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(D) Given that $|\overline{A}|=3, |\overline{B}|=4, |\overline{C}|=5$. Since $\overline{A} \perp (\overline{B}+\overline{C})$,we have $\overline{A} \c

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$5 \sqrt{2}$

Vedclass · Answer

(D) Given that $|\overline{A}|=3, |\overline{B}|=4, |\overline{C}|=5$. Since $\overline{A} \perp (\overline{B}+\overline{C})$,we have $\overline{A} \cdot (\overline{B}+\overline{C}) = 0 \Rightarrow \overline{A} \cdot \overline{B} + \overline{A} \cdot \overline{C} = 0$. Since $\overline{B} \perp (\overline{C}+\overline{A})$,we have $\overline{B} \cdot (\overline{C}+\overline{A}) = 0 \Rightarrow \overline{B} \cdot \overline{C} + \overline{B} \cdot \overline{A} = 0$. Since $\overline{C} \perp (\overline{A}+\overline{B})$,we have $\overline{C} \cdot (\overline{A}+\overline{B}) = 0 \Rightarrow \overline{C} \cdot \overline{A} + \overline{C} \cdot \overline{B} = 0$. Adding these three equations,we get $2(\overline{A} \cdot \overline{B} + \overline{B} \cdot \overline{C} + \overline{C} \cdot \overline{A}) = 0$. Now,consider $|\overline{A}+\overline{B}+\overline{C}|^2 = |\overline{A}|^2 + |\overline{B}|^2 + |\overline{C}|^2 + 2(\overline{A} \cdot \overline{B} + \overline{B} \cdot \overline{C} + \overline{C} \cdot \overline{A})$. Substituting the values,$|\overline{A}+\overline{B}+\overline{C}|^2 = 3^2 + 4^2 + 5^2 + 0 = 9 + 16 + 25 = 50$. Therefore,$|\overline{A}+\overline{B}+\overline{C}| = \sqrt{50} = 5 \sqrt{2}$.

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