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

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(D) The scalar triple product is defined as $[\bar{a} - 2\bar{b}, \bar{b} - 3\bar{c}, \bar{c} - 4\bar{a}] = ((\bar{a} - 2\bar{b}) 	imes (\bar{b} - 3\

Vedclass · Accepted Answer

$-345$

Vedclass · Answer

(D) The scalar triple product is defined as $[\bar{a} - 2\bar{b}, \bar{b} - 3\bar{c}, \bar{c} - 4\bar{a}] = ((\bar{a} - 2\bar{b}) 	imes (\bar{b} - 3\bar{c})) \cdot (\bar{c} - 4\bar{a})$.Expanding the cross product: $(\bar{a} 	imes \bar{b} - 3(\bar{a} 	imes \bar{c}) - 2(\bar{b} 	imes \bar{b}) + 6(\bar{b} 	imes \bar{c})) \cdot (\bar{c} - 4\bar{a})$.Since $\bar{b} 	imes \bar{b} = 0$,this simplifies to $(\bar{a} 	imes \bar{b} - 3(\bar{a} 	imes \bar{c}) + 6(\bar{b} 	imes \bar{c})) \cdot (\bar{c} - 4\bar{a})$.Distributing the dot product:$= (\bar{a} 	imes \bar{b}) \cdot \bar{c} - 4(\bar{a} 	imes \bar{b}) \cdot \bar{a} - 3(\bar{a} 	imes \bar{c}) \cdot \bar{c} + 12(\bar{a} 	imes \bar{c}) \cdot \bar{a} + 6(\bar{b} 	imes \bar{c}) \cdot \bar{c} - 24(\bar{b} 	imes \bar{c}) \cdot \bar{a}$.Since the scalar triple product is zero if any two vectors are the same,terms like $(\bar{a} 	imes \bar{b}) \cdot \bar{a}$ and $(\bar{a} 	imes \bar{c}) \cdot \bar{c}$ vanish.$= [\bar{a}, \bar{b}, \bar{c}] - 24[\bar{b}, \bar{c}, \bar{a}] = [\bar{a}, \bar{b}, \bar{c}] - 24[\bar{a}, \bar{b}, \bar{c}] = -23[\bar{a}, \bar{b}, \bar{c}]$.Given $\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular,$[\bar{a}, \bar{b}, \bar{c}] = |\bar{a}| |\bar{b}| |\bar{c}| = 1 	imes 3 	imes 5 = 15$.Therefore,$-23 	imes 15 = -345$.

If $\bar{a}, \bar{b},$ and $\bar{c}$ are mutually perpendicular vectors such that $|\bar{a}| = 1, |\bar{b}| = 3,$ and $|\bar{c}| = 5,$ then find the value of $[\bar{a} - 2\bar{b}, \bar{b} - 3\bar{c}, \bar{c} - 4\bar{a}]$.

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