If vec{a} and vec{b} are two unit vectors wit | vedclass

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(A) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$. Given $|\vec{a} - \vec{b}| = 1$,squaring both sides gi

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$3$

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(A) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$. Given $|\vec{a} - \vec{b}| = 1$,squaring both sides gives $|\vec{a} - \vec{b}|^2 = 1$. $|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} = 1$. $1 + 1 - 2\vec{a} \cdot \vec{b} = 1$,which implies $2\vec{a} \cdot \vec{b} = 1$,so $\vec{a} \cdot \vec{b} = \frac{1}{2}$. Now,$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 1 + 1 + 2 	imes \frac{1}{2} = 3$. Thus,$|\vec{a} + \vec{b}| = \sqrt{3}$. Since $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 	heta = \frac{1}{2}$,we have $\cos 	heta = \frac{1}{2}$. Using the identity $\cos 	heta = 2 \cos^2 \frac{	heta}{2} - 1$,we get $2 \cos^2 \frac{	heta}{2} - 1 = \frac{1}{2}$,so $2 \cos^2 \frac{	heta}{2} = \frac{3}{2}$,which means $\cos^2 \frac{	heta}{2} = \frac{3}{4}$. Therefore,$\cos \frac{	heta}{2} = \frac{\sqrt{3}}{2}$. Finally,$2|\vec{a} + \vec{b}| \cos \frac{	heta}{2} = 2 	imes \sqrt{3} 	imes \frac{\sqrt{3}}{2} = 3$.

If $\vec{a}$ and $\vec{b}$ are two unit vectors with $(\vec{a}, \vec{b}) = \theta$ and $|\vec{a} - \vec{b}| = 1$,then $2|\vec{a} + \vec{b}| \cos \frac{\theta}{2} =$

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