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(B) Given that $|\overline{a}| = 2, |\overline{b}| = 3, |\overline{c}| = 4$.Since $\overline{a} \perp (\overline{b}+\overline{c})$,we have $\overline{

Vedclass · Accepted Answer

$\sqrt{29}$

Vedclass · Answer

(B) Given that $|\overline{a}| = 2, |\overline{b}| = 3, |\overline{c}| = 4$.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:$2(\overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a}) = 0 \Rightarrow \overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a} = 0$.Now,consider the magnitude squared of the sum:$|\overline{a}+\overline{b}+\overline{c}|^2 = |\overline{a}|^2 + |\overline{b}|^2 + |\overline{c}|^2 + 2(\overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a})$.Substituting the known values:$|\overline{a}+\overline{b}+\overline{c}|^2 = 2^2 + 3^2 + 4^2 + 2(0) = 4 + 9 + 16 = 29$.Therefore,$|\overline{a}+\overline{b}+\overline{c}| = \sqrt{29}$.

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