Find Angle Between Two Vectors...2

[nycmathguy]

Section 6.4

Find the angle theta (in radians) between the two vectors.

Please, work out (38) in step by step fashion. I will then do all odd numbers when time allows and check the answers for myself in the back of the book. Thank you.

 

[MathLover1]

38.

The angle theta between two vectors u and v is given by the formula of the dot product:

u *v =|u|*|v|cos(θ) where |u| and |v| are the absolute values/magnitudes

given:

u=cos(pi/4)i +sin(pi/4)j
u=(1/sqrt(2))i+(1/sqrt(2))j
u=(sqrt(2)/2)i+(sqrt(2)/2)j

v=cos(5pi/4)i +sin(5pi/4)j
v=-(1/sqrt(2))i -(1/sqrt(2))j
v= -(sqrt(2)/2)i-(sqrt(2)/2)j

u*v=- (j + i)^2/2
u*v=- (j ^2+2ij+ i^2)/2
u*v=- j ^2/2+2ij/2+ i^2/2
u*v=- j ^2/2+ij/+ i^2/2

By the dot product, i*i=j*j=1 and i*j=0

u*v=(sqrt(2)/2)*(-(sqrt(2)/2))i^2 +(sqrt(2)/2)(-(sqrt(2)/2)j^2

u*v=(-1/2)*1+ (-1/2)*1

u*v=-1

the magnitude of a vector u and v is:

u=sqrt(((sqrt(2)/2))^2+((sqrt(2)/2))^2)=1

v=sqrt((-(sqrt(2)/2))^2+(-(sqrt(2)/2))^2)=1

.u *v =|u|*|v|*cos(θ)

cos(θ)=-1/(1*1)

cos(θ)=-1

θ= cos^-1(-1)

θ= π (result in radians)
θ=180°

 

[nycmathguy]

38.

The angle theta between two vectors u and v is given by the formula of the dot product:

u *v =|u|*|v|cos(θ) where |u| and |v| are the absolute values/magnitudes

given:

u=cos(pi/4)i +sin(pi/4)j
u=(1/sqrt(2))i+(1/sqrt(2))j
u=(sqrt(2)/2)i+(sqrt(2)/2)j

v=cos(5pi/4)i +sin(5pi/4)j
v=-(1/sqrt(2))i -(1/sqrt(2))j
v= -(sqrt(2)/2)i-(sqrt(2)/2)j

u*v=- (j + i)^2/2
u*v=- (j ^2+2ij+ i^2)/2
u*v=- j ^2/2+2ij/2+ i^2/2
u*v=- j ^2/2+ij/+ i^2/2

By the dot product, i*i=j*j=1 and i*j=0

u*v=(sqrt(2)/2)*(-(sqrt(2)/2))i^2 +(sqrt(2)/2)(-(sqrt(2)/2)j^2

u*v=(-1/2)*1+ (-1/2)*1

u*v=-1

the magnitude of a vector u and v is:

u=sqrt(((sqrt(2)/2))^2+((sqrt(2)/2))^2)=1

v=sqrt((-(sqrt(2)/2))^2+(-(sqrt(2)/2))^2)=1

.u *v =|u|*|v|*cos(θ)

cos(θ)=-1/(1*1)

cos(θ)=-1

θ= cos^-1(-1)

θ= π (result in radians)
θ=180°

Thank you. I will try 37 and return here if I get stuck or my answer is wrong.

 

[nycmathguy]

38.

The angle theta between two vectors u and v is given by the formula of the dot product:

u *v =|u|*|v|cos(θ) where |u| and |v| are the absolute values/magnitudes

given:

u=cos(pi/4)i +sin(pi/4)j
u=(1/sqrt(2))i+(1/sqrt(2))j
u=(sqrt(2)/2)i+(sqrt(2)/2)j

v=cos(5pi/4)i +sin(5pi/4)j
v=-(1/sqrt(2))i -(1/sqrt(2))j
v= -(sqrt(2)/2)i-(sqrt(2)/2)j

u*v=- (j + i)^2/2
u*v=- (j ^2+2ij+ i^2)/2
u*v=- j ^2/2+2ij/2+ i^2/2
u*v=- j ^2/2+ij/+ i^2/2

