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

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) Given that $\overline{a}, \overline{b}, \overline{c}$ are mutually perpendicular vectors,we have $\overline{a} \cdot \overline{b} = 0, \overline{b} \cdot \overline{c} = 0, \overline{c} \cdot \overline{a} = 0$ and magnitudes $|\overline{a}|=1, |\overline{b}|=2, |\overline{c}|=3$. The scalar triple product is defined as $[\overline{a}+\overline{b}+\overline{c} \quad \overline{b}-\overline{a} \quad \overline{c}] = (\overline{a}+\overline{b}+\overline{c}) \cdot ((\overline{b}-\overline{a}) 	imes \overline{c})$. Expanding the cross product: $(\overline{b}-\overline{a}) 	imes \overline{c} = \overline{b} 	imes \overline{c} - \overline{a} 	imes \overline{c}$. Now,substitute this back: $(\overline{a}+\overline{b}+\overline{c}) \cdot (\overline{b} 	imes \overline{c} - \overline{a} 	imes \overline{c})$. Since $\overline{a}, \overline{b}, \overline{c}$ are mutually perpendicular,$\overline{a} \cdot (\overline{b} 	imes \overline{c}) = |\overline{a}| |\overline{b}| |\overline{c}| = 1 	imes 2 	imes 3 = 6$. Also,$\overline{b} \cdot (\overline{b} 	imes \overline{c}) = 0$ and $\overline{c} \cdot (\overline{b} 	imes \overline{c}) = 0$. Similarly,$\overline{a} \cdot (\overline{a} 	imes \overline{c}) = 0, \overline{b} \cdot (\overline{a} 	imes \overline{c}) = 0, \overline{c} \cdot (\overline{a} 	imes \overline{c}) = 0$. The expression simplifies to: $\overline{a} \cdot (\overline{b} 	imes \overline{c}) - \overline{a} \cdot (\overline{a} 	imes \overline{c}) + \overline{b} \cdot (\overline{b} 	imes \overline{c}) - \overline{b} \cdot (\overline{a} 	imes \overline{c}) + \overline{c} \cdot (\overline{b} 	imes \overline{c}) - \overline{c} \cdot (\overline{a} 	imes \overline{c})$. This reduces to $\overline{a} \cdot (\overline{b} 	imes \overline{c}) - (-\overline{b} \cdot (\overline{a} 	imes \overline{c})) = [\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{a} \overline{c}] = [\overline{a} \overline{b} \overline{c}] + [\overline{a} \overline{b} \overline{c}] = 2[\overline{a} \overline{b} \overline{c}]$. Since $[\overline{a} \overline{b} \overline{c}] = 1 	imes 2 	imes 3 = 6$,the final value is $2 	imes 6 = 12$.

If $\overline{a}, \overline{b}, \overline{c}$ are mutually perpendicular vectors having magnitudes $1, 2, 3$ respectively,then the value of $[\overline{a}+\overline{b}+\overline{c} \quad \overline{b}-\overline{a} \quad \overline{c}]$ is

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