The document covers fundamental concepts in vector mathematics, including the definition and manipulation of vectors, operations like addition and subtraction, unit vectors, and both scalar and vector products. It provides exercises to reinforce understanding of vector magnitude, direction, and properties such as the dot and cross products. Additionally, it discusses the geometric interpretation of vectors and offers practical examples and configurations for calculating results.

1. Vectors

2. Vocabulary
Magnitude
Direction
Location
Scalar
Vector
Perpendicular
Unit
Graphical
Geometric
Component
Size, amount
Place
Measurement with only magnitude
Measurement with magnitude and direction

3. Graphical form
Unit vectors (a) have a magnitude of 1
Vectors are equal if theyhave:
– Same magnitude
– Same direction
– Not necessarily the same position
Magnitude Head
Tail
AB
^
A
B
a
Vector: =a =a
a

4. Multiplying (by a scalar)
Product is a:
Vector
v

5. Adding Vectors
Triangle Law
v
u
u+v

6. Subtracting Vectors
u – v
= u + (-v)
v – u
v
u
u-v

7. Ex. 26.2
1. Given that vector of magnitude 2, state
a) A vector of magnitude 4 in the direction of v
b) A vector of magnitude 0.5 in the direction of v
c) A vector of magnitude 6 in the opposite direction of
v
d) A unit vector in the direction of v
e) A unit vector in the opposite direction of v

8. S.E. 26.3
2. Which of the following are true and which are false?
a) a + b is the same as a + b
b) b – a is the same as a – b
c) a + b + c is the same as c + a + b
Explain the term ‘resultant as applied to two vectors

9. Ex 26.3
• Using the diagram below, find:
a) AB+BC b) AC+CD+ DE c) AC – EC
d) DE + EA + AC e) AB-CB f) BC-AC
g) EC+CB+BA
B
D
EA
C

10. 1 2 3 4 5 6 7 8
9
8
7
6
5
4
3
2
1
P
AO
i
j
Component form
OP =
i =
j =
OA =
OP =
OP =
OA + AP
Unit vector (x)
Unit vector (y)
6i
8j
6i + 8j

11. S.E. 26.4
1. State a unit vector:
(a)In the x direction (b) in the ydirection
1. State a unit vector: (a) In the negative x direction (b) in the negative ydirection
2. What is the angle between
3. a) i and j b) i and i c) i and –i
4. State an expression for the magnitude of xi + yj
5. State a vector of magnitude 6 in the x direction
6. State a vector of magnitude 2 in the negative ydirection

12. Adding
a = a
1
i + a
2
j
b = b
1
i + b
2
j
a + b = (a
1
+ b
1
)i + (a
2
+ b
2
)j
eg:
a = 7i + 3j
b = i + 2j
a + b = (7+1)i + (3+2)j
= 8i + 5j
1 2 3 4 5 6 7 8
9
8
7
6
5
4
3
2
1
i
j

13. Multiplying (by a scalar)
v = v
1
i + v
2
j
av =a(v
1
i + v
2
j)
av =av
1
i + av
2
j
eg:
v = 2i + 3j
a = 3
av = 3(2i + 3j)
= 6i + 9j
1 2 3 4 5 6 7 8
9
8
7
6
5
4
3
2
1
i
j

14. Ex 26.4
1. Calculate the magnitude of:
a) i + 3j b) 3i + 4j c) –i d)1/2(i + j)
1. Calculate the vector from
a) (0,0) to (3,7)
b) 3,7) to (5,2)
c) (5,2) to (-1,-2)
1. Given a = 3i – 2j, b= - i – 3j, c = 2 i + 5j, find
a) 4a b) 3b + 2c c) 2a – 3b – 4c
d) IaI e) I4aI f) I3b + 2cI

15. Dot (scalar) product
• A scalar product of two vectors
• The product shows the magnitude of the influence that one vector has on another.
Eg
• Running a race
• Pushing a piano
• Mario kart zooms

16. Component form
a = a
1
i + a
2
j
b = b
1
i + b
2
j
a.b = a
1
b
1
+ a
2
b
2
For two equal vectors:
a.a = a
1
a
1
+ a
2
= a
1
2 + a
2
2
= IaI2
i
j
IaI2
a1
2
a2
2

17. Cosine Rule:
c2 = a2 +b2 – 2ab.cosC
c =
c.c = (a-b).(a-b)
= a.a + b.b – 2a.b
IcI2 = IaI2 + IbI2 – 2a.b
a.b = IaIIbIcosC
Geometric form
a – b

18. Comparing the component and
geometric forms
Show:
• i.i=1
• i.j=0
a.b
• i.i = I1II1IcosC = 1
• i.j = I1II1IcosC = 0
= (a
1
i + a
2
j).(b
1
i+ b
2
j)
=a
1
b
1
i.i + a
2
b
2
j.j + a
2
b
1
i.j + a
1
b
2
i.j
= a
1
b
1
+ a
2
b
2
…as shown before

19. Exercise 26.5
1. Find the dot product of the following pairs of vectors:
(a) 2i-j, 3i+2j (b) –i+3j, 4i-j c) i-j, -2i-j
2. Find the angles between the following pairs of vectors:
(a) 3i+4j, i+2j (b) i+j, 2i-3j c) i-j, i-j (d) j,-j
3. Show that the vectors a=2i-j and b=-i-2j are perpendicular
4. Points A, B and C have coordinates (-6,2), (3,8) and (-3,-2) respectively.
(a) Find AB and BC
(b) Byusing the dot product, calculate the angle between AB and BC

20. Cross (vector) product
• A vector product produced by two vectors
• It also measures the interaction between the 2 vectors
• The vector should be the same irrespective of the vector set’s orientation
• Therefore, we need a vector which remains constant irrespective of orientation
• Which direction should the vector point in order to satisfy this requirement?

