MATHEMATICS Assignment – 4
1. Recitation Problems
1.1. Chapter 9
1.1.1. Three computers, A, B, and C, have the numerical features listed below:
We may imagine these values as defining a vector for each computer; for instance, A’s
vector is [3.06, 500, 6]. We can compute the cosine distance between any two of the
vectors, but if we do not scale the components, then the disk size will dominate and make
differences in the other components essentially invisible. Let us use 1 as the scale factor for
processor speed, α for the disk size, and β for the main memory size.
Feature Processor Speed Disk Size Main-Memory Size
A 3.06 500α 6β
B 2.68 320α 4β
C 2.92 640α 6β
(a) In terms of α and β, compute the cosines of the angles between the vectors for each pair
of the three computers.
A=[3.06, 500α, 6β]
B=[2.68, 320α, 4β]
Cos θ = 8.2008 + 160000α2 + 24β2
√(9.36 + 250000α2 + 36β2) * √(7.18 + 102400α2 + 16β2)
A=[3.06, 500α, 6β]
C=[2.92, 640α, 6β]
Cos θ = 8.9352 + 320000α2 + 36β2
√(9.36 + 250000α2 + 36β2) * √(8.53 + 409600α2 + 36β2)
B=[2.68, 320α, 4β]
, C=[2.92, 640α, 6β]
Cos θ = 7.8256 + 204800α2 + 24β2
√(7.18 + 102400α2 + 16β2) * √(8.53 + 409600α2 + 36β2)
(b) What are the angles between the vectors if α = β = 1?
A=[3.06, 500α, 6β]
B=[2.68, 320α, 4β]
Cos θ = 8.2008 + 160000α2 + 24β2
√(9.36 + 250000α2 + 36β2) * √(7.18 + 102400α2 + 16β2)
= 8.2008 + 160000(1)2 + 24(1)2
√(9.36 + 250000(1)2 + 36(1)2) * √(7.18 + 102400(1)2 + 16(1)2)
= 8.2008 + 160000 + 24
√(9.36 + 250000 + 36) * √(7.18 + 102400 + 16)
= 0.99
The angle between vector A and B is approximately 0◦
A=[3.06, 500α, 6β]
C=[2.92, 640α, 6β]
Cos θ = 8.9352 + 320000α2 + 36β2
√(9.36 + 250000α2 + 36β2) * √(8.53 + 409600α2 + 36β2)
= 8.9352 + 320000(1)2 + 36(1)2
√(9.36 + 250000(1)2 + 36(1)2) * √(8.53 + 409600(1)2 + 36(1)2)
= 8.9352 + 320000 + 36
√(9.36 + 250000 + 36) * √(8.53 + 409600 + 36)
= 0.99
The angle between vector A and C is approximately 0◦
B=[2.68, 320α, 4β]
1. Recitation Problems
1.1. Chapter 9
1.1.1. Three computers, A, B, and C, have the numerical features listed below:
We may imagine these values as defining a vector for each computer; for instance, A’s
vector is [3.06, 500, 6]. We can compute the cosine distance between any two of the
vectors, but if we do not scale the components, then the disk size will dominate and make
differences in the other components essentially invisible. Let us use 1 as the scale factor for
processor speed, α for the disk size, and β for the main memory size.
Feature Processor Speed Disk Size Main-Memory Size
A 3.06 500α 6β
B 2.68 320α 4β
C 2.92 640α 6β
(a) In terms of α and β, compute the cosines of the angles between the vectors for each pair
of the three computers.
A=[3.06, 500α, 6β]
B=[2.68, 320α, 4β]
Cos θ = 8.2008 + 160000α2 + 24β2
√(9.36 + 250000α2 + 36β2) * √(7.18 + 102400α2 + 16β2)
A=[3.06, 500α, 6β]
C=[2.92, 640α, 6β]
Cos θ = 8.9352 + 320000α2 + 36β2
√(9.36 + 250000α2 + 36β2) * √(8.53 + 409600α2 + 36β2)
B=[2.68, 320α, 4β]
, C=[2.92, 640α, 6β]
Cos θ = 7.8256 + 204800α2 + 24β2
√(7.18 + 102400α2 + 16β2) * √(8.53 + 409600α2 + 36β2)
(b) What are the angles between the vectors if α = β = 1?
A=[3.06, 500α, 6β]
B=[2.68, 320α, 4β]
Cos θ = 8.2008 + 160000α2 + 24β2
√(9.36 + 250000α2 + 36β2) * √(7.18 + 102400α2 + 16β2)
= 8.2008 + 160000(1)2 + 24(1)2
√(9.36 + 250000(1)2 + 36(1)2) * √(7.18 + 102400(1)2 + 16(1)2)
= 8.2008 + 160000 + 24
√(9.36 + 250000 + 36) * √(7.18 + 102400 + 16)
= 0.99
The angle between vector A and B is approximately 0◦
A=[3.06, 500α, 6β]
C=[2.92, 640α, 6β]
Cos θ = 8.9352 + 320000α2 + 36β2
√(9.36 + 250000α2 + 36β2) * √(8.53 + 409600α2 + 36β2)
= 8.9352 + 320000(1)2 + 36(1)2
√(9.36 + 250000(1)2 + 36(1)2) * √(8.53 + 409600(1)2 + 36(1)2)
= 8.9352 + 320000 + 36
√(9.36 + 250000 + 36) * √(8.53 + 409600 + 36)
= 0.99
The angle between vector A and C is approximately 0◦
B=[2.68, 320α, 4β]
