Angle between equal-magnitude vectors when |A+B| is 100 times |A-B|

[abhikesbhat]

Homework Statement

Ok I did all the chapter 3 questions in Fundamentals of Physics, but I could not get 2 of them.

53. Two vectors A and B have precisely equal magnitudes. For the magnitude of A+B to be 100 times greater that the magnitude of A-B, what must be the angle between them?

54. Two vectors A and B have precisely equal magnitudes. For the magnitude of A+B to be n times greater that the magnitude of A-B, what must be the angle between them?

Homework Equations

Law of Cosines.

The Attempt at a Solution

I am pretty sure the angle has to be small for 53. I know that the magnitude of A is equal to the magnitude of B. Using the Law of Cosines I get cos=1-mag(A-B)/2A^2. For A+B I get cos=1-50mag(A-B)/A^2. 180-cos=cos. I don't know where to go after this. I messed around with the equations, but I can't ever get rid of A or B. The second one I have not tried yet, because if I can't the first, then how am I supposed to get the second. THANKS for the help!

 

[slider142]

Do you know the parallelogram geometric view of addition and subtraction of vectors? Note that in the case that all four sides are equal, you are dealing with a rhombus, for which the diagonals bisect each other in a right angle.

 

[abhikesbhat]

Oh got it so, that makes a right triangle with legs 50mag(A-B) and another leg mag(A-B)/2. aTan(1/100)=.5729. Doubling that i get 1.1458 degrees. Right?

For the second one I get aTan(1/n). Right?

Thanks slider142!

 

[slider142]

Yep. Good job!