Let hat{a} and hat{b} be two unit vectors suc | vedclass

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(C) Given $|(\hat{a}+\hat{b})+2(\hat{a} 	imes \hat{b})|=2$. Since $(\hat{a}+\hat{b}) \perp (\hat{a} 	imes \hat{b})$,we have $|\hat{a}+\hat{b}|^2 + 4

Vedclass · Accepted Answer

Both $(S_{1})$ and $(S_{2})$ are true

Vedclass · Answer

(C) Given $|(\hat{a}+\hat{b})+2(\hat{a} 	imes \hat{b})|=2$. Since $(\hat{a}+\hat{b}) \perp (\hat{a} 	imes \hat{b})$,we have $|\hat{a}+\hat{b}|^2 + 4|\hat{a} 	imes \hat{b}|^2 = 4$.Using $|\hat{a}+\hat{b}|^2 = 2+2\cos	heta$ and $|\hat{a} 	imes \hat{b}|^2 = \sin^2	heta = 1-\cos^2	heta$,we get:$2+2\cos	heta + 4(1-\cos^2	heta) = 4$$2+2\cos	heta + 4 - 4\cos^2	heta = 4$$4\cos^2	heta - 2\cos	heta - 2 = 0 \implies 2\cos^2	heta - \cos	heta - 1 = 0$.$(2\cos	heta+1)(\cos	heta-1) = 0$.Since $	heta \in (0, \pi)$,$\cos	heta = -\frac{1}{2}$,so $	heta = \frac{2\pi}{3}$.For $(S_{1})$: $2|\hat{a} 	imes \hat{b}| = 2\sin(\frac{2\pi}{3}) = 2(\frac{\sqrt{3}}{2}) = \sqrt{3}$.$|\hat{a}-\hat{b}| = \sqrt{1+1-2\cos(\frac{2\pi}{3})} = \sqrt{2-2(-\frac{1}{2})} = \sqrt{3}$. Thus $(S_{1})$ is true.For $(S_{2})$: Projection of $\hat{a}$ on $(\hat{a}+\hat{b})$ is $\frac{\hat{a} \cdot (\hat{a}+\hat{b})}{|\hat{a}+\hat{b}|} = \frac{1+\cos	heta}{\sqrt{2+2\cos	heta}} = \frac{1-\frac{1}{2}}{\sqrt{2-1}} = \frac{1/2}{1} = \frac{1}{2}$. Thus $(S_{2})$ is true.

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