Let |vec{a}| = |vec{b}| = |vec{a} - vec{b}| = | vedclass

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(B) Given that $|\vec{a}| = 1$,$|\vec{b}| = 1$,and $|\vec{a} - \vec{b}| = 1$. We know the identity $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2

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$\frac{\pi}{3}$

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(B) Given that $|\vec{a}| = 1$,$|\vec{b}| = 1$,and $|\vec{a} - \vec{b}| = 1$. We know the identity $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$. Since $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta$,where $	heta$ is the angle between $\vec{a}$ and $\vec{b}$,we have: $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}| |\vec{b}| \cos 	heta$. Substituting the given values: $1^2 = 1^2 + 1^2 - 2(1)(1) \cos 	heta$ $1 = 1 + 1 - 2 \cos 	heta$ $1 = 2 - 2 \cos 	heta$ $2 \cos 	heta = 1$ $\cos 	heta = \frac{1}{2}$ Since $\cos 	heta = \frac{1}{2}$,the angle $	heta = \frac{\pi}{3}$.

Let $|\vec{a}| = |\vec{b}| = |\vec{a} - \vec{b}| = 1$,then the angle between $\vec{a}$ and $\vec{b}$ is:

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