Physics..

So.. Errr, yeah. -confused-

-Note- AP Physics problems. FUN TIMES AHEAD!

52) Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A+B to be larger than the magnitude of A-B by the factor n, what must be the angle between them?

He had given us a few hints, but I can't seem the piece together how I'd actually use these hints in practice.

Hint 1: Law of Cosines
Hint 2: Trig Identities

So if anyone could explain this problem, that would be dandy.

Edit: Oh never mind, I somehow overlooked the word "factor". Big difference there. D:

Wouldn't 'by a factor of n' mean you multiply (A-B) by n to get (A+B)?

Wouldn't 'by a factor of n' mean you multiply (A-B) by n to get (A+B)?

That is correct.

Code:

|[B]A[/B]-[B]B[/B]|*n = |[B]A[/B]+[B]B[/B]|
|[B]A[/B]| = |[B]B[/B]|
a*[B][I]i[/I][/B] + b*[B][I]j[/I][/B] = [B]A[/B]
x*[B][I]i[/I][/B] + y*[B][I]j[/I][/B] = [B]B[/B]

I'm thinking rhombi and triangles.

If n = 1, then theta = 90 degrees. That much I can tell you immediately.

 Workspace

Code:

Law of Cosines: C^2 = A^2 + B^2 - 2ABcos(c), where C is the side of a triangle opposite of angle c, and A, B are the other sides of the triangle.

Assume C1 = |[B]A[/B]-[B]B[/B]| and C2 = |[B]A[/B]+[B]B[/B]|.  Using substitution, we get...
|[B]A[/B]-[B]B[/B]|*n = |[B]A[/B]+[B]B[/B]|
C1*n = C2
[|[B]A[/B]|^2 + |[B]B[/B]|^2 - 2|[B]A[/B]||[B]B[/B]|cos(θ)]n^2 = |[B]A[/B]|^2 + |[B]B[/B]|^2 - 2|[B]A[/B]||[B]B[/B]|cos(180-θ)
The reason why I put 180-θ for the 2nd angle for C2 is due to the property of parallelograms in general.

Further simplification...
(n^2 - 1) * [|[B]A[/B]|^2 - |[B]A[/B]|^2 + |[B]B[/B]|^2 - |[B]B[/B]|^2] - 2|[B]A[/B]||[B]B[/B]|cos(θ)*n^2 + 2|[B]A[/B]||[B]B[/B]|cos(180-θ) = 0
- 2|[B]A[/B]||[B]B[/B]|cos(θ)*n^2 + 2|[B]A[/B]||[B]B[/B]|cos(180-θ) = 0
2|[B]A[/B]||[B]B[/B]|cos(180-θ) = 2|[B]A[/B]||[B]B[/B]|cos(θ)*n^2
cos(180-θ) = cos(θ)*n^2

I don't really get there the (n^2 - 1) came from.

So, the answer, as told from someone in my class who refused to explain it to me, is:

θ = 2tan[SIZE="1"]-1[/SIZE](1/n)

e_e;

Oops, I think I jumped too far ahead and was being a naughty boy >.< It should have actually been this:

Code:

|[B]A[/B]|^2*n^2 - |[B]A[/B]|^2 = (n^2 - 1) * |[B]A[/B]|^2

This might get a bit random until I finish...
(n^2 - 1) * (|[B]A[/B]|^2 + |[B]B[/B]|^2) - 2|[B]A[/B]||[B]B[/B]|cos(θ)*n^2 + 2|[B]A[/B]||[B]B[/B]|cos(180-θ) = 0
(n^2 - 1) * (|[B]A[/B]|^2 + |[B]B[/B]|^2) + 2|[B]A[/B]||[B]B[/B]|cos(180-θ) = 2|[B]A[/B]||[B]B[/B]|cos(θ)*n^2
(n^2 - 1) * (|[B]A[/B]|^2 + |[B]B[/B]|^2) + 2|[B]A[/B]||[B]B[/B]| * (cos(180-θ) - cos(θ)*n^2) = 0
(n^2 - 1) * (|[B]A[/B]|^2 + |[B]B[/B]|^2) / (2|[B]A[/B]||[B]B[/B]|) + cos(180-θ) - cos(θ)*n^2 = 0
(n^2 - 1) * (|[B]A[/B]|^2 + |[B]B[/B]|^2) / (2|[B]A[/B]||[B]B[/B]|) = cos(θ)*n^2 / cos(180-θ)
(1 - 1/n^2) * (|[B]A[/B]|^2 + |[B]B[/B]|^2) / (2|[B]A[/B]||[B]B[/B]|) = cos(θ)/cos(180-θ)

cos(180-θ) = cos(180)cos(θ) + sin(180)sin(θ) = -cos(θ)

(1 - 1/n^2) * (|[B]A[/B]|^2 + |[B]B[/B]|^2) / (2|[B]A[/B]||[B]B[/B]|) = -1

Crap... this isn't right...

