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Question
For any two vectors \[\vec{a} \text{ and } \vec{b}\] , prove that \[\left| \vec{a} \times \vec{b} \right|^2 = \begin{vmatrix}\vec{a} . \vec{a} & & \vec{a} . \vec{b} \\ \vec{b} . \vec{a} & & \vec{b} . \vec{b}\end{vmatrix}\]
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Solution
\[RHS=\begin{vmatrix}\vec{a} . \vec{a} & \vec{a} . \vec{b} \\ \vec{b} . \vec{a} & \vec{b} . \vec{b}\end{vmatrix}\]
\[ =\begin{vmatrix}\left| \vec{a} \right|^2 & \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta \\ \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta & \left| \vec{b} \right|^2\end{vmatrix}\]
\[ = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 - \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \cos^2 \theta\]
\[ = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \left( 1 - \cos^2 \theta \right)\]
\[ = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \sin^2 \theta\]
\[ = \left( \left| \vec{a} \right|\left| \vec{b} \right| \sin \theta \right)^2 \]
\[ = \left| \vec{a} \times \vec{b} \right|^2 \]
\[ = LHS\]
\[ \text{ Hence proved } . \]
