If |vec{a}| = 3, |vec{b}| = 4, |vec{c}| = 5 s | vedclass

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(A) Given that each vector is perpendicular to the sum of the other two:$\vec{a} \cdot (\vec{b} + \vec{c}) = 0 \implies \vec{a} \cdot \vec{b} + \vec{a

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$5\sqrt{2}$

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(A) Given that each vector is perpendicular to the sum of the other two:$\vec{a} \cdot (\vec{b} + \vec{c}) = 0 \implies \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0$$\vec{b} \cdot (\vec{a} + \vec{c}) = 0 \implies \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{c} = 0$$\vec{c} \cdot (\vec{a} + \vec{b}) = 0 \implies \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0$Adding these equations, we get $2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$, so $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = 0$.Now, $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.$|\vec{a} + \vec{b} + \vec{c}|^2 = 3^2 + 4^2 + 5^2 + 2(0) = 9 + 16 + 25 = 50$.Therefore, $|\vec{a} + \vec{b} + \vec{c}| = \sqrt{50} = 5\sqrt{2}$.

If $|\vec{a}| = 3, |\vec{b}| = 4, |\vec{c}| = 5$ such that each vector is perpendicular to the sum of the other two, then $|\vec{a} + \vec{b} + \vec{c}|$ is equal to

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