One thing that I should have done from the very start was saying |A| = |B| so I wouldn't have to deal with the clutter of A's and B's.

Uhhh I haven't taken physics AP but don't you just do law of cosines?

Assume |A| = |B| = 1 (problem doesn't state otherwise, so why not)

Then, by Law of Cosines:
|A+B| = sqrt(1 + 1 - 2 cos θ = sqrt(2 - 2 cos θ
|A-B| = sqrt(1 + 1 - 2 cos (180-θ) = sqrt(2 + 2 cos θ

|A+B| = n*|A-B|
sqrt(2 - 2 cos θ = n*(sqrt(2 + 2 cos θ)
2 - 2 cos θ = 2n^2 + 2n^2 cos θ
2 - 2n^2 = (2 + 2n^2) cos θ
cos θ = (1 - n^2)/(1 + n^2)

Probably able to be simplified, but I can't be bothered.

Uhhh I haven't taken physics AP but don't you just do law of cosines?

Assume |A| = |B| = 1 (problem doesn't state otherwise, so why not)

That's a no-no. Only reason why it doesn't matter this time is because they completely cancel each other out.

 Alright, 2nd attempt, or so help me jebus!

Code:

[|A|^2 + |B|^2 - 2|A||B|cos(θ)]n^2 = |A|^2 + |B|^2 - 2|A||B|cos(180-θ)
[|A|^2 + |B|^2 - 2|A||B|cos(θ)]n^2 = |A|^2 + |B|^2 + 2|A||B|cos(θ)
[2|A|^2 - 2|A|^2*cos(θ)]n^2 = 2|A|^2 + 2|A|^2cos(θ)
[1 - cos(θ)]n^2 = 1 + cos(θ)
n^2 = 1 + cos(θ) + cos(θ)*n^2
n^2 = 1 + (1 + n^2)*cos(θ)
(n^2 - 1) / (n^2 + 1) = cos(θ)

tbh, I'm not sure how the other guy managed to get θ = 2*arctan(1/n) @_@ I like his answer though.

That's a no-no. Only reason why it doesn't matter this time is because they completely cancel each other out.

 Alright, 2nd attempt, or so help me jebus!

Code:

[|A|^2 + |B|^2 - 2|A||B|cos(θ)]n^2 = |A|^2 + |B|^2 - 2|A||B|cos(180-θ)
[|A|^2 + |B|^2 - 2|A||B|cos(θ)]n^2 = |A|^2 + |B|^2 + 2|A||B|cos(θ)
[2|A|^2 - 2|A|^2*cos(θ)]n^2 = 2|A|^2 + 2|A|^2cos(θ)
[1 - cos(θ)]n^2 = 1 + cos(θ)
n^2 = 1 + cos(θ) + cos(θ)*n^2
n^2 = 1 + (1 + n^2)*cos(θ)
(n^2 - 1) / (n^2 + 1) = cos(θ)

Well the way the problem is stated, it either doesn't matter, or the answer has to be in terms of |A| as well as n. Intuitively the magnitude of the vectors can change and the angles are still the same, so I just assumed they were unit vectors to make it cleaner. Probably not the most proper, but it gets the answer.

Hm, seems either you or I dropped a negative sign somewhere. I think it might've been me.

Edit: Oh, I see. I designated theta as the other angle.

Edit: Oh, I see. I designated theta as the other angle.

Yup. The only way I can see it being simplified further is doing partial fractions for your approach, but even then that doesn't get it into the nice compact form of arctan.

I hate trig >_>

That's a no-no. Only reason why it doesn't matter this time is because they completely cancel each other out.

 Alright, 2nd attempt, or so help me jebus!

Code:

[|A|^2 + |B|^2 - 2|A||B|cos(θ)]n^2 = |A|^2 + |B|^2 - 2|A||B|cos(180-θ)
[|A|^2 + |B|^2 - 2|A||B|cos(θ)]n^2 = |A|^2 + |B|^2 + 2|A||B|cos(θ)
[2|A|^2 - 2|A|^2*cos(θ)]n^2 = 2|A|^2 + 2|A|^2cos(θ)
[1 - cos(θ)]n^2 = 1 + cos(θ)
n^2 = 1 + cos(θ) + cos(θ)*n^2
n^2 = 1 + (1 + n^2)*cos(θ)
(n^2 - 1) / (n^2 + 1) = cos(θ)

tbh, I'm not sure how the other guy managed to get θ = 2*arctan(1/n) @_@ I like his answer though.

Nicely done. Only thing left to do is to use half angle formula for cosine, and you'll get θ = 2*arctan(1/n)

 Spoiler

cos(θ = [1-(tan(θ/2))^2]/[1+(tan(θ/2))^2]

Edit: I completely missed the part where you defined your θ as the angle between vector A and -B.

The angle between vector A and B would be equal to [180 - 2*arctan(1/n)] or simply 2*arctan(n)