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(B) Let the magnitude of vectors be $X = Y = A$.The magnitude of the difference vector is $|\overrightarrow{X} - \overrightarrow{Y}| = \sqrt{A^2 + A^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the magnitude of vectors be $X = Y = A$.The magnitude of the difference vector is $|\overrightarrow{X} - \overrightarrow{Y}| = \sqrt{A^2 + A^2 - 2A^2 \cos 	heta} = \sqrt{2A^2(1 - \cos 	heta)}$.The magnitude of the sum vector is $|\overrightarrow{X} + \overrightarrow{Y}| = \sqrt{A^2 + A^2 + 2A^2 \cos 	heta} = \sqrt{2A^2(1 + \cos 	heta)}$.According to the problem,$|\overrightarrow{X} - \overrightarrow{Y}| = n |\overrightarrow{X} + \overrightarrow{Y}|$.Substituting the expressions: $\sqrt{2A^2(1 - \cos 	heta)} = n \sqrt{2A^2(1 + \cos 	heta)}$.Squaring both sides: $2A^2(1 - \cos 	heta) = n^2 \cdot 2A^2(1 + \cos 	heta)$.$1 - \cos 	heta = n^2(1 + \cos 	heta)$.$1 - \cos 	heta = n^2 + n^2 \cos 	heta$.$1 - n^2 = \cos 	heta(1 + n^2)$.$\cos 	heta = \frac{1 - n^2}{1 + n^2} = \frac{n^2 - 1}{-n^2 - 1}$.Therefore,$	heta = \cos^{-1}\left(\frac{n^2 - 1}{-n^2 - 1}ight)$.

Two vectors $\overrightarrow{X}$ and $\overrightarrow{Y}$ have equal magnitude. The magnitude of $(\overrightarrow{X}-\overrightarrow{Y})$ is $n$ times the magnitude of $(\overrightarrow{X}+\overrightarrow{Y})$. The angle between $\overrightarrow{X}$ and $\overrightarrow{Y}$ is:

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