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Question

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

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(C) The volume of a tetrahedron with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by $V_{tetrahedron} = \frac{1}{6} |[\vec{a} \, \vec{b} \, \vec{c}]|$.Given $V_{tetrahedron} = 3$,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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