Find Angle Between Two Vectors

[nycmathguy]

Section 6.4

Please, work out (30) 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]

30.

u=<3,2>
v=<4,0>

u *v =|u|*|v|cos(θ)
cos(θ) =(u *v)/(|u|*|v|)


u *v =<3,2>*<4,0>=3*4+4*0=12

|u|=sqrt(3^2+2^2)=sqrt(13)

|v|=sqrt(4^2+0^2)=sqrt(16)=4

cos(θ) =(u *v ) /(|u|*|v|)=12/(sqrt(13)*4)=3/(sqrt(13)

θ=cos^-1(3/(sqrt(13))

θ=0.5880026 (result in radians)

θ=33.69°

 

[nycmathguy]

30.

u=<3,2>
v=<4,0>

u *v =|u|*|v|cos(θ)
cos(θ) =(u *v)/(|u|*|v|)


u *v =<3,2>*<4,0>=3*4+4*0=12

|u|=sqrt(3^2+2^2)=sqrt(13)

|v|=sqrt(4^2+0^2)=sqrt(16)=4

cos(θ) =(u *v ) /(|u|*|v|)=12/(sqrt(13)*4)=3/(sqrt(13)

θ=cos^-1(3/(sqrt(13))

θ=0.5880026 (result in radians)

θ=33.69°

Perfect.

How do you convert θ = 0.5880026 (result in radians) to θ = 33.69°?

 

[nycmathguy]

30.

u=<3,2>
v=<4,0>

u *v =|u|*|v|cos(θ)
cos(θ) =(u *v)/(|u|*|v|)


u *v =<3,2>*<4,0>=3*4+4*0=12

|u|=sqrt(3^2+2^2)=sqrt(13)

|v|=sqrt(4^2+0^2)=sqrt(16)=4

cos(θ) =(u *v ) /(|u|*|v|)=12/(sqrt(13)*4)=3/(sqrt(13)

θ=cos^-1(3/(sqrt(13))

θ=0.5880026 (result in radians)

θ=33.69°

Unfortunately, I will need to post my work for you to check because the answers in the back of the book are expressed differently. I just want to know that my work is correct.

Here is my work:

 

[MathLover1]

29. is correct, pi/2 is answer in radians which is same as 90°, but question ask you to find angle in radians
31.
cos^(-1)(-4/5)=2.4980915447965=2.50 (result in radians)

 

[nycmathguy]

29. is correct, pi/2 is answer in radians which is same as 90°, but question ask you to find angle in radians
31.
cos^(-1)(-4/5)=2.4980915447965=2.50 (result in radians)

I see my error.