By the dot product, i*i=j*j=1 and i*j=0

u*v=(sqrt(2)/2)*(-(sqrt(2)/2))i^2 +(sqrt(2)/2)(-(sqrt(2)/2)j^2

u*v=(-1/2)*1+ (-1/2)*1

u*v=-1

the magnitude of a vector u and v is:

u=sqrt(((sqrt(2)/2))^2+((sqrt(2)/2))^2)=1

v=sqrt((-(sqrt(2)/2))^2+(-(sqrt(2)/2))^2)=1

.u *v =|u|*|v|*cos(θ)

cos(θ)=-1/(1*1)

cos(θ)=-1

θ= cos^-1(-1)

θ= π (result in radians)
θ=180°

Here is my work for 33.

 

[nycmathguy]

38.

The angle theta between two vectors u and v is given by the formula of the dot product:

u *v =|u|*|v|cos(θ) where |u| and |v| are the absolute values/magnitudes

given:

u=cos(pi/4)i +sin(pi/4)j
u=(1/sqrt(2))i+(1/sqrt(2))j
u=(sqrt(2)/2)i+(sqrt(2)/2)j

v=cos(5pi/4)i +sin(5pi/4)j
v=-(1/sqrt(2))i -(1/sqrt(2))j
v= -(sqrt(2)/2)i-(sqrt(2)/2)j

u*v=- (j + i)^2/2
u*v=- (j ^2+2ij+ i^2)/2
u*v=- j ^2/2+2ij/2+ i^2/2
u*v=- j ^2/2+ij/+ i^2/2

By the dot product, i*i=j*j=1 and i*j=0

u*v=(sqrt(2)/2)*(-(sqrt(2)/2))i^2 +(sqrt(2)/2)(-(sqrt(2)/2)j^2

u*v=(-1/2)*1+ (-1/2)*1

u*v=-1

the magnitude of a vector u and v is:

u=sqrt(((sqrt(2)/2))^2+((sqrt(2)/2))^2)=1

v=sqrt((-(sqrt(2)/2))^2+(-(sqrt(2)/2))^2)=1

.u *v =|u|*|v|*cos(θ)

cos(θ)=-1/(1*1)

cos(θ)=-1

θ= cos^-1(-1)

θ= π (result in radians)
θ=180°

Here is my work for 35.

 

[nycmathguy]

38.

The angle theta between two vectors u and v is given by the formula of the dot product:

u *v =|u|*|v|cos(θ) where |u| and |v| are the absolute values/magnitudes

given:

u=cos(pi/4)i +sin(pi/4)j
u=(1/sqrt(2))i+(1/sqrt(2))j
u=(sqrt(2)/2)i+(sqrt(2)/2)j

v=cos(5pi/4)i +sin(5pi/4)j
v=-(1/sqrt(2))i -(1/sqrt(2))j
v= -(sqrt(2)/2)i-(sqrt(2)/2)j

u*v=- (j + i)^2/2
u*v=- (j ^2+2ij+ i^2)/2
u*v=- j ^2/2+2ij/2+ i^2/2
u*v=- j ^2/2+ij/+ i^2/2

By the dot product, i*i=j*j=1 and i*j=0

u*v=(sqrt(2)/2)*(-(sqrt(2)/2))i^2 +(sqrt(2)/2)(-(sqrt(2)/2)j^2

u*v=(-1/2)*1+ (-1/2)*1

u*v=-1

the magnitude of a vector u and v is:

u=sqrt(((sqrt(2)/2))^2+((sqrt(2)/2))^2)=1

v=sqrt((-(sqrt(2)/2))^2+(-(sqrt(2)/2))^2)=1

.u *v =|u|*|v|*cos(θ)

cos(θ)=-1/(1*1)

cos(θ)=-1

θ= cos^-1(-1)

θ= π (result in radians)
θ=180°

Here is my work for 37.

 

[MathLover1]

33. correct
35. correct
37. ->you put 38 but actually did 37 and it is correct

 

[nycmathguy]

33. correct
35. correct
37. ->you put 38 but actually did 37 and it is correct

Thank you. I see that there's no getting around posting my work here. Textbooks are not always the best companion for learning, or in my case, revisiting a course taken in the Spring 1993 semester at Lehman College.