . b = 10√2
a . b = |a| |b| cos θ
 a . b = 4 × 5 cos 90o
 a . b = 20 × 0
 a . b = 0
When θ = 90o

27. This equilateral triangle has sides of 3 units. p . q
The Scalar Product
This equilateral triangle has sides of 3 units. p . q
p . q = |p| |q| cos θ
 p . q = 3 × 3 cos 60o
 p . q = 9 × 1/2
 p . q = 41/2 27

28. This equilateral triangle has sides of 3 units. p . (q + r)
The Scalar Product
This equilateral triangle has sides of 3 units. p . (q + r) r r
p . (q + r)= p . q + p . r
60o
 p . q = 3 × 3 cos 60o
p and r are not divergent so move r
 p . q = 3 × 3 × ½
 p . q = 41/2
 p . r = 3 × 3 cos 60o
 p . r = 9  ½
p . (q + r) = p . q + p . r = 9
 p . r = 4½ 28

29. If a and b are perpendicular then
a . b = 0

30. a . b = a1b1 + a2b2 + a3b3 Component Form Scalar Product a1 b1
If a = a and b = b2
a b3
a . b = a1b1 + a2b2 + a3b3

31. Angle between Vectors
To find the angle between two vectors we simply use the scalar product formulae rearranged
a . b = |a| |b| cos θ
a . b = a1b1 + a2b2 + a3b3
a . b
|a| |b|
a1b1 + a2b2 + a3b3
|a| |b|
cos θ =
cos θ =

32. Find the angle between the two vectors below.
p = 3i + 2j + 5k and q = 4i + j + 3k 3
p = 2 5 4
q = 1 3
|p| = √( )
|q| = √( )
|q| = √26
|p| = √38
a1b1 + a2b2 + a3b3
|a| |b|
a . b = a1b1 + a2b2 + a3b3
cos θ =
= 3×4 + 2×1 + 5×3
29 √38 × √26 =
= 0∙923
= 29
θ = cos-1 0∙923 = 22∙7o

33. Perpendicular Vectors
a . b = | a | | b | cosƟ
If a and b are perpendicular then a . b = 0
cos 90o = 0
If a and b are perpendicular then a1b1 + a2b2 + a3b3 = 0

34. If a and b are perpendicular then a . b = 0
Show that
a and b are perpendicular
if a = and b =
a . b = a1b1 + a2b2 + a3b3
 a . b = 3×1 + 2×2 + (-1)×7
 a . b = – 7
 a . b = 0
 a and b are perpendicular 34

35. Two properties that you need to be aware of
Properties of a Scalar Product
Two properties that you need to be aware of
a . b = b . a
a .( b + c)= a . b + a . c

36. If | p | = 5 and | q | = 4, find p . (p + q)
60o q
p . (p + q) = p . p + p . q
= | p | × | p |cos 0o + | p | × | q |cos60o
= 5 × 5 × × 4 × ½ =
= 35 36

37. Find the acute angle between the diagonals of PQRS
T divides PR in the ratio 5:4
P(-2,-1,-4)
Show that Q, T and S are collinear, and find the ratio in which T divides QS
S(7,2,17)
Q(1,5,-7)
Find the acute angle between the diagonals of PQRS
R(7,8,5)
TS = 2QT, vectors are parallel through the common point T, so Q , T , S are collinear
T(-2+5,-1+5,-4+5)
T(3 , 4 , 1)
QT : TS = 1 : 2 37

38. Find the acute angle between the diagonals of PQRS38

39. Hence vectors are perpendicular
Vectors u and v are defined by u = 3i + 2j and v = 2i – 3j + 4k
Determine whether or not u and v are perpendicular.
3 2
u . v =
 u . v = 3×2 + 2×(-3) + 0×4
 u . v = 6 –
 u . v = 0
Hence vectors are perpendicular

40. u . v = 0 if vectors perpendicular
For what value of t are the vectors u and v perpendicular ?
u . v = 0 if vectors perpendicular
t 2
u . v = t
 u . v = t ×2 + (-2)×10 + 3×t
 u . v = 5t – 20
u . v = 0  5t – 20 = 0
t = 4

