Question

Two vectors $\overrightarrow{ A }$ and $\overrightarrow{ B }$ have equal magnitudes. If magnitudes of $\overrightarrow{ A }+\overrightarrow{ B }$ is equal to $n$ times the magnitude of $\overrightarrow{ A }-\overrightarrow{ B },$ then the angle between $\overrightarrow{ A }$ and $\overrightarrow{ B }$ is:

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

Answer 1

Answer: (C)

According to the question
$|\overrightarrow{ A }+\overrightarrow{ B }|= n |\overrightarrow{ A }-\overrightarrow{ B }|$
Squaring both side
$A^{2}+B^{2}+2 A B \cos \theta=n^{2}\left(A^{2}+B^{2}-2 A B \cos \theta\right)$
$A^{2}+B^{2}+2 A B \cos \theta=n^{2} A^{2}+n^{2} B^{2}-n^{2} \times 2 A B \cos \theta$
$2 A^{2}+2 A^{2} \cos \theta=2 n^{2} A^{2}-2 n^{2} A^{2} \cos \theta$
$2 A^{2}(1+\cos \theta)=2 n^{2} A^{2}(1-\cos \theta)$
$1+\cos \theta=n^{2}-n^{2} \cos \theta|\vec{A}+\vec{B}|=\sqrt{A^{2}+B^{2}+2 A B \cos \theta}$
$1+\cos \theta+n^{2} \cos \theta=n^{2} \quad|\vec{A}-\vec{B}|=\sqrt{A^{2} B^{2}-2 A B \cos \theta}$
$\cos \theta=\frac{n^{2}-1}{n^{2}+1} \quad|\vec{A}|=|\vec{B}|$
$\theta=\cos ^{-1}\left(\frac{n^{2}-1}{n^{2}+1}\right)$