Let vec{a}, vec{b}, vec{c} be three vectors s | vedclass

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(B) Given $\vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = \vec{c} \cdot \vec{c} = 5$. This implies $|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 5$. We

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$-\frac{5}{2}$

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(B) Given $\vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = \vec{c} \cdot \vec{c} = 5$. This implies $|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 5$. We know that $|\vec{x} + \vec{y} + \vec{z}|^2 = |\vec{x}|^2 + |\vec{y}|^2 + |\vec{z}|^2 + 2(\vec{x} \cdot \vec{y} + \vec{y} \cdot \vec{z} + \vec{z} \cdot \vec{x})$. Expanding each term: $|\vec{a} + \vec{b} - \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{c} - \vec{c} \cdot \vec{a}) = 15 + 2(\vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{c} - \vec{c} \cdot \vec{a})$. $|\vec{b} + \vec{c} - \vec{a}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + |\vec{a}|^2 + 2(\vec{b} \cdot \vec{c} - \vec{c} \cdot \vec{a} - \vec{a} \cdot \vec{b}) = 15 + 2(\vec{b} \cdot \vec{c} - \vec{c} \cdot \vec{a} - \vec{a} \cdot \vec{b})$. $|\vec{c} + \vec{a} - \vec{b}|^2 = |\vec{c}|^2 + |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{c} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{c}) = 15 + 2(\vec{c} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{c})$. Adding these three equations: $50 = 45 + 2(\vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{c} - \vec{c} \cdot \vec{a} + \vec{b} \cdot \vec{c} - \vec{c} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{c})$. $50 = 45 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$. $5 = -2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$. $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{5}{2}$.

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