41. VABCD is a pyramid with rectangular base ABCD.
The vectors AB, AD and AV are given by
AB = 8i + 2j + 2k, AD = -2i + 10j – 2k and
AV = i + 7j + 7k Express CV in component form.
Look for an alternative route from C to V along vectors you know Note BC = AD and AB = DC
CV = CB + BA + AV 8
AB = 2 2 -2
AD = 10 1
AV = 7 7
CV = -AD – AB + AV 2
-10 = 8 - + 1 7
CV = –5i – 5j + 7k -5 7 =

42. The diagram shows two vectors a and b, with | a | = 3 and | b | = 22.
These vectors are inclined at an angle of 45° to each other.
a) Evaluate
i) a.a ii) b.b iii) a.b
b) Another vector p is defined by p = 2a + 3b
Evaluate p.p and hence write down | p |. i)
ii)
iii) b) p2

43. Vectors p, q and r are defined by p = i + j – k, q = i + 4k, r = 4i – 3j
a) Express p – q + 2r in component form
b) Calculate p.r
c) Find |r| a) b) c)

44. hence there are 2 points on the x axis that are 7 units from P
The diagram shows a point P with co-ordinates
(4, 2, 6) and two points S and T which lie on the x-axis.
If P is 7 units from S and 7 units from T,
find the co-ordinates of S and T.
hence there are 2 points on the x axis that are 7 units from P
i.e. S and T
and

45. The position vectors of the points P and Q are
p = –i + 3j + 4k and q = 7i – j + 5k respectively.
a) Express PQ in component form.
b) Find the length of PQ.

46. 120o so, a is perpendicular to b + cQ
60° a b c
PQR is an equilateral triangle of side 2 units.
Evaluate a.(b + c) and hence identify
two vectors which are perpendicular.
120o
NB for a.c vectors must diverge ( so angle is 120° )
so, a is perpendicular to b + c
Hence

47. Calculate the length of the vector 2i – 3j + 3k

48. Find the value of k for which the vectors and are perpendicular
Put Scalar product = 0

49. A is the point (2, –1, 4), B is (7, 1, 3) and C is (–6, 4, 2).
If ABCD is a parallelogram, find the co-ordinates of D.
The journey B to A is the same as the journey from C to D
That journey from C gives:
D(–6 + [-5], 4 + [-2], 2 + 1)
D(–11, 2 , 3)

50. b + c is perpendicular to a
The vectors a, b and c are defined as follows:
a = 2i – k, b = i + 2j + k, c = –j + k
a) Evaluate a.b + a.c
b) From your answer to (a), make a deduction about the vector b + c a) b)
b + c is perpendicular to a

51. Triangular faces are all equilateral
In the square based pyramid,
all the eight edges are of length 3 units.
Evaluate p.(q + r)
Triangular faces are all equilateral

52. (-1,-3,2) (2,-1,1)  D is (-1 + 9, -3 + 6, 2 – 3)  D is (8, 3, –1)
A and B are the points (-1, -3, 2)
and (2, -1, 1) respectively.
B and C are the points of trisection of AD.
That is, AB = BC = CD.
Find the coordinates of D
(-1,-3,2)
(2,-1,1)
 D is (-1 + 9, , 2 – 3)
 D is (8, 3, –1)

53. Find the co-ordinates of Q.
The point Q divides the line joining P(–1, –1, 0) to R(5, 2 –3) in the ratio 2:1.
Find the co-ordinates of Q. P Q R 2 1
(5,2,-3)
(-1,-1,0)
 Q is (-1 + 4, , 0 – 2)
 Q is (3, 1, –2)

54. VABCD is a pyramid with rectangular base ABCD.
VA = – 7i – 13j – 11k, AB = 6i + 6j – 6k, AD = 8i – 4j – 4k ; K divides BC in the ratio 1:3.
Find VK in component form.
+ ¼BC
= VA
+ AB VK
+ ¼AD
= VA
+ AB
= – 7i – 13j – 11k
6i + 6j – 6k
2i – j – k
= i – 8j – 18k

55. AB and BC are scalar multiples, so are parallel.
A is the point (2, –5, 6), B is (6, –3, 4) and C is (12, 0, 1).
Show that A, B and C are collinear
and determine the ratio in which B divides AC
AB and BC are scalar multiples, so are parallel.
B is common.
A, B, C are collinear A B C 2 3
B divides AB in ratio 2 : 3