Let bar{a}, bar{b}, bar{c} be three vectors s | vedclass

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(B) Given that $\bar{a}+\bar{b}+\bar{c}=\bar{0}$.Taking the dot product of the sum with itself: $(\bar{a}+\bar{b}+\bar{c}) \cdot (\bar{a}+\bar{b}+\bar

Vedclass · Accepted Answer

$-25$

Vedclass · Answer

(B) Given that $\bar{a}+\bar{b}+\bar{c}=\bar{0}$.Taking the dot product of the sum with itself: $(\bar{a}+\bar{b}+\bar{c}) \cdot (\bar{a}+\bar{b}+\bar{c}) = \bar{0} \cdot \bar{0} = 0$.Expanding the dot product: $|\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) = 0$.Substitute the given magnitudes: $3^2 + 4^2 + 5^2 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) = 0$.Calculate the squares: $9 + 16 + 25 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) = 0$.$50 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) = 0$.$2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) = -50$.Therefore,$\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a} = -25$.

Let $\bar{a}, \bar{b}, \bar{c}$ be three vectors such that $\bar{a}+\bar{b}+\bar{c}=\bar{0}$,$|\bar{a}|=3$,$|\bar{b}|=4$,and $|\bar{c}|=5$. Then,find the value of $\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}$.

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