The angle between two vectors
P1
8
● As you work through the proof in this section, make a list of all the results that
you are assuming.
The angle between two vectors
To find the angle θ between the y
B
two vectors (b1, b2)
→ A
OA = a = a1i + a2j
(a1, a2)
and
a b
→
OB = b = b1i + b2j
start by applying the cosine rule to θ
triangle OAB in figure 8.21. O x
OA 2 +OB2 – AB2 Figure 8.21
cos θ =
2OA × OB
→
→
→
In this, OA, OB and AB are the lengths of the vectors OA, OB and AB, and so
OA = | a | = a 12 + a22 and OB = | b | = b 12 + b22 .
→
The vector AB = b − a = (b1i + b2j) − (a1i + a2j)
= (b1 − a1)i + (b2 − a2)j
and so its length is given by
AB = | b − a | = (b1 – a1)2 + (b2 – a2)2.
Substituting for OA, OB and AB in the cosine rule gives
(a12 + a 22) + (b 12 + b 22) – [(b1 – a1)2 + (b2 – a2)2]
cos θ =
2 a 12 + a 22 × b 12 + b 22
a 2 + a 2 + b 12 + b 22 – (b 12 – 2a1b1 + a12 + b 22 – 2a2b2 + a 22 )
= 1 2
2a b
This simplifies to
2a1b1 + 2a2b2 a1b1 + a2b2
cos θ = =
2 a b a b
The expression on the top line, a1b1 + a2b2, is called the scalar product (or dot
product) of the vectors a and b and is written a . b. Thus
cos θ = a . b .
a b
This result is usually written in the form
271
a . b = | a | | b | cos θ.
, The next example shows you how to use it to find the angle between two vectors
P1 given numerically.
8
Find the angle between the vectors and
3 5
.
–12
EXAMPLE 8.11
4
Vectors
SOLUTION
3
Let a= ⇒ | a | = 32 + 42 = 5
4
5
b= ⇒ | b | = 52 + (–12)2 = 13.
–12
and
The scalar product
3 5 = 3 × 5 + 4 × (−12)
4 . –12
= 15 − 48
= −33.
Substituting in a . b = | a | | b | cos θ gives
−33 = 5 × 13 × cos θ
cos θ = –33
65
⇒ θ = 120.5°.
Perpendicular vectors
Since cos 90° = 0, it follows that if vectors a and b are perpendicular then
a . b = 0.
Conversely, if the scalar product of two non-zero vectors is zero, they are
perpendicular.
2 6
EXAMPLE 8.12 Show that the vectors a = and b = are perpendicular.
4 –3
SOLUTION
The scalar product of the vectors is
2 6
a.b = .
4 –3
= 2 × 6 + 4 × (−3)
= 12 − 12 = 0.
Therefore the vectors are perpendicular.
272
P1
8
● As you work through the proof in this section, make a list of all the results that
you are assuming.
The angle between two vectors
To find the angle θ between the y
B
two vectors (b1, b2)
→ A
OA = a = a1i + a2j
(a1, a2)
and
a b
→
OB = b = b1i + b2j
start by applying the cosine rule to θ
triangle OAB in figure 8.21. O x
OA 2 +OB2 – AB2 Figure 8.21
cos θ =
2OA × OB
→
→
→
In this, OA, OB and AB are the lengths of the vectors OA, OB and AB, and so
OA = | a | = a 12 + a22 and OB = | b | = b 12 + b22 .
→
The vector AB = b − a = (b1i + b2j) − (a1i + a2j)
= (b1 − a1)i + (b2 − a2)j
and so its length is given by
AB = | b − a | = (b1 – a1)2 + (b2 – a2)2.
Substituting for OA, OB and AB in the cosine rule gives
(a12 + a 22) + (b 12 + b 22) – [(b1 – a1)2 + (b2 – a2)2]
cos θ =
2 a 12 + a 22 × b 12 + b 22
a 2 + a 2 + b 12 + b 22 – (b 12 – 2a1b1 + a12 + b 22 – 2a2b2 + a 22 )
= 1 2
2a b
This simplifies to
2a1b1 + 2a2b2 a1b1 + a2b2
cos θ = =
2 a b a b
The expression on the top line, a1b1 + a2b2, is called the scalar product (or dot
product) of the vectors a and b and is written a . b. Thus
cos θ = a . b .
a b
This result is usually written in the form
271
a . b = | a | | b | cos θ.
, The next example shows you how to use it to find the angle between two vectors
P1 given numerically.
8
Find the angle between the vectors and
3 5
.
–12
EXAMPLE 8.11
4
Vectors
SOLUTION
3
Let a= ⇒ | a | = 32 + 42 = 5
4
5
b= ⇒ | b | = 52 + (–12)2 = 13.
–12
and
The scalar product
3 5 = 3 × 5 + 4 × (−12)
4 . –12
= 15 − 48
= −33.
Substituting in a . b = | a | | b | cos θ gives
−33 = 5 × 13 × cos θ
cos θ = –33
65
⇒ θ = 120.5°.
Perpendicular vectors
Since cos 90° = 0, it follows that if vectors a and b are perpendicular then
a . b = 0.
Conversely, if the scalar product of two non-zero vectors is zero, they are
perpendicular.
2 6
EXAMPLE 8.12 Show that the vectors a = and b = are perpendicular.
4 –3
SOLUTION
The scalar product of the vectors is
2 6
a.b = .
4 –3
= 2 × 6 + 4 × (−3)
= 12 − 12 = 0.
Therefore the vectors are perpendicular.
272
