Cross Product between a Dyad and a Vector

1. For calculating Cross Product between a Dyad and a Vector the Order of the Dyad and Number of Elements in the Vector (i.e. Dimension of the Vector) must be 3. The result of Cross Product between a Dyad and a Vector is another Dyad.
2. Calculating Cross Product between a Dyad and a Vector is same as Multiplication of Dyad Matrix with the Skew-Symmetric Matrix Corresponding to Vector. Calculating Cross Product between a Vector and a Dyad is same as Multiplication of Skew-Symmetric Matrix Corresponding to Vector with the Dyad Matrix.
3. Given 3 Vectors \(\vec{A}\), \(\vec{B}\) and \(\vec{C}\), the following assosiative relations exist between Tensor Product (that results in a Dyad) and Cross Product
   
   \(\vec{C} \times (\vec{A} \otimes_{T} \vec{B})=\vec{C} \times \overleftrightarrow{AB} = (\vec{C} \times \vec{A}) \otimes_{T} \vec{B}\)
   
   \((\vec{A} \otimes_{T} \vec{B}) \times \vec{C} =\overleftrightarrow{AB} \times \vec{C}= \vec{A} \otimes_{T} (\vec{B} \times \vec{C})\)
   
   
4. The Cross Product between a Dyad and a Vector is Non-Commutative, that is for Any Non Identity Dyad/Dyadic \(\overleftrightarrow{AB}\) and Vector \(\vec{C}\)
   
   \(\overleftrightarrow{AB} \times \vec{C} \neq \vec{C} \times \overleftrightarrow{AB}\)
5. Following examples demonstrates the calculation Cross Product between a Dyad and a Vector and assosiative relations that exist between Tensor Product (that results in a Dyad) and Cross Product for Vectors \(\vec{A}\), \(\vec{B}\) and \(\vec{C}\) as given below
   
   \(\vec{A}=\begin{bmatrix}5\\1\\2\end{bmatrix}\hspace{.5cm}\vec{B}=\begin{bmatrix}-4\\-3\\2\end{bmatrix}\hspace{.5cm}\vec{C}=\begin{bmatrix}-17\\2\\-4\end{bmatrix}\)
   
   \(\overleftrightarrow{AB}=\vec{A} \otimes_{T} \vec{B}=\begin{bmatrix}5\\1\\2\end{bmatrix} \otimes_{T} \begin{bmatrix}-4 \\ -3 \\ 2\end{bmatrix}=\vec{A} \otimes \vec{B}^{T}=\begin{bmatrix}5\\1\\2\end{bmatrix} \otimes \begin{bmatrix}-4 & -3 & 2\end{bmatrix}=\begin{bmatrix}-20 & -15 & 10 \\ -4 & -3 & 2 \\ -8 & -6 & 4\end{bmatrix}\)
   
   \(\vec{C}_{\mathbf{\times}}=\begin{bmatrix}0 & 4 & 2 \\ -4 & 0 & 17 \\ -2 & -17 & 0\end{bmatrix}\)
   
   Now,
   
   \(\vec{C} \times (\vec{A} \otimes_{T} \vec{B})=\vec{C} \times \overleftrightarrow{AB}= [\vec{C}_{\mathbf{\times}}]\overleftrightarrow{AB}=\begin{bmatrix}0 & 4 & 2 \\ -4 & 0 & 17 \\ -2 & -17 & 0\end{bmatrix}\begin{bmatrix}-20 & -15 & 10 \\ -4 & -3 & 2 \\ -8 & -6 & 4\end{bmatrix}=\begin{bmatrix}-32 & -24 & 16 \\ -56 & -42 & 28 \\ 108 & 51 & -54\end{bmatrix}\)   ...(1)
   
   Also,
   
   \((\vec{C} \times \vec{A}) \otimes_{T} \vec{B}=(\begin{bmatrix}-17\\2\\-4\end{bmatrix} \times \begin{bmatrix}5\\1\\2\end{bmatrix})\otimes_{T}\begin{bmatrix}-4\\-3\\2\end{bmatrix}=([\vec{C}_{\mathbf{\times}}]\vec{A}) \otimes \vec{B}^{T}=(\begin{bmatrix}0 & 4 & 2 \\ -4 & 0 & 17 \\ -2 & -17 & 0\end{bmatrix}\begin{bmatrix}5\\1\\2\end{bmatrix})\otimes \begin{bmatrix}-4&-3&2\end{bmatrix}= \begin{bmatrix}8 \\ 14 \\ -27\end{bmatrix} \otimes \begin{bmatrix}-4&-3&2\end{bmatrix} =\begin{bmatrix}-32 & -24 & 16 \\ -56 & -42 & 28 \\ 108 & 51 & -54\end{bmatrix}\)   ...(2)
   
   From equations (1) and (2) we get
   
   \(\vec{C} \times (\vec{A} \otimes_{T} \vec{B})=\vec{C} \times \overleftrightarrow{AB} = (\vec{C} \times \vec{A}) \otimes_{T} \vec{B}\)
   
   Similarly,
   
   \((\vec{A} \otimes_{T} \vec{B}) \times \vec{C} =\overleftrightarrow{AB} \times \vec{C} = \overleftrightarrow{AB}[\vec{C}_{\mathbf{\times}}]=\begin{bmatrix}-20 & -15 & 10 \\ -4 & -3 & 2 \\ -8 & -6 & 4\end{bmatrix}\begin{bmatrix}0 & 4 & 2 \\ -4 & 0 & 17 \\ -2 & -17 & 0\end{bmatrix}=\begin{bmatrix}40 & -250 & -295 \\ 8 & -50 & -59 \\ 16 & -100 & -118\end{bmatrix}\)   ...(3)
   
   Also,
   
   \(\vec{A} \otimes_{T} (\vec{B}\times \vec{C})=\begin{bmatrix}5\\1\\2\end{bmatrix} \otimes_{T} (\begin{bmatrix}-4\\-3\\2\end{bmatrix}\times \begin{bmatrix}-17\\2\\-4\end{bmatrix})=\vec{A} \otimes {(-[\vec{C}_{\mathbf{\times}}]\vec{B})}^{T}=\begin{bmatrix}5\\1\\2\end{bmatrix} \otimes {(\begin{bmatrix}0 & -4 & -2 \\ 4 & 0 & -17 \\ 2 & 17 & 0\end{bmatrix}\begin{bmatrix}-4\\-3\\2\end{bmatrix})}^{T}=\begin{bmatrix}5\\1\\2\end{bmatrix} \otimes \begin{bmatrix}8 & -50 & -59\end{bmatrix}=\begin{bmatrix}40 & -250 & -295 \\ 8 & -50 & -59 \\ 16 & -100 & -118\end{bmatrix}\)   ...(4)
   
   From equations (3) and (4) we get
   
   \((\vec{A} \otimes_{T} \vec{B}) \times \vec{C} =\overleftrightarrow{AB} \times \vec{C}= \vec{A} \otimes_{T} (\vec{B} \times \vec{C})\)