A) \[{{\cos }^{-1}}\left[ \frac{{{n}^{2}}-1}{{{n}^{2}}+1} \right]\]
B) \[{{\cos }^{-1}}\left[ \frac{n-1}{n+1} \right]\]
C) \[si{{n}^{-1}}\left[ \frac{{{n}^{2}}-1}{{{n}^{2}}+1} \right]\]
D) \[si{{n}^{-1}}\left[ \frac{n-1}{n+1} \right]\]
Correct Answer: A
Solution :
| \[|\overrightarrow{A}+\overrightarrow{B}|=n|\overrightarrow{A}-\overrightarrow{B}|\] |
| \[\Rightarrow \] \[{{A}^{2}}+{{B}^{2}}+2AB\cos \theta \]\[={{n}^{2}}({{A}^{2}}+{{B}^{2}}-2ABcos\theta )\]\[\Rightarrow \] \[\cos \theta (1+{{n}^{2}})=\frac{2{{a}^{2}}({{n}^{2}}-1)}{2{{a}^{2}}}\] \[[A=B=a]\] |
| \[\cos \theta =\frac{{{n}^{2}}-1}{{{n}^{2}}+1}\] |
You need to login to perform this action.
You will be redirected in
3 sec
