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°
