# Homework Help: Angle Between Vectors given Ratio of Resultants, why is this wrong?

1. Sep 26, 2011

### Ordain

1. The problem statement, all variables and given/known data

Vectors A and B have the same magnitude. Given that the magnitude of A + B is 75 times greater than the magnitude of A - B, find the angle between them?

2. Relevant equations

We know that A=B, so:

2AB+2ABCos$\theta$=75(2AB-2ABCos$\theta$)

Given that A=B

2A2+2A2Cos$\theta$=75(2A2-2A2Cos$\theta$)

3. Attempt at a solution

2A2+2A2Cos$\theta$=75(2A2-2A2Cos$\theta$)

Couldn't get latex to work for the fraction so:

2A2+2A2Cos$\theta$ divided by (2A2-2A2Cos$\theta$)=75

Factor out 2A2:

$\frac{1+Cos}{1-cos}=75 1+Cos[itex]\theta$=75-75$\theta$
-74=-76Cos$\theta$
Cos$\theta$=74/76
$\theta$=Cos-1(74/76)
$\theta$=13.2 deg

Apparently this is wrong, how so?

2. Sep 26, 2011

### danielakkerma

Hey,
Actually, the way I see it, verifying your logic and procedure, it seems fine!
It might also help, to get abit of order in your calculations, to mark it as this:
Assuming the angle between the two is theta(like you said), but the ratio, for the sake of generality, we'll call alpha, that gives us:
$\large |A+B|=\alpha |A-B|, |A|=|B| \Longrightarrow A^2+2A^2\cos{\theta}+A^2 = \alpha(A^2-2A^2\cos{\theta}+A^2)$
Dividing by 2*A^2:
$\large 1+\cos{\theta} = \alpha(1-\cos{\theta}) \Longrightarrow \cos{\theta} = \frac{\alpha-1}{\alpha+1}$
Check to see that you have the proper answer key.
Are there any other considerations you may have overlooked?
Daniel

3. Sep 26, 2011

### Ordain

Thanks for that, noted for next time. This is the question in exact form, I can't find anything I'm missing.

4. Sep 26, 2011

### danielakkerma

Try entering it in radians, namely: 0.229 ~ 0.23.
What's the typical accuracy, digit significance asked? are two decimal places usually sufficient?
If so, try either 0.229, 0.23.
Daniel