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Wedge Product in Arbitrary Non Standard Basis

1. Just like any other kind of Product of Vectors Wedge Product of Vectors can be calculated for Vectors given in Any Arbitrary Non Standard Basis.
2. Following example demonstrates the calculation of the Wedge Product of Vectors \(\vec{A}\) and \(\vec{B}\) represented in Non Standard Basis \(\vec{e_1}\), \(\vec{e_2}\) and \(\vec{e_3}\) as give below
   
   \(\vec{e_1}=\begin{bmatrix}2\\-2\\4\end{bmatrix}\hspace{.5cm}\vec{e_2}=\begin{bmatrix}1\\-1\\-2\end{bmatrix}\hspace{.5cm}\vec{e_3}=\begin{bmatrix}6\\-3\\3\end{bmatrix}\)
   
   \(\vec{A}=2\vec{e_1} + 3\vec{e_2} + 7\vec{e_3}=2\begin{bmatrix}2\\-2\\4\end{bmatrix} + 3\begin{bmatrix}1\\-1\\-2\end{bmatrix} + 7\begin{bmatrix}6\\-3\\3\end{bmatrix}\)
   
   \(\vec{B}=-5\vec{e_1} + 1\vec{e_2} + -3\vec{e_3}=-5\begin{bmatrix}2\\-2\\4\end{bmatrix} + 1\begin{bmatrix}1\\-1\\-2\end{bmatrix} - 3\begin{bmatrix}6\\-3\\3\end{bmatrix}\)
   
   \(\vec{A}=2\vec{e_1} + 3\vec{e_2} + 7\vec{e_3},\hspace{.5cm}\vec{B}=-5\vec{e_1} + 1\vec{e_2} - 3\vec{e_3},\hspace{.5cm}\vec{A_3}=4\vec{e_1} + 2\vec{e_2} + 3\vec{e_3}\)
   
   \(\vec{A}\wedge\vec{B}=(2\vec{e_1} + 3\vec{e_2} + 7\vec{e_3}) \wedge (-5\vec{e_1} + 1\vec{e_2} - 3\vec{e_3})\)
   
   \(\Rightarrow \vec{A}\wedge\vec{B}=2(\vec{e_1}\wedge\vec{e_2}) - 6(\vec{e_1}\wedge\vec{e_3}) - 15(\vec{e_2}\wedge\vec{e_1}) - 9(\vec{e_2}\wedge\vec{e_3}) - 35(\vec{e_3}\wedge\vec{e_1}) + 7(\vec{e_3}\wedge\vec{e_2}) )\)
   
   \(\Rightarrow \vec{A}\wedge\vec{B}=17(\vec{e_1}\wedge\vec{e_2}) + 29(\vec{e_1}\wedge\vec{e_3}) - 16(\vec{e_2}\wedge\vec{e_3})\)
   
   Similarly \(\vec{e_1}\wedge\vec{e_2}\), \(\vec{e_1}\wedge\vec{e_3}\) and \(\vec{e_2}\wedge\vec{e_3}\) can be calculated as follows
   
   \(\vec{e_1}\wedge\vec{e_2}=(2\hat{e_1} - 2\hat{e_2} + 4\hat{e_3}) \wedge (1\hat{e_1} - 1\hat{e_2} + 2\hat{e_3})= 0(\hat{e_1}\wedge\hat{e_2}) + 0(\hat{e_1}\wedge\hat{e_3}) + 0(\hat{e_2}\wedge\hat{e_3})\)
   
   \(\vec{e_1}\wedge\vec{e_3}=(2\hat{e_1} - 2\hat{e_2} + 4\hat{e_3}) \wedge (6\hat{e_1} - 3\hat{e_2} + 3\hat{e_3})= 6(\hat{e_1}\wedge\hat{e_2}) - 18(\hat{e_1}\wedge\hat{e_3}) + 6(\hat{e_2}\wedge\hat{e_3})\)
   
   \(\vec{e_2}\wedge\vec{e_3}=(1\hat{e_1} - 1\hat{e_2} + 2\hat{e_3}) \wedge (6\hat{e_1} - 3\hat{e_2} + 3\hat{e_3})= 3(\hat{e_1}\wedge\hat{e_2}) - 9(\hat{e_1}\wedge\hat{e_3}) + 3(\hat{e_2}\wedge\hat{e_3})\)
   
   Hence \(\vec{A}\wedge\vec{B}\) can be given as follows
   
   \(\vec{A}\wedge\vec{B}=17(\vec{e_1}\wedge\vec{e_2}) + 29(\vec{e_1}\wedge\vec{e_3}) - 16(\vec{e_2}\wedge\vec{e_3})=17\begin{bmatrix}0\\0\\0\end{bmatrix} + 29\begin{bmatrix}6\\-18\\6\end{bmatrix} - 16\begin{bmatrix}3\\-9\\3\end{bmatrix}\)