A) \[{{V}^{3}}\]
B) \[3V\]
C) \[{{V}^{2}}\]
D) \[2V\]
Correct Answer: A
Solution :
We have, \[|[\overrightarrow{\mathbf{a}}\overrightarrow{\mathbf{b}}\overrightarrow{\mathbf{c}}]|\,=V\] Let\[{{V}_{1}}\]be the volume of parallelepiped formed by the vectors\[\vec{\alpha },\,\,\vec{\beta }\]and\[\vec{\gamma }\]. Then, \[{{V}_{1}}=|[\overset{\to }{\mathop{\alpha }}\,\overset{\to }{\mathop{\beta }}\,\overset{\to }{\mathop{\gamma }}\,]|\] \[\Rightarrow \] \[{{V}_{1}}=\left| \begin{matrix} \mathbf{\vec{a}}\cdot \mathbf{\vec{a}} & \mathbf{\vec{a}}\cdot \mathbf{\vec{b}} & \mathbf{\vec{a}}\cdot \mathbf{\vec{c}} \\ \mathbf{\vec{a}}\cdot \mathbf{\vec{b}} & \mathbf{\vec{b}}\cdot \mathbf{\vec{b}} & \mathbf{\vec{b}}\cdot \mathbf{\vec{c}} \\ \mathbf{\vec{a}}\cdot \mathbf{\vec{c}} & \mathbf{\vec{b}}\cdot \mathbf{\vec{c}} & \mathbf{\vec{c}}\cdot \mathbf{\vec{c}} \\ \end{matrix} \right|[\mathbf{\vec{a}\vec{b}\vec{c}}]\] \[\Rightarrow \] \[{{V}_{1}}={{[\mathbf{\vec{a}\vec{b}\vec{c}}]}^{2}}[\mathbf{\vec{a}\vec{b}\vec{c}}]\] \[\Rightarrow \] \[{{V}_{1}}={{[\mathbf{\vec{a}\vec{b}\vec{c}}]}^{3}}\] \[\Rightarrow \,\,\,\,\,\,\,\,{{V}_{1}}={{V}^{3}}\]
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