Three vectors vec{a}, vec{b}, vec{c} satisfy | vedclass

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(C) 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

Vedclass · Accepted Answer

$-13$

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(C) 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 $|\vec{a}|=1, |\vec{b}|=3, |\vec{c}|=4$:$(1)^2+(3)^2+(4)^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = 0$.$1+9+16+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = 0$.$26+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = 0$.$2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = -26$.Therefore,$\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} = -13$.

Three vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy the condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. If $|\vec{a}|=1, |\vec{b}|=3, |\vec{c}|=4$,then find the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$.

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