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

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(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors,we have $|\vec{a}|=1, |\vec{b}|=2, |\vec{c}|=3$ and $\vec{a} \cdot \vec{

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$12$

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(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors,we have $|\vec{a}|=1, |\vec{b}|=2, |\vec{c}|=3$ and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$.The scalar triple product is defined as $[\vec{x} \quad \vec{y} \quad \vec{z}] = (\vec{x} 	imes \vec{y}) \cdot \vec{z}$.We need to evaluate $[\vec{a}+\vec{b}+\vec{c} \quad \vec{b}-\vec{a} \quad \vec{c}] = ((\vec{a}+\vec{b}+\vec{c}) 	imes (\vec{b}-\vec{a})) \cdot \vec{c}$.Expanding the cross product: $(\vec{a}+\vec{b}+\vec{c}) 	imes (\vec{b}-\vec{a}) = \vec{a} 	imes \vec{b} - \vec{a} 	imes \vec{a} + \vec{b} 	imes \vec{b} - \vec{b} 	imes \vec{a} + \vec{c} 	imes \vec{b} - \vec{c} 	imes \vec{a}$.Since $\vec{x} 	imes \vec{x} = 0$ and $\vec{b} 	imes \vec{a} = -(\vec{a} 	imes \vec{b})$,we get $\vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{b} + \vec{c} 	imes \vec{b} - \vec{c} 	imes \vec{a} = 2(\vec{a} 	imes \vec{b}) + \vec{c} 	imes \vec{b} - \vec{c} 	imes \vec{a}$.Now,taking the dot product with $\vec{c}$: $(2(\vec{a} 	imes \vec{b}) + \vec{c} 	imes \vec{b} - \vec{c} 	imes \vec{a}) \cdot \vec{c} = 2(\vec{a} 	imes \vec{b}) \cdot \vec{c} + (\vec{c} 	imes \vec{b}) \cdot \vec{c} - (\vec{c} 	imes \vec{a}) \cdot \vec{c}$.Since the scalar triple product is zero if any two vectors are the same,$(\vec{c} 	imes \vec{b}) \cdot \vec{c} = 0$ and $(\vec{c} 	imes \vec{a}) \cdot \vec{c} = 0$.Thus,the expression simplifies to $2(\vec{a} 	imes \vec{b}) \cdot \vec{c} = 2[\vec{a} \quad \vec{b} \quad \vec{c}]$.Since $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular,$[\vec{a} \quad \vec{b} \quad \vec{c}] = |\vec{a}| |\vec{b}| |\vec{c}| = 1 	imes 2 	imes 3 = 6$.Therefore,the result is $2 	imes 6 = 12$.

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

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