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(B) Given that $\bar{a}$ and $\bar{b}$ are unit vectors,so $|\bar{a}| = 1$ and $|\bar{b}| = 1$. Consider the magnitude of the difference vector: $|\ba

Vedclass · Accepted Answer

$\frac{|\bar{a}-\bar{b}|}{|\bar{a}+\bar{b}|}$

Vedclass · Answer

(B) Given that $\bar{a}$ and $\bar{b}$ are unit vectors,so $|\bar{a}| = 1$ and $|\bar{b}| = 1$. Consider the magnitude of the difference vector: $|\bar{a} - \bar{b}|^2 = |\bar{a}|^2 + |\bar{b}|^2 - 2(\bar{a} \cdot \bar{b}) = 1 + 1 - 2(1)(1)\cos 	heta = 2 - 2\cos 	heta = 2(1 - \cos 	heta) = 2(2\sin^2(	heta/2)) = 4\sin^2(	heta/2)$. Thus,$|\bar{a} - \bar{b}| = 2\sin(	heta/2)$. Similarly,for the sum vector: $|\bar{a} + \bar{b}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + 2(\bar{a} \cdot \bar{b}) = 1 + 1 + 2(1)(1)\cos 	heta = 2 + 2\cos 	heta = 2(1 + \cos 	heta) = 2(2\cos^2(	heta/2)) = 4\cos^2(	heta/2)$. Thus,$|\bar{a} + \bar{b}| = 2\cos(	heta/2)$. Now,dividing the two magnitudes: $\frac{|\bar{a} - \bar{b}|}{|\bar{a} + \bar{b}|} = \frac{2\sin(	heta/2)}{2\cos(	heta/2)} = 	an(	heta/2)$. Therefore,$	an(	heta/2) = \frac{|\bar{a} - \bar{b}|}{|\bar{a} + \bar{b}|}$.

If $\bar{a}$ and $\bar{b}$ are unit vectors and $\theta$ is the angle between them,then $\tan(\theta/2) =$

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