How Do I Calculate the Angle Between Vectors in Vector Projection Problems?

[wicked1977]

Homework Statement

http://imgur.com/a/Yq8pW

Homework Equations

projection u onto v: ((u x v)/(||v||^2)) x v
Unit vector: u/||u||

The Attempt at a Solution

For number 2, I absolutely do not know how to set up the problem. I do not know what vectors to use, I assumed F vector to be <0.00375, 0.00625> and v vector to be <3 , 5> and plug them into the projection formula- projection u onto v: ((u x v)/(||v||^2)) x v-but that seems wrong since the teacher addressed that I find the vector of V, magnitude of vector V, and find unit vector so it should be <3, 5>/(sqrt 34) but how do I find the angle off of that?

 

[jedishrfu]

Do you know about the vector dot product and what it represents?

 

[wicked1977]

Do you know about the vector dot product and what it represents?

I can find the angle between the vectors. Yes? But what I am confused about is what vectors to use.

 

[jedishrfu]

You have F in i,j,k form and you should have v in i,j,k form.

 

[wicked1977]

You have F in i,j,k form and you should have v in i,j,k form.

I think I got it! Except that the question states that the vector V is in the xy plane:
cos^-1((<3 , 5> x <0.00375 , 0.00625>)/(sqrt34 x sqrt 50))=89.94

 

[jedishrfu]

Vector F=<3,5,4>/sqrt(3^2 + 5^2 + 4^2) * 800

so the unit vector for F is: Funit=<3,5,4> / sqrt(3^2 + 5^2 + 4^2) = <3,5,4> / (5*sqrt(2))

and unit vector Vunit = <3,5,0> / sqrt(3^2 + 5^2 + 0^2) = <3,5,0> / sqrt(34)

Funit . Vunit = (3^2 + 5^2 + 0*4) / (5*sqrt(2)*sqrt(34))

I didn't get 89.94 degrees for the angle.

 

[wicked1977]

Vector F=<3,5,4>/sqrt(3^2 + 5^2 + 4^2) * 800

so the unit vector for F is: Funit=<3,5,4> / sqrt(3^2 + 5^2 + 4^2) = <3,5,4> / (5*sqrt(2))

and unit vector Vunit = <3,5,0> / sqrt(3^2 + 5^2 + 0^2) = <3,5,0> / sqrt(34)

Funit . Vunit = (3^2 + 5^2 + 0*4) / (5*sqrt(2)*sqrt(34))

I didn't get 89.94 degrees for the angle.

Did not realize I had to multiply my F vector by 800. And based on your solution, the angle is found to be 34.4. Thanks much!

 

[jedishrfu]

In this case, there was a shortcut way too. If you notice that F, v and the 4k z component of F form a right triangle with F as the hypotenuse so that the sin of the angle must be 4/magnitude(F) = 4/sqrt(50) and hence its 34.4 degrees.