3-1.Vectors
hard

બે સદિશ $\vec X$ અને $\vec Y$ સમાન માન ધરાવે છે. $(\vec X - \vec Y)$ નું માન એ $(\vec X + \vec Y)$ ના માન કરતા $n$ ગણું છે. $\vec X$ અને $\vec Y$ વચ્ચેનો ખૂણો કેટલો હશે?

A

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

B

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

C

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

D

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

(JEE MAIN-2021)

Solution

Given $X=Y$

$\sqrt{ X ^{2}+ Y ^{2}-2 \times Y \cos \theta}$

$= n \sqrt{ X ^{2}+ Y ^{2}+2 \times Y \cos \theta}$

Square both sides

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

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

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

$\theta=\cos ^{-1}\left[\frac{ n ^{2}-1}{- n ^{2}-1}\right]$

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.