Volume of a parallelepiped with coterminous e | vedclass

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(A) The volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $[\vec{a} \vec{b} \vec{c}]

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 12$.The volume of a tetrahedron with coterminous edges $\vec{u}, \vec{v}, \vec{w}$ is given by $V = \frac{1}{6} |[\vec{u} \vec{v} \vec{w}]|$.Here,$\vec{u} = \vec{a} - \vec{b}$,$\vec{v} = \vec{b} - \vec{c}$,and $\vec{w} = \vec{a} + \vec{b} - \vec{c}$.Calculating the scalar triple product $[\vec{u} \vec{v} \vec{w}] = [(\vec{a} - \vec{b}) (\vec{b} - \vec{c}) (\vec{a} + \vec{b} - \vec{c})]$.Using the properties of the scalar triple product,this simplifies to:$[\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{a} + \vec{b} - \vec{c}] = [\vec{a} \vec{b} \vec{c}] = 12$.Therefore,the volume of the tetrahedron is $\frac{1}{6} 	imes 12 = 2$ cubic units.

Volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is $12$ cubic units. The volume of a tetrahedron with coterminous edges $\vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{a} + \vec{b} - \vec{c}$ will be ............. cubic units.

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