Magnitudes of vectors vec a, vec b, vec c are | 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 $\vec c \cdot (\vec a + \vec b) = 0$.This implies:$(i) \q

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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 $\vec c \cdot (\vec a + \vec b) = 0$.This implies:$(i) \quad \vec a \cdot \vec b + \vec a \cdot \vec c = 0$$(ii) \quad \vec b \cdot \vec c + \vec b \cdot \vec a = 0$$(iii) \quad \vec c \cdot \vec a + \vec c \cdot \vec b = 0$Adding these three equations,we get $2(\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a) = 0$,which means $\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a = 0$.Now,$|\vec a + \vec b + \vec c|^2 = |\vec a|^2 + |\vec b|^2 + |\vec c|^2 + 2(\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a)$.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 find the magnitude of $|\vec a + \vec b + \vec c|$.

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