The volume of the parallelepiped determined b | vedclass

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(D) The volume of a parallelepiped determined by vectors $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $|[\vec{u} \quad \vec{v} \q

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) The volume of a parallelepiped determined by vectors $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $|[\vec{u} \quad \vec{v} \quad \vec{w}]|$.Given $[\vec{a}+\vec{b} \quad \vec{b}+\vec{c} \quad \vec{c}+\vec{a}] = 4$.We know that $[\vec{a}+\vec{b} \quad \vec{b}+\vec{c} \quad \vec{c}+\vec{a}] = 2[\vec{a} \quad \vec{b} \quad \vec{c}]$.So,$2[\vec{a} \quad \vec{b} \quad \vec{c}] = 4 \Rightarrow [\vec{a} \quad \vec{b} \quad \vec{c}] = 2$.The volume of the parallelepiped determined by $\vec{a} 	imes \vec{b}, \vec{b} 	imes \vec{c}$ and $\vec{c} 	imes \vec{a}$ is given by $[\vec{a} 	imes \vec{b} \quad \vec{b} 	imes \vec{c} \quad \vec{c} 	imes \vec{a}]$.Using the property $[\vec{a} 	imes \vec{b} \quad \vec{b} 	imes \vec{c} \quad \vec{c} 	imes \vec{a}] = [\vec{a} \quad \vec{b} \quad \vec{c}]^2$.Substituting the value,we get $2^2 = 4$.

The volume of the parallelepiped determined by the vectors $\vec{a} + \vec{b}, \vec{b} + \vec{c}$ and $\vec{c} + \vec{a}$ is $4$. Then the volume of the parallelepiped determined by the vectors $\vec{a} \times \vec{b}, \vec{b} \times \vec{c}$ and $\vec{c} \times \vec{a}$ is:

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