a, b and c are three vectors with magnitudes | vedclass

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(B) Given: $|a| = 4, |b| = 4, |c| = 2$.Since $a \perp (b + c)$,we have $a \cdot (b + c) = 0 \Rightarrow a \cdot b + a \cdot c = 0$ $(i)$.Since $b \per

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$6$

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(B) Given: $|a| = 4, |b| = 4, |c| = 2$.Since $a \perp (b + c)$,we have $a \cdot (b + c) = 0 \Rightarrow a \cdot b + a \cdot c = 0$ $(i)$.Since $b \perp (c + a)$,we have $b \cdot (c + a) = 0 \Rightarrow b \cdot c + b \cdot a = 0$ $(ii)$.Since $c \perp (a + b)$,we have $c \cdot (a + b) = 0 \Rightarrow c \cdot a + c \cdot b = 0$ $(iii)$.Adding $(i), (ii),$ and $(iii)$,we get $2(a \cdot b + b \cdot c + c \cdot a) = 0$,which 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: $|a + b + c|^2 = 4^2 + 4^2 + 2^2 + 2(0) = 16 + 16 + 4 = 36$.Therefore,$|a + b + c| = \sqrt{36} = 6$.

$a, b$ and $c$ are three vectors with magnitudes $|a| = 4, |b| = 4, |c| = 2$ such that $a$ is perpendicular to $(b + c)$,$b$ is perpendicular to $(c + a)$,and $c$ is perpendicular to $(a + b)$. It follows that $|a + b + c|$ is equal to

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