For vectors vec{a}, vec{b} and vec{c},if vec{ | 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}) = \vec{0} \cdot

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$-19$

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(B) Given that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$.Squaring both sides,we get $(\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c}) = \vec{0} \cdot \vec{0}$.Expanding the dot product,we have $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.Substituting the given magnitudes $|\vec{a}|=2, |\vec{b}|=3, |\vec{c}|=5$:$2^2 + 3^2 + 5^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.$4 + 9 + 25 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.$38 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -38$.Therefore,$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -19$.

For vectors $\vec{a}, \vec{b}$ and $\vec{c}$,if $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and $|\vec{a}|=2, |\vec{b}|=3, |\vec{c}|=5$,then the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ is . . . . . . .

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