The volume of a tetrahedron with coterminus e | vedclass

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

Vedclass · Accepted Answer

$256$

Vedclass · Answer

(C) The volume of a tetrahedron with coterminus edges $\bar{a}, \bar{b}, \bar{c}$ is given by $V_{tetra} = \frac{1}{6} |[\bar{a} \bar{b} \bar{c}]|$.Given $V_{tetra} = \frac{64}{3}$,we have $\frac{1}{6} |[\bar{a} \bar{b} \bar{c}]| = \frac{64}{3}$,which implies $|[\bar{a} \bar{b} \bar{c}]| = 128$.The volume of a parallelepiped with coterminus edges $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$ is given by the scalar triple product $|[(\bar{a}+\bar{b}) (\bar{b}+\bar{c}) (\bar{c}+\bar{a})]|$.Using the property of scalar triple products,$[(\bar{a}+\bar{b}) (\bar{b}+\bar{c}) (\bar{c}+\bar{a})] = 2 [\bar{a} \bar{b} \bar{c}]$.Thus,the volume is $|2 [\bar{a} \bar{b} \bar{c}]| = 2 	imes 128 = 256$ cubic units.

The volume of a tetrahedron with coterminus edges $\bar{a}, \bar{b}, \bar{c}$ is $\frac{64}{3}$ cubic units. Then,the volume of a parallelepiped with coterminus edges given by the vectors $\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}$ is ... cubic units.

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