Cross Product in Arbitrary Non Standard Basis

1. Just like Inner Product, the Cross Product of 2 Vectors can be calculated for Vectors given in Any Arbitrary Non Standard Basis.
2. The following demonstrates the calculation of Cross Product for 2 3-Dimensional Vectors \(\vec{A}\) and \(\vec{B}\) represented in any Arbitrary Non Standard Basis \(\vec{e_1}\), \(\vec{e_2}\) and \(\vec{e_3}\)
   
   \(\vec{A}=A_1\vec{e_1} + A_2\vec{e_2} + A_3\vec{e_3}\hspace{.5cm}\vec{B}=B_1\vec{e_1} + B_2\vec{e_2} + B_3\vec{e_3}\)
   
   \(\vec{A} \times \vec{B} =(A_1\vec{e_1} + A_2\vec{e_2} + A_3\vec{e_3}) \times (B_1\vec{e_1} + B_2\vec{e_2} + B_3\vec{e_3})\)
   
   \(\Rightarrow \vec{A} \times \vec{B} =A_1B_1(\vec{e_1} \times \vec{e_1}) + A_1B_2(\vec{e_1} \times \vec{e_2}) + A_1B_3(\vec{e_1} \times \vec{e_3}) + A_2B_1(\vec{e_2} \times \vec{e_1}) + A_2B_2(\vec{e_2} \times \vec{e_2}) + A_2B_3(\vec{e_2} \times \vec{e_3}) + A_3B_1(\vec{e_3} \times \vec{e_1}) + A_3B_2(\vec{e_3} \times \vec{e_2}) + A_3B_3(\vec{e_3} \times \vec{e_3}) \)
   
   Since \(\vec{e_1} \times \vec{e_1}=\vec{e_2} \times \vec{e_2}=\vec{e_3} \times \vec{e_3}=0 \),   \(\vec{e_2} \times \vec{e_1}=-\vec{e_1} \times \vec{e_2}\),   \(\vec{e_3} \times \vec{e_2}=-\vec{e_2} \times \vec{e_3}\)   and   \(\vec{e_3} \times \vec{e_1}=-\vec{e_1} \times \vec{e_3}\),   the above equation can be written as
   
   \(\vec{A} \times \vec{B} =A_1B_2(\vec{e_1} \times \vec{e_2}) - A_2B_1(\vec{e_1} \times \vec{e_2}) + A_2B_3(\vec{e_2} \times \vec{e_3}) - A_3B_2(\vec{e_2} \times \vec{e_3}) + A_1B_3(\vec{e_1} \times \vec{e_3}) - A_3B_1(\vec{e_1} \times \vec{e_3})\)
   
   \(\Rightarrow \vec{A} \times \vec{B}=(A_1B_2- A_2B_1)(\vec{e_1} \times \vec{e_2}) + (A_2B_3 - A_3B_2)(\vec{e_2} \times \vec{e_3}) + (A_1B_3 - A_3B_1)(\vec{e_1} \times \vec{e_3})\)
   
   Setting \(\vec{e_4}=\vec{e_1} \times \vec{e_2}\),   \(\vec{e_5}=\vec{e_2} \times \vec{e_3}\),   \(\vec{e_6}=\vec{e_1} \times \vec{e_3}\) we get
   
   \(\vec{A} \times \vec{B}=(A_1B_2- A_2B_1)\vec{e_4} + (A_2B_3 - A_3B_2)\vec{e_5} + (A_1B_3 - A_3B_1)\vec{e_6}\)
3. Following examples demonstrates the calculation of the Cross 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
   
   \(\vec{e_1}=\begin{bmatrix}2\\-2\\4\end{bmatrix}\hspace{.5cm}\vec{e_2}=\begin{bmatrix}1\\-1\\-2\end{bmatrix}\hspace{.5cm}\vec{e_3}=\begin{bmatrix}6\\-3\\3\end{bmatrix}\)
   
   \(\vec{A}=1\vec{e_1} + 2\vec{e_2} + 3\vec{e_3}=1\begin{bmatrix}2\\-2\\4\end{bmatrix} + 2\begin{bmatrix}1\\-1\\-2\end{bmatrix} + 3\begin{bmatrix}6\\-3\\3\end{bmatrix}\)
   
   \(\vec{B}=7\vec{e_1} + 8\vec{e_2} + 9\vec{e_3}=7\begin{bmatrix}2\\-2\\4\end{bmatrix} + 8\begin{bmatrix}1\\-1\\-2\end{bmatrix} + 9\begin{bmatrix}6\\-3\\3\end{bmatrix}\)
   
   \(\vec{A} \times \vec{B} =(1\vec{e_1} + 2\vec{e_2} + 3\vec{e_3}) \times (7\vec{e_1} + 8\vec{e_2} + 9\vec{e_3})\)
   
   \(\Rightarrow \vec{A} \times \vec{B}=(1 \cdot 8 - 2 \cdot 7)(\vec{e_1} \times \vec{e_2}) + (2 \cdot 9 - 3 \cdot 8)(\vec{e_2} \times \vec{e_3}) + (1 \cdot 9 - 3 \cdot 7)(\vec{e_1} \times \vec{e_3})\)
   
   \(\Rightarrow \vec{A} \times \vec{B}=-6(\vec{e_1} \times \vec{e_2}) - 6(\vec{e_2} \times \vec{e_3}) - 12(\vec{e_1} \times \vec{e_3})\)
   
   Now,
   
   \(\vec{e_4}=\vec{e_1} \times \vec{e_2}=\begin{vmatrix}\mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}}\\2& -2 & 4 \\ 1 & -1 & -2\end{vmatrix}=8\mathbf{\hat{i}} + 8\mathbf{\hat{j}} + 0\mathbf{\hat{k}}=\begin{bmatrix}8\\8\\0\end{bmatrix}\)
   
   \(\vec{e_5}=\vec{e_2} \times \vec{e_3}=\begin{vmatrix}\mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}}\\1& -1 & -2 \\ 6 & -3 & 3\end{vmatrix}=-9\mathbf{\hat{i}} - 15\mathbf{\hat{j}} + 3\mathbf{\hat{k}}=\begin{bmatrix}-9\\-15\\3\end{bmatrix}\)
   
   \(\vec{e_6}=\vec{e_1} \times \vec{e_3}=\begin{vmatrix}\mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}}\\2& -2 & 4 \\ 6 & -3 & 3\end{vmatrix}=6\mathbf{\hat{i}} + 18\mathbf{\hat{j}} + 6\mathbf{\hat{k}}=\begin{bmatrix}6\\18\\6\end{bmatrix}\)
   
   Therefore,
   
   \( \vec{A} \times \vec{B}=-6\vec{e_4} - 6\vec{e_5} - 12\vec{e_6}=-6\begin{bmatrix}8\\8\\0\end{bmatrix} - 6\begin{bmatrix}-9\\-15\\3\end{bmatrix} - 12\begin{bmatrix}6\\18\\6\end{bmatrix}\)