Question

two vectors A & B have equal magnitude if magnitude of A + B is n times the magnitude of A-B then what will be the angle between vector A & B

Answer 1

You can further solve if you want more simplified value.

Answer 2

Answer:  

The angle between the ‘two vectors’ is  

$\bold{\theta=\cos ^{-1}\left(\frac{n^{2}-1}{n^{2}+1}\right)}$

Solution:

Let the magnitude of A+B be represented by $\bold{\mathrm{R}_{1}}$ and the ‘magnitude’ of A-B be represented as $\bold{\mathrm{R}_{2}}$.

The resultant $\bold{\mathrm{R}_{1}}$ is  

$R_{1}=\sqrt{A^{2}+A^{2}+2 A^{2} \cos \theta} \rightarrow(1)$

The resultant $\bold{\mathrm{R}_{2}}$ is  

$R_{2}=\sqrt{A^{2}+A^{2}-2 A^{2} \cos \theta} \rightarrow(2)$

It is given that

$R_{1}=n R_{2}$

$\sqrt{2 A^{2}+2 A^{2} \cos \theta}=n \times \sqrt{2 A^{2}-2 A^{2} \cos \theta}$

Squaring on both sides, we get

$2 A^{2}(1+\cos \theta)=n^{2} \times 2 A^{2}(1-\cos \theta)$

$(1+\cos \theta)=n^{2} \times(1-\cos \theta)$

$(1+\cos \theta)=n^{2}-n^{2} \cos \theta$

$\cos \theta+n^{2} \cos \theta=n^{2}-1$

$\cos \theta\left(1+n^{2}\right)=n^{2}-1$

$\cos \theta=\frac{n^{2}-1}{n^{2}+1}$

Thus,

$\theta=\cos ^{-1}\left(\frac{n^{2}-1}{n^{2}+1}\right)$

Thus the angle between the two vectors is $\bold{\theta=\cos ^{-1}\left(\frac{n^{2}-1}{n^{2}+1}\right)}$.