bar{a}, bar{b}, bar{c} are vectors such that | vedclass

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(D) Given that $\bar{a} \cdot (\bar{b} + \bar{c}) = 0$,$\bar{b} \cdot (\bar{c} + \bar{a}) = 0$,and $\bar{c} \cdot (\bar{a} + \bar{b}) = 0$. Expanding

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$50$

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(D) Given that $\bar{a} \cdot (\bar{b} + \bar{c}) = 0$,$\bar{b} \cdot (\bar{c} + \bar{a}) = 0$,and $\bar{c} \cdot (\bar{a} + \bar{b}) = 0$. Expanding these,we get: $\bar{a} \cdot \bar{b} + \bar{a} \cdot \bar{c} = 0$ . . . $(1)$ $\bar{b} \cdot \bar{c} + \bar{b} \cdot \bar{a} = 0$ . . . $(2)$ $\bar{c} \cdot \bar{a} + \bar{c} \cdot \bar{b} = 0$ . . . $(3)$ Adding equations $(1)$,$(2)$,and $(3)$,we get: $2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) = 0$. We know that $|\bar{a} + \bar{b} + \bar{c}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a})$. Substituting the given values $|\bar{a}| = 5, |\bar{b}| = 4, |\bar{c}| = 3$ and the sum of dot products as $0$: $|\bar{a} + \bar{b} + \bar{c}|^2 = (5)^2 + (4)^2 + (3)^2 + 0 = 25 + 16 + 9 = 50$.

$\bar{a}, \bar{b}, \bar{c}$ are vectors such that $|\bar{a}| = 5, |\bar{b}| = 4, |\bar{c}| = 3$ and each is perpendicular to the sum of the other two,then $|\bar{a} + \bar{b} + \bar{c}|^2 = $

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