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Vectors
VocabularyMagnitude
Direction
Location
Scalar
Vector
Perpendicular
Unit
Graphical
Geometric
Component
Size, amount
Place
Measurement with only magnitude
Measurement with magnitude and direction
Graphical form
Unit vectors (a) have a magnitude of 1
Vectors are equal if they have:
– Same magnitude– Same direction– Not necessarily the same position
Magnitude Head
TailAB
^
A
B
aVector: =a =aa
Multiplying (by a scalar)
Product is a:
Vector
a
v
Adding Vectors
Triangle Law
av
u
u+v
Subtracting Vectors
u – v
= u + (-v)
v – ua
v
u
u-v
Ex. 26.21. Given that vector of magnitude 2, state
a) A vector of magnitude 4 in the direction of vb) A vector of magnitude 0.5 in the direction of vc) A vector of magnitude 6 in the opposite direction of
vd) A unit vector in the direction of ve) A unit vector in the opposite direction of v
S.E. 26.32. 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
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
1 2 3 4 5 6 7 8
9
8
7
6
5
4
3
2
1
P
AOi
j
Component formOP =
i =
j =
OA =
OP =
OP =
OA + APUnit vector (x)Unit vector (y)6i8j
6i + 8j
S.E. 26.41. State a unit vector:
(a)In the x direction (b) in the y direction
1. State a unit vector: (a) In the negative x direction (b) in the negative y direction
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 y direction
Addinga = a
1i + a
2 j
b = b1
i + b2
j
a + b = (a1
+ 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
1i
j
Multiplying (by a scalar)v = v
1i + v
2 j
av =a(v1
i + v2
j)
av =av1
i + av2
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
1i
j
Ex 26.41. Calculate the magnitude of:
a) i + 3j b) 3i + 4j c) –i d)1/2(i + j)
2. 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 – 4cd) IaI e) I4aI f) I3b + 2cI
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
Component forma = a
1i + a
2 j
b = b1
i + b2
j
a.b = a1
b1
+ a2
b2
For two equal vectors:
a.a = a1
a1
+ a2
= a1
2 + a2
2
= IaI2 ij
IaI2
a12
a22
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
Comparing the component and geometric forms
Show:
• i.i=1
• i.j=0
a.b
• i.i = I1II1IcosC = 1
• i.j = I1II1IcosC = 0
= (a1
i + a2
j).(b1
i+ b2
j)
=a1
b1
i.i + a2
b2
j.j + a2
b1
i.j + a1
b2
i.j
= a1
b1
+ a2
b2
…as shown before
Exercise 26.51. 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) By using the dot product, calculate the angle between AB and BC
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?
Right hand rulea x b
a. 1st term, 1st finger
b. 2nd term, 2nd finger
x product. Thumb
j
i
If k = i x j,which direction is k?
k
-k
ExerciseFind:
• 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
kij
===
0
Component formu = u
1i + u
2j + u
3k
v = v1
i + v2
j + v3
k
u x v =
= (u1
i + u2
j + u3
k)(v1
i + v2
j + v3
k)
= u1
iv1
i + u1
i v2
j + u1
iv3
k + u2
jv1
i + u2
jv2
j + u2
j v3
k + u3
kv1
i + u3
kv2
j +
u3
k v3
k
= u1
i v2
j + u1
iv3
k + u2
jv1
i + u2
j v3
k + u3
kv1
i + u3
kv2
j
u x v = (u2
v3
– u3
v2
)i + (u3
v1
– u
1v3
)j + (u1
v2
– u2
v1
)k
ixj = -jxi = kjxk = -kxj = ikxi = -ixk =j
Magnitude of cross productu x v = (u
2v3
– u3
v2
) + (u3
v1
– u
1v3
) + (u1
v2
– u2
v1
)
From Pythagoras’ Theorem:
IuxvI2 = (u2
v3
– u3
v2
)2 + (u3
v1
– u
1v3
)2 + (u1
v2
– u2
v1
)2
= u2
2v3
2 – 2u
2u
3v2
v3
+ u3
2v2
2
+ u3
2v1
2 – 2u
1u
3v1
v3
+ u1
2v3
2
+ u1
2v2
2 – 2u
1u
2v1
v2
+ u2
2v1
2
= u1
2(v2
2+v3
2) + u2
2(v1
2+v3
2) + u3
2(v1
2+v2
2)
- 2(u2
u3
v2
v3
+ u1
u3
v1
v3
+ u1
u2
v1
v2
) 1
IuIIvIcosC = u.v = u1
v1
+ u2
v2
+ u3
v3
IuI2IvI2cos2C = (u.v)2 = (u1
v1
+ u2
v2
+ u3
v3
)2
= u1
2v1
2 u1
u2
v1
v2
u1
u3
v1
v3
u2
u3
v2
v3
u2
2v2
2 + u1
u2
v1
v2
+ u1
u3
v1
v3
+ u2
u3
v2
v3
u3
2v3
2
= u1
2v1
2 + u2
2v2
2 + u3
2v3
2
+ 2(u1
u2
v1
v2
+ u1
u3
v1
v3
+ u2
u3
v2
v3
) 2
+
= IuxvI + IuI2IvI2cos2C
= u1
2(v2
2+v3
2) + u2
2(v1
2+v3
2) + u3
2(v1
2+v2
2)
- 2(u2
u3
v2
v3
+ u1
u3
v1
v3
+ u1
u2
v1
v2
)
u1
2v1
2 + u2
2v2
2 + u3
2v3
2
+ 2(u1
u2
v1
v2
+ u1
u3
v1
v3
+ u2
u3
v2
v3
)
= u1
2(v2
2+v3
2) + u2
2(v1
2+v3
2) + u3
2(v1
2+v2
2)
+ u1
2v1
2 + u2
2v2
2 + u3
2v3
2
= u1
2(v1
2+v2
2+v3
2) + u2
2(v1
2+v2
2+v3
2) + u3
2(v1
2+v2
2+v3
2)
= (u1
2 + u2
2 + u3
2 )(v1
2+v2
2+v3
2)
1 2
By pythagoras’ Theorem:
(u1
2 + u2
2 + u3
2 )(v1
2+v2
2+v3
2)
= IuI2IvI2
In summary:
+ = Iu x vI2 + IuI2IvI2cos2C = IuI2IvI2
Iu x vI2 = IuI2IvI2 - IuI2IvI2cos2C
= IuI2IvI2 (1- cos2C)
= IuI2IvI2 (sin2C)
Iuxvi2 = IuI2IvI2 sin2C
1 2
Magnitude as a parallelogram
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