Answer to Question #244559 in Physics for Asmar

Question

Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A + B to be one 

hundred times larger than the magnitude of A - B, what must be the angle between them?

Answer 1

The magnitude of A+B:

$R_1=\sqrt{(A+B\cos\theta)^2+(0+B\sin\theta)^2}=\\ =\sqrt{(A+B\cos\theta)^2+B^2\sin^2\theta}.$

The Magnitude of A-B:

$R_2=\sqrt{(A-B\cos\theta)^2+(0-B\sin\theta)^2}=\\ =\sqrt{(A-B\cos\theta)^2+B^2\sin^2\theta}.$

The ratio:

$\frac{R_1}{R_2}=100,\\\space\\ 100=\frac{\sqrt{(A+B\cos\theta)^2+B^2\sin^2\theta}}{\sqrt{(A-B\cos\theta)^2+B^2\sin^2\theta}},\\\space\\ 10^4=\frac{(A+B\cos\theta)^2+B^2\sin^2\theta}{(A-B\cos\theta)^2+B^2\sin^2\theta},\\\space\\ 10^4=\frac{A^2+B^2\cos^2\theta+2AB\cos\theta+B^2-B^2\cos^2\theta}{A^2+B^2\cos^2\theta-2AB\cos\theta+B^2-B^2\cos^2\theta},\\\space\\ 10^4=\frac{A^2+2AB\cos\theta+B^2}{A^2-2AB\cos\theta+B^2}.$

Since A and B have "precisely equal magnitudes," then, putting A=B=V:

$10^4=\frac{2V^2+2V^2\cos\theta}{2V^2-2V^2\cos\theta},\\\space\\ 10^4-10^4\cos\theta=1+\cos\theta,\\\space\\ \cos\theta=\frac{10^4-1}{10^4+1},\\\space\\ \theta=\arccos\frac{10^4-1}{10^4+1}=1.15°.$