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(D) Given that $\bar{a}$ is perpendicular to $\bar{b}+\bar{c}$,$\bar{b}$ is perpendicular to $\bar{c}+\bar{a}$,and $\bar{c}$ is perpendicular to $\bar

Vedclass · Accepted Answer

$\sqrt{29}$

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(D) Given that $\bar{a}$ is perpendicular to $\bar{b}+\bar{c}$,$\bar{b}$ is perpendicular to $\bar{c}+\bar{a}$,and $\bar{c}$ is perpendicular to $\bar{a}+\bar{b}$.Therefore,we have:$\bar{a} \cdot (\bar{b} + \bar{c}) = 0 \implies \bar{a} \cdot \bar{b} + \bar{a} \cdot \bar{c} = 0 \quad (1)$$\bar{b} \cdot (\bar{c} + \bar{a}) = 0 \implies \bar{b} \cdot \bar{c} + \bar{b} \cdot \bar{a} = 0 \quad (2)$$\bar{c} \cdot (\bar{a} + \bar{b}) = 0 \implies \bar{c} \cdot \bar{a} + \bar{c} \cdot \bar{b} = 0 \quad (3)$Adding equations $(1), (2),$ and $(3)$,we get:$2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) = 0 \implies \bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a} = 0$.Now,consider the magnitude squared of the sum:$|\bar{a} + \bar{b} + \bar{c}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a})$.Substituting the given values $|\bar{a}|=2, |\bar{b}|=3, |\bar{c}|=4$ and the result $0$ for the dot product sum:$|\bar{a} + \bar{b} + \bar{c}|^2 = 2^2 + 3^2 + 4^2 + 2(0) = 4 + 9 + 16 = 29$.Thus,$|\bar{a} + \bar{b} + \bar{c}| = \sqrt{29}$.

If $\bar{a}, \bar{b}, \bar{c}$ are three vectors which are perpendicular to $\bar{b}+\bar{c}, \bar{c}+\bar{a}$,and $\bar{a}+\bar{b}$ respectively,such that $|\bar{a}|=2, |\bar{b}|=3, |\bar{c}|=4$,then $|\bar{a}+\bar{b}+\bar{c}|=$

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