# Vector Problem

1. Sep 15, 2004

### Bertuzzi

Vector Problem URGENT

Two vectors A and B have precisly equal magnitudes. In order for the magnitude of A +B to be larger then the magnitude of A - B by the factor n, what must be the anle between them?

There is the question i need help on this quickly thank you for nehelp i know u have to use the cosine rule

2. Sep 15, 2004

### zefram_c

Have you worked out the magnitudes of A+B and A - B ?

3. Sep 15, 2004

### Bertuzzi

no there not given you just know they are equal

4. Sep 15, 2004

### zefram_c

I didn't mean the magnitudes of the original vectors. Let's say |A|=|B|=1 (the answer shouldn't depend on this number). If you let C = A+B and D = A-B , what are the magnitudes of C and D? Hint: to get the magnitude of an arbitrary vector, you take its dot product with....

5. Sep 15, 2004

### Bertuzzi

sorry i still dont get it you must think im stupid but i really have no idea how this will get a solution

6. Sep 15, 2004

### wm

the angle is twice the angle whose tan is 1/(n+1)

the angle is twice the angle whose tan is 1/(n+1).

Last edited: Sep 15, 2004
7. Sep 15, 2004

### Bertuzzi

Please tell me how you got that thank you so much.

8. Sep 15, 2004

### wm

Let o be the desired angle; let the magnitudes of A and B be |A| = a and |B| = b; then let a and b = r, the radius of a circle. So |A-B| is the related chord of the circle: c = 2rsin(o/2). And |A+B| is twice the distance from the centre of the circle to the midpoint of the chord (call it x): x = rcos(o/2) =(n+1)c/2 [because you want |A+B| = (n+1)|A-B|]. So c/x = 2tan(o/2) = 2/(n+1); whence o = 2tan^(-1) [1/(n+1)]. I hope!

PS: Note that I've read the question fairly strictly -- perhaps too strict?
"In order for the magnitude of A +B to be larger then the magnitude of A - B by the factor n"

To me this is not the same as saying:
"In order for the magnitude of A +B to be n-times the magnitude of A - B." If that's what you meant then its 1/n you use.

Last edited: Sep 15, 2004
9. Sep 15, 2004

### Bertuzzi

Thank you * infinity