If vec{A}, vec{B} and vec{C} are vectors havi | vedclass

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(B) Given that $\vec{A} + \vec{B} + \vec{C} = \vec{0}$.Squaring both sides,we get:$(\vec{A} + \vec{B} + \vec{C}) \cdot (\vec{A} + \vec{B} + \vec{C}) =

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$-1.5$

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(B) Given that $\vec{A} + \vec{B} + \vec{C} = \vec{0}$.Squaring both sides,we get:$(\vec{A} + \vec{B} + \vec{C}) \cdot (\vec{A} + \vec{B} + \vec{C}) = 0$$|\vec{A}|^2 + |\vec{B}|^2 + |\vec{C}|^2 + 2(\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A}) = 0$Since the vectors have unit magnitude,$|\vec{A}| = |\vec{B}| = |\vec{C}| = 1$.Substituting these values:$1^2 + 1^2 + 1^2 + 2(\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A}) = 0$$3 + 2(\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A}) = 0$$2(\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A}) = -3$$\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A} = -1.5$

If $\vec{A}, \vec{B}$ and $\vec{C}$ are vectors having unit magnitude. If $\vec{A} + \vec{B} + \vec{C} = \vec{0}$,then $\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A}$ will be:

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