A) \[\left| \,\begin{matrix} \mathbf{a}\,\,.\,\,\mathbf{b} & \mathbf{a}\,\,.\,\,\mathbf{a} \\ \mathbf{b}\,\,.\,\,\mathbf{b} & \mathbf{b}\,\,.\,\,\mathbf{a} \\ \end{matrix}\, \right|\]
B) \[\left| \,\begin{matrix} \mathbf{a}\,\,.\,\,\mathbf{a} & \mathbf{a}\,\,.\,\,\mathbf{b} \\ \mathbf{b}\,\,.\,\,\mathbf{a} & \mathbf{b}\,\,.\,\,\mathbf{b} \\ \end{matrix}\, \right|\]
C) \[\left| \,\begin{matrix} \mathbf{a}\,\,.\,\,\mathbf{b} \\ \mathbf{b}\,\,.\,\,\mathbf{a} \\ \end{matrix}\, \right|\]
D) None of these
Correct Answer: B
Solution :
\[{{(\mathbf{a}\times \mathbf{b})}^{2}}={{a}^{2}}{{b}^{2}}-{{(\mathbf{a}\,.\,\mathbf{b})}^{2}}=\left| \,\begin{matrix} \mathbf{a}\,.\,\mathbf{a} & \mathbf{a}\,.\,\mathbf{b} \\ \mathbf{a}\,.\,\mathbf{b} & \mathbf{b}\,.\,\mathbf{b} \\ \end{matrix}\, \right|.\]You need to login to perform this action.
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