If bar{a}, bar{b}, bar{c} are mutually perpen | vedclass

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(A) Given that $\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular vectors,we have $\bar{a} \cdot \bar{b} = 0, \bar{b} \cdot \bar{c} = 0, \bar{c} \

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$12$

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(A) Given that $\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular vectors,we have $\bar{a} \cdot \bar{b} = 0, \bar{b} \cdot \bar{c} = 0, \bar{c} \cdot \bar{a} = 0$ and magnitudes $|\bar{a}|=1, |\bar{b}|=2, |\bar{c}|=3$. The scalar triple product is defined as $[\bar{a}+\bar{b}+\bar{c} \quad \bar{b}-\bar{a} \quad \bar{c}] = (\bar{a}+\bar{b}+\bar{c}) \cdot ((\bar{b}-\bar{a}) 	imes \bar{c})$. Expanding the cross product: $(\bar{b}-\bar{a}) 	imes \bar{c} = (\bar{b} 	imes \bar{c}) - (\bar{a} 	imes \bar{c})$. Now,the dot product becomes: $(\bar{a}+\bar{b}+\bar{c}) \cdot ((\bar{b} 	imes \bar{c}) - (\bar{a} 	imes \bar{c}))$. Since $\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular,$\bar{b} 	imes \bar{c}$ is parallel to $\bar{a}$ and $\bar{a} 	imes \bar{c}$ is parallel to $\bar{b}$. Specifically,$\bar{a} \cdot (\bar{b} 	imes \bar{c}) = [\bar{a} \bar{b} \bar{c}]$ and $\bar{b} \cdot (\bar{a} 	imes \bar{c}) = -[\bar{a} \bar{b} \bar{c}]$. The expression simplifies to: $\bar{a} \cdot (\bar{b} 	imes \bar{c}) - \bar{a} \cdot (\bar{a} 	imes \bar{c}) + \bar{b} \cdot (\bar{b} 	imes \bar{c}) - \bar{b} \cdot (\bar{a} 	imes \bar{c}) + \bar{c} \cdot (\bar{b} 	imes \bar{c}) - \bar{c} \cdot (\bar{a} 	imes \bar{c})$. Terms like $\bar{a} \cdot (\bar{a} 	imes \bar{c}) = 0$ and $\bar{c} \cdot (\bar{b} 	imes \bar{c}) = 0$. Thus,we are left with $\bar{a} \cdot (\bar{b} 	imes \bar{c}) - \bar{b} \cdot (\bar{a} 	imes \bar{c}) = [\bar{a} \bar{b} \bar{c}] - (-[\bar{a} \bar{b} \bar{c}]) = 2[\bar{a} \bar{b} \bar{c}]$. Since they are mutually perpendicular,$[\bar{a} \bar{b} \bar{c}] = |\bar{a}| |\bar{b}| |\bar{c}| = 1 	imes 2 	imes 3 = 6$. Therefore,$2 	imes 6 = 12$.

If $\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular vectors having magnitudes $1, 2, 3$ respectively,then $[\bar{a}+\bar{b}+\bar{c} \quad \bar{b}-\bar{a} \quad \bar{c}] = $

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