The volume of the tetrahedron whose co-termin | vedclass

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(A) The volume of a tetrahedron with co-terminus edges $\bar{a}, \bar{b}, \bar{c}$ is given by $V = \frac{1}{6} |[\bar{a} \bar{b} \bar{c}]|$.Given $V

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$18$

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(A) The volume of a tetrahedron with co-terminus edges $\bar{a}, \bar{b}, \bar{c}$ is given by $V = \frac{1}{6} |[\bar{a} \bar{b} \bar{c}]|$.Given $V = 12$,we have $\frac{1}{6} |[\bar{a} \bar{b} \bar{c}]| = 12$,which implies $|[\bar{a} \bar{b} \bar{c}]| = 72$.The scalar triple product is defined as $[\bar{a} \bar{b} \bar{c}] = \bar{a} \cdot (\bar{b} 	imes \bar{c})$.The scalar projection of $\bar{a}$ on $\bar{b} 	imes \bar{c}$ is given by $\frac{\bar{a} \cdot (\bar{b} 	imes \bar{c})}{|\bar{b} 	imes \bar{c}|} = 4$.Let $X = |\bar{b} 	imes \bar{c}|$. Then $\frac{[\bar{a} \bar{b} \bar{c}]}{X} = 4$,so $[\bar{a} \bar{b} \bar{c}] = 4X$.Substituting this into the volume equation: $|4X| = 72$.Therefore,$4X = 72$,which gives $X = \frac{72}{4} = 18$.Thus,$|\bar{b} 	imes \bar{c}| = 18$.

The volume of the tetrahedron whose co-terminus edges are $\bar{a}, \bar{b}, \bar{c}$ is $12$ cubic units. If the scalar projection of $\bar{a}$ on $\bar{b} \times \bar{c}$ is $4$,then $|\bar{b} \times \bar{c}|=$

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