a, b, c are three vectors such that |a|=3, |b | vedclass

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(B) Given that $|a|=3, |b|=5, |c|=7$. Since $a$ is perpendicular to $(b+c)$,we have $a \cdot (b+c) = 0 \implies a \cdot b + a \cdot c = 0$ ... $(i)$ S

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$9$

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(B) Given that $|a|=3, |b|=5, |c|=7$. Since $a$ is perpendicular to $(b+c)$,we have $a \cdot (b+c) = 0 \implies a \cdot b + a \cdot c = 0$ ... $(i)$ Since $b$ is perpendicular to $(c+a)$,we have $b \cdot (c+a) = 0 \implies b \cdot c + b \cdot a = 0$ ... $(ii)$ Since $c$ is perpendicular to $(a+b)$,we have $c \cdot (a+b) = 0 \implies c \cdot a + c \cdot b = 0$ ... $(iii)$ Adding equations $(i), (ii),$ and $(iii)$,we get: $2(a \cdot b + b \cdot c + c \cdot a) = 0 \implies a \cdot b + b \cdot c + c \cdot a = 0$. Now,we calculate $|a+b+c|^2$: $|a+b+c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a)$ $|a+b+c|^2 = (3)^2 + (5)^2 + (7)^2 + 2(0) = 9 + 25 + 49 = 83$. Finally,we evaluate the required expression: $\sqrt{(a+b+c)^2 - 2} = \sqrt{83 - 2} = \sqrt{81} = 9$.

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

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