21. Right hand rule
a x b
a. 1st term, 1st finger
b. 2nd term, 2nd finger
xproduct. Thumb
j
i
If k = i x j,
which
direction
is k?
k
-k

22. Exercise
Find:
• i x j =
• i x k =
• k x i =
Notice:
u x v = -v x u
u x u = -u x u
u = -u =
i x i = j x j = k x k = 0
-j x i
-k x j
-i x k
k
i
j
=
=
=
0

23. Component form
u = u
1
i + u
2
j + u
3
k
v = v
1
i + v
2
j + v
3
k
u x v =
= (u
1
i + u
2
j + u
3
k)(v
1
i + v
2
j + v
3
k)
= u
1
iv
1
i + u
1
i v
2
j + u
1
iv
3
k + u
2
jv
1
i + u
2
jv
2
j + u
2
j v
3
k + u
3
kv
1
i + u
3
kv
2
j + u
3
k v
3
k
= u
1
i v
2
j + u
1
iv
3
k + u
2
jv
1
i + u
2
j v
3
k + u
3
kv
1
i + u
3
kv
2
j
u x v = (u
2
v
3
– u
3
v
2
)i + (u
3
v
1
– u
1
v
3
)j + (u
1
v
2
– u
2
v
1
)k
ixj = -jxi = k
jxk = -kxj = i
kxi = -ixk =j

24. Magnitude of cross product
u x v = (u
2
v
3
– u
3
v
2
) + (u
3
v
1
– u
1
v
3
) + (u
1
v
2
– u
2
v
1
)
From Pythagoras’ Theorem:
IuxvI2 = (u
2
v
3
– u
3
v
2
)2 + (u
3
v
1
– u
1
v
3
)2 + (u
1
v
2
– u
2
v
1
)2
= u
2
2v
3
2 – 2u
2
u
3
v
2
v
3
+ u
3
2v
2
2
+ u
3
2v
1
2 – 2u
1
u
3
v
1
v
3
+ u
1
2v
3
2
+ u
1
2v
2
2 – 2u
1
u
2
v
1
v
2
+ u
2
2v
1
2
= u
1
2(v
2
2+v
3
2) + u
2
2(v
1
2+v
3
2) + u
3
2(v
1
2+v
2
2)
- 2(u
2
u
3
v
2
v
3
+ u
1
u
3
v
1
v
3
+ u
1
u
2
v
1
v
2
)
1

25. IuIIvIcosC = u.v = u
1
v
1
+ u
2
v
2
+ u
3
v
3
IuI2IvI2cos2C = (u.v)2 = (u
1
v
1
+ u
2
v
2
+ u
3
v
3
)2
= u
1
2v
1
2 u
1
u
2
v
1
v
2
u
1
u
3
v
1
v
3
u
2
u
3
v
2
v
3
u
2
2v
2
2 + u
1
u
2
v
1
v
2
+ u
1
u
3
v
1
v
3
+ u
2
u
3
v
2
v
3
u
3
2v
3
2
= u
1
2v
1
2 + u
2
2v
2
2 + u
3
2v
3
2
+ 2(u
1
u
2
v
1
v
2
+ u
1
u
3
v
1
v
3
+ u
2
u
3
v
2
v
3
)
2

26. +
= IuxvI + IuI2IvI2cos2C
= u
1
2(v
2
2+v
3
2) + u
2
2(v
1
2+v
3
2) + u
3
2(v
1
2+v
2
2)
- 2(u
2
u
3
v
2
v
3
+ u
1
u
3
v
1
v
3
+ u
1
u
2
v
1
v
2
)
u
1
2v
1
2 + u
2
2v
2
2 + u
3
2v
3
2
+ 2(u
1
u
2
v
1
v
2
+ u
1
u
3
v
1
v
3
+ u
2
u
3
v
2
v
3
)
= u
1
2(v
2
2+v
3
2) + u
2
2(v
1
2+v
3
2) + u
3
2(v
1
2+v
2
2)
+ u
1
2v
1
2 + u
2
2v
2
2 + u
3
2v
3
2
= u
1
2(v
1
2+v
2
2+v
3
2) + u
2
2(v
1
2+v
2
2+v
3
2) + u
3
2(v
1
2+v
2
2+v
3
2)
= (u 2 + u 2 + u 2 )(v 2+v 2+v 2)
1 2

27. Bypythagoras’ Theorem:
(u
1
2 + u
2
2 + u
3
2 )(v
1
2+v
2
2+v
3
2)
= IuI2IvI2
In summary:
+ = Iu x vI2 + IuI2IvI2cos2C = IuI2IvI2
Iu x vI2 = IuI2IvI2 - IuI2IvI2cos2C
= IuI2IvI2 (1- cos2C)
= IuI2IvI2 (sin2C)
Iuxvi2 = IuI2IvI2 sin2C
1 2

28. Magnitude as a parallelogram