vec{a}, vec{b}, text{ and } vec{c} are three | vedclass

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(B) Given,$|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$. Since $\vec{a} \perp (\vec{b}+\vec{c})$,we have $\vec{a} \cdot (\vec{b}+\vec{c}) = 0$,which implies

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$9$

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(B) Given,$|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$. Since $\vec{a} \perp (\vec{b}+\vec{c})$,we have $\vec{a} \cdot (\vec{b}+\vec{c}) = 0$,which implies $\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0 \dots (i)$. Similarly,$\vec{b} \cdot (\vec{c}+\vec{a}) = 0 \implies \vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{a} = 0 \dots (ii)$. And $\vec{c} \cdot (\vec{a}+\vec{b}) = 0 \implies \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0 \dots (iii)$. Solving these equations,we get $\vec{a} \cdot \vec{b} = 0, \vec{b} \cdot \vec{c} = 0, 	ext{ and } \vec{c} \cdot \vec{a} = 0$. 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 values,$|\vec{a}+\vec{b}+\vec{c}|^2 = 3^2 + 5^2 + 7^2 + 2(0 + 0 + 0) = 9 + 25 + 49 = 83$. Finally,$\sqrt{|\vec{a}+\vec{b}+\vec{c}|^2 - 2} = \sqrt{83 - 2} = \sqrt{81} = 9$.

$\vec{a}, \vec{b}, \text{ and } \vec{c}$ are three vectors such that $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$. If $\vec{a}, \vec{b}, \vec{c}$ are perpendicular to the vectors $\vec{b}+\vec{c}, \vec{c}+\vec{a}, \vec{a}+\vec{b}$ respectively,then $\sqrt{|\vec{a}+\vec{b}+\vec{c}|^2-2} = $

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