Question

Two vectors A and B have equal magnitudes. The magnitude of (A + B) is ‘n’ times the magnitude of (A – B). The angle between A and B is
1. \({\rm{si}}{{\rm{n}}^{ - 1}}\left( {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right)\)
2. \({\rm{si}}{{\rm{n}}^{ - 1}}\left( {\frac{{n - 1}}{{n + 1}}} \right)\)
3. \({\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right)\)
4. \({\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{n - 1}}{{n + 1}}} \right)\)

Answer 1

Correct Answer - Option 3 : \({\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right)\)

Concept:

Given,

Magnitude of A = Magnitude of B

|A| = |B|

Or A = B      ----(1)

Let magnitude of (A + B) is R and for (A – B) is R'.

Given, magnitude of R is n times the magnitude of R'

Thus, R = nR'

Now,

R = A + B

R² = A² + B² + 2AB cos θ

[∵using equation (1)]

R² = A² + A² + 2AA cos θ

R² = 2A² + 2A² cos θ      ----(2)

Again,

R' = A – B

\({\left( {R'} \right)^2} = {A^2} + {B^2} - 2AB\cos \theta\) 

[∵ using equation (1)]

\({\left( {R'} \right)^2} = {A^2} + {A^2} - 2AA\cos \theta\)

\({\left( {R'} \right)^2} = 2{A^2} - 2{A^2}\cos \theta\)       ----(3)

Calculation:

Given

R = nR'

\({\left( {\frac{R}{{R'}}} \right)^2} = {n^2}\) 

Dividing equation (ii) with equation (iii), we get

\({n^2} = \frac{{2{A^2} + 2{A^2}\cos \theta }}{{2{A^2} - 2{A^2}\cos \theta }}\) 

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

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

Thus,

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

The value of \(\frac{{{n^2} - 1}}{{{n^2} + 1}}\) will be,

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

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

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

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

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

The angle between A and B is \({\cos ^{ - 1}}\left( {\frac{{{n^2} - 1}}{{{n^2} + 1}}} \right)\;\)