It is given that a, b, c are vectors of lengt | vedclass

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(D) Given: $|a|=6, |b|=8, |c|=10$.Also, $a \cdot (b+c) = 0$, $b \cdot (c+a) = 0$, and $c \cdot (a+b) = 0$.Expanding these, we get:$a \cdot b + a \cdot

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$10$

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(D) Given: $|a|=6, |b|=8, |c|=10$.Also, $a \cdot (b+c) = 0$, $b \cdot (c+a) = 0$, and $c \cdot (a+b) = 0$.Expanding these, we get:$a \cdot b + a \cdot c = 0$ $(i)$$b \cdot c + b \cdot a = 0$ (ii)$c \cdot a + c \cdot b = 0$ (iii)Adding $(i)$, (ii), and (iii):$2(a \cdot b + b \cdot c + c \cdot a) = 0 \implies a \cdot b + b \cdot c + c \cdot a = 0$.Now, $|a+b+c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a)$.Substituting the values:$|a+b+c|^2 = 6^2 + 8^2 + 10^2 + 2(0) = 36 + 64 + 100 = 200$.Therefore, $|a+b+c| = \sqrt{200} = 10 \sqrt{2}$.

It is given that $a, b, c$ are vectors of lengths $6, 8, 10$ respectively. If $a$ is perpendicular to $(b+c)$, $b$ is perpendicular to $(c+a)$, and $c$ is perpendicular to $(a+b)$, then the length of the vector $a+b+c$ is (in $\sqrt{2}$)

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