Two vectors `vec A and vecB` have equal magnitudes.If magnitude of `(vecA+vecB)` is equal to n times of the magnitude of `(vecA-vecB)` then the angle between `vecA and vecB` is :-

Options

A. `cos^(-1) ((n - 1)/(n + 1))`
B. `cos^(-1) ((n^(2) - 1)/(n^(2) + 1))`
C. `sin^(-1) ((n - 1)/(n + 1))`
D. `sin^(-1) ((n^(2) - 1)/(n^(2) + 1))`

Answer

B

Explanation

Let `theta` be the angle between `vec(A)` and `vec(B)`
`|vec(A) + vec(B)| = n|vec(A) - vec(B)| rArr sqrt(A^(2) + B^(2) + 2 AB cos theta)`
`= n sqrt(A^(2) + B^(2) + 2 AB cos (180 - theta))`
`|vec(A)| = |vec(B)| = A = B = x`
`2x^(2) ( 1 + cos theta) = n^(2) . 2x^(2) ( 1 - cos theta)`
`1 + cos theta = n^(2) . n^(2) cos theta`
`(1 + n^(2)) cos theta = n^(2) - 1`
`cos theta = (n^(2) - 1)/(n^(2) + 1)`
`theta = cos^(-1) ((n^(2) - 1)/(n^(2) + 1))`