Let vec{a}, vec{b}, vec{c} be three unit vect | vedclass

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(A) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.Consider the magnitude of the sum of the vector

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$\frac{1}{2}$

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(A) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.Consider the magnitude of the sum of the vectors: $|\vec{a} - \vec{b} + \vec{c}|^2 \ge 0$.$|\vec{a} - \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 - 2\vec{a} \cdot \vec{b} - 2\vec{b} \cdot \vec{c} + 2\vec{a} \cdot \vec{c} = 1 + 1 + 1 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} - \vec{a} \cdot \vec{c}) = 3 - 2(\frac{3}{2}) = 3 - 3 = 0$.Since $|\vec{a} - \vec{b} + \vec{c}|^2 = 0$,we have $\vec{a} - \vec{b} + \vec{c} = \vec{0}$,which implies $\vec{b} = \vec{a} + \vec{c}$.Squaring both sides: $|\vec{b}|^2 = |\vec{a} + \vec{c}|^2 \implies 1 = |\vec{a}|^2 + |\vec{c}|^2 + 2\vec{a} \cdot \vec{c} \implies 1 = 1 + 1 + 2\vec{a} \cdot \vec{c}$.Thus,$2\vec{a} \cdot \vec{c} = -1$,so $\vec{a} \cdot \vec{c} = -\frac{1}{2}$.Now,substitute this into the given equation: $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} = \frac{3}{2} + \vec{a} \cdot \vec{c} = \frac{3}{2} - \frac{1}{2} = 1$.Finally,$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = 1 + (-\frac{1}{2}) = \frac{1}{2}$.

Let $\vec{a}, \vec{b}, \vec{c}$ be three unit vectors such that $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} - \vec{a} \cdot \vec{c} = \frac{3}{2}$. Then the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ is:

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