Two vectors vec A and vec B have equal magnit | vedclass

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

Vedclass · Answer

(A) Let the magnitude of vectors be $|\vec A| = |\vec B| = A$.Given that $|\vec A + \vec B| = n |\vec A - \vec B|$.Squaring both sides,we get $|\vec A + \vec B|^2 = n^2 |\vec A - \vec B|^2$.Using the vector identity $|\vec A \pm \vec B|^2 = A^2 + B^2 \pm 2AB \cos 	heta$,we have:$A^2 + A^2 + 2A^2 \cos 	heta = n^2 (A^2 + A^2 - 2A^2 \cos 	heta)$.$2A^2 (1 + \cos 	heta) = n^2 [2A^2 (1 - \cos 	heta)]$.Dividing both sides by $2A^2$:$1 + \cos 	heta = n^2 (1 - \cos 	heta)$.$1 + \cos 	heta = n^2 - n^2 \cos 	heta$.$\cos 	heta (1 + n^2) = n^2 - 1$.$\cos 	heta = \frac{n^2 - 1}{n^2 + 1}$.Therefore,$	heta = \cos^{-1} \left[ \frac{n^2 - 1}{n^2 + 1} ight]$.

Two vectors $\vec A$ and $\vec B$ have equal magnitudes. The magnitude of $(\vec A + \vec B)$ is $n$ times the magnitude of $(\vec A - \vec B)$. The angle between $\vec A$ and $\vec B$ is

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