Magnitudes of vectors vec{a}, vec{b}, vec{c} | vedclass

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

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$5 \sqrt{2}$

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(C) Given that $\vec{a} \cdot (\vec{b} + \vec{c}) = 0$,$\vec{b} \cdot (\vec{c} + \vec{a}) = 0$,and $\vec{c} \cdot (\vec{a} + \vec{b}) = 0$. Expanding these,we get: $(i)$ $\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0$ (ii) $\vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} = 0$ (iii) $\vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0$ Adding these three equations: $2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = 0$. Now,we calculate the magnitude squared of the sum: $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$. Substituting the given magnitudes $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=5$: $|\vec{a} + \vec{b} + \vec{c}|^2 = 3^2 + 4^2 + 5^2 + 2(0) = 9 + 16 + 25 = 50$. Therefore,$|\vec{a} + \vec{b} + \vec{c}| = \sqrt{50} = 5\sqrt{2}$.

Magnitudes of vectors $\vec{a}, \vec{b}, \vec{c}$ are $3, 4, 5$ respectively. If $\vec{a}$ and $\vec{b} + \vec{c}$,$\vec{b}$ and $\vec{c} + \vec{a}$,and $\vec{c}$ and $\vec{a} + \vec{b}$ are mutually perpendicular,then the magnitude of $\vec{a} + \vec{b} + \vec{c}$ is:

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