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(C) The volume of a tetrahedron with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by $V_{tetra} = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]|$.G

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(C) The volume of a tetrahedron with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by $V_{tetra} = \frac{1}{6} |[\vec{a} \vec{b} \vec{c}]|$.Given $V_{tetra} = 3$,we have $\frac{1}{6} |[\vec{a} \vec{b} \vec{c}]| = 3$,which implies $|[\vec{a} \vec{b} \vec{c}]| = 18$.The volume of a parallelepiped with coterminous edges $\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}$ is given by the scalar triple product $|[(\vec{a} + \vec{b}) (\vec{b} + \vec{c}) (\vec{c} + \vec{a})]|$.Using the property of scalar triple products,$[(\vec{a} + \vec{b}) (\vec{b} + \vec{c}) (\vec{c} + \vec{a})] = 2 [\vec{a} \vec{b} \vec{c}]$.Therefore,the volume is $2 	imes 18 = 36$.

The volume of the tetrahedron formed by the coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is $3$. Then the volume of the parallelepiped formed by the coterminous edges $\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}$ is

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