For any three vectors vec{a}, vec{b} and vec{ | vedclass

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(D) 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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$-\frac{29}{2}$

Vedclass · Answer

(D) Given that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$.Squaring both sides,we get $(\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c}) = \vec{0} \cdot \vec{0}$.This expands to $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.Substituting the given magnitudes $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=2$:$3^2 + 4^2 + 2^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.$9 + 16 + 4 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.$29 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.Therefore,$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{29}{2}$.

For any three vectors $\vec{a}, \vec{b}$ and $\vec{c}$,if $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=2$,then $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} = $ . . . . . . .

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