Question

Answer this or you officially lose your NNN

Answer 1

Answer:

1) A

2) B, 6.0°

Answer 2

Answer:

1.) The dot product of the vectors a and b is:

a . b = 30 (Fourth option)

2.) The angle between the vectors is:

ϴ = 6.0° (Second option)

Step-by-step explanation:

1.) a=<5,2> = <a1, a2> → a1=5, a2=2

b=<4,5> = <b1, b2> → b1=4, b2=5

The dot product of the vectors a and b is:

a . b = a1 b1 + a2 b2

a . b = (5)(4)+(2)(5)

a . b = 20 + 10

a . b = 30

2.) u=<2,-4> = <u1, u2> → u1=2, u2=-4

v=<3,-8> = <v1, v2> → v1=3, v2=-8

u . v = │u││v│cos ϴ     (1)

where:

u . v is the dot product of the vectors u and v

│u│is the module of vector u

│v│is the module of vector v

ϴ is the angle between the vectors u and v

The dot product of the vectors u and v is:

u . v = u1 v1 + u2 v2

u . v = (2)(3)+(-4)(-8)

u . v = 6 + 32

u . v = 38

│u│=sqrt( u1^2+u2^2)

│u│=sqrt[ (2)^2+(-4)^2 ]

│u│=sqrt(4+16)

│u│=sqrt(20)

│u│=sqrt(4*5)

│u│=sqrt(4) sqrt(5)

│u│= 2 sqrt(5)

│v│=sqrt( v1^2+v2^2)

│v│=sqrt[ (3)^2+(-8)^2 ]

│v│=sqrt(9+64)

│v│=sqrt(73)

Replacing the calculated values in equation (1):

(1) u . v = │u││v│cos ϴ

38 = 2 sqrt(5) sqrt(73) cos ϴ

38 = 2 sqrt(5*73) cos ϴ

38 = 2 sqrt(365) cos ϴ

Solving for cos ϴ: Dividing both sides of the equation by 2 sqrt(365):

38 / [2 sqrt(365)] = 2 sqrt(365) cos ϴ / [2 sqrt(365)]

19/sqrt(365) =cos ϴ

cos ϴ = 19/sqrt(365)

Solving for ϴ :

ϴ = cos^(-1) [19/sqrt(365)]

ϴ = cos^(-1) (19/19.10497317)

ϴ = cos^(-1) (0.994505453)

ϴ = 6.009005915°

Rounding to the nearest tenth of a degree:

ϴ = 6.0°