Cross Product of Vectors

1. Cross Product of Vectors is carried out Between 2 Vectors of 3 Dimensions each.
2. Cross Product of 2 Vectors is also known as Vector Products of Vectors as it generates another Vector.
3. Cross Product of a Vector with itself or between any 2 Parallel Vectors is a NULL Vector.
4. Cross Product of 2 Vectors is Anti-Commutative. That is, for any 2 Non-Parallel Vectors \(\vec{A}\) and \(\vec{B}\)
   
   \(\vec{A} \times \vec{B} = - \vec{B} \times \vec{A}\)
5. Resultant Vector of a Cross Product of 2 Vectors is Perpendicular to both the participating Vectors. This property of Cross Products can be used to find Mutually Orthogonal Basis Vector Sets in 3-Dimension.
6. In 3-Dimensional Cartesian Coordinate System any Vector is by default Directionally Represented on the Basis of the Identity Orthonormal Basis Vectors given by Unit Vectors \(\mathbf{\hat{i}}\), \(\mathbf{\hat{j}}\) and \(\mathbf{\hat{k}}\). Because of Perpendicular Result Vector property of Cross Products
   
   \(\hat{i} \times \hat{j} = \hat{k},\hspace{.5cm} \hat{j} \times \hat{k} = \hat{i},\hspace{.5cm} \hat{k} \times \hat{i} = \hat{j}\)
   
   Also, because of Anti Commutative property of Cross Product
   
   \(\hat{j} \times \hat{i} = -\hat{k},\hspace{.5cm} \hat{k} \times \hat{j} = -\hat{i},\hspace{.5cm} \hat{i} \times \hat{k} = -\hat{j}\)
   
   And, since Cross Product of a Vector with itself is Null Vector, therefore
   
   \(\hat{i} \times \hat{i} = 0,\hspace{.5cm} \hat{j} \times \hat{j} = 0,\hspace{.5cm} \hat{k} \times \hat{k} = 0\)
7. The following demonstrates the calculation of Cross Product for 2 3-Dimensional Vectors \(\vec{A}\) and \(\vec{B}\) Represented in Identity Orthonormal Basis in Cartesian Coordinate System
   
   \(\vec{A}=A_1\hat{i} + A_2\hat{j} + A_3\hat{k}\hspace{.5cm}\vec{B}=B_1\hat{i} + B_2\hat{j} + B_3\hat{k}\)
   
   \(\vec{A} \times \vec{B} =(A_1\hat{i} + A_2\hat{j} + A_3\hat{k}) \times (B_1\hat{i} + B_2\hat{j} + B_3\hat{k})\)
   
   \(\Rightarrow \vec{A} \times \vec{B} =A_1B_1(\hat{i} \times \hat{i}) + A_1B_2(\hat{i} \times \hat{j}) + A_1B_3(\hat{i} \times \hat{k}) + A_2B_1(\hat{j} \times \hat{i}) + A_2B_2(\hat{j} \times \hat{j}) + A_2B_3(\hat{j} \times \hat{k}) + A_3B_1(\hat{k} \times \hat{i}) + A_3B_2(\hat{k} \times \hat{j}) + A_3B_3(\hat{k} \times \hat{k}) \)
   
   Replacing the result of Cross Product of Unit Vectors \(\mathbf{\hat{i}}\), \(\mathbf{\hat{j}}\) and \(\mathbf{\hat{k}}\) in above equation we get
   
   \(\vec{A} \times \vec{B} =A_1B_2\hat{k} - A_1B_3\hat{j} - A_2B_1\hat{k} + A_2B_3\hat{i} + A_3B_1\hat{j} - A_3B_2\hat{i}\)
   
   \(\Rightarrow \vec{A} \times \vec{B} = (A_2B_3- A_3B_2)\hat{i} + (A_3B_1- A_1B_3)\hat{j} + (A_1B_2-A_2B_1)\hat{k}\)
   
   The calculation of Cross Product given above can be given in form of Determinant as the following
   
   \(\vec{A} \times \vec{B} =\begin{vmatrix}\mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}}\\A_1 & A_2 & A_3\\B_1 & B_2 & B_3\end{vmatrix}\)
   
   
8. The Cross Product \(\vec{A} \times \vec{B}\) between Vectors \(\vec{A}\) and \(\vec{B}\) (as defined above) can also be calculated as a Matrix Product between Skew-Symmetric Matrix Corresponding to Vector \(\vec{A}\) and Vector \(\vec{B}\). Similarly, the Cross Product \(\vec{B} \times \vec{A}\) between Vectors \(\vec{A}\) and \(\vec{B}\) can also be calculated as a Matrix Product between Skew-Symmetric Matrix Corresponding to Vector \(\vec{B}\) and Vector \(\vec{A}\). These are demonstrated below
   
   \(\vec{A} \times \vec{B} =[\vec{A}_{\mathbf{\times}}]\vec{B}\hspace{.8cm}\vec{B} \times \vec{A} =[\vec{B}_{\mathbf{\times}}]\vec{A}\)
   
   In the above equations \(\vec{A}_{\mathbf{\times}}\) and \(\vec{B}_{\mathbf{\times}}\) are Skew-Symmetric Matrices corresponding to Vectors \(\vec{A}\) and \(\vec{B}\) respectively and are given as
   
   \(\vec{A}_{\mathbf{\times}}=\begin{bmatrix}0 & -A_3 & A_2\\A_3 & 0 & -A_1\\-A_2 & A_1 & 0\end{bmatrix}\hspace{.5cm}\vec{B}_{\mathbf{\times}}=\begin{bmatrix}0 & -B_3 & B_2\\B_3 & 0 & -B_1\\-B_2 & B_1 & 0\end{bmatrix}\)
   
   Hence, the Matrix Product \([\vec{A}_{\mathbf{\times}}]\vec{B}\) for Cross Product \(\vec{A} \times \vec{B}\) is calculated as
   
   \(\vec{A} \times \vec{B}=[\vec{A}_{\mathbf{\times}}]\vec{B}=\begin{bmatrix}0 & -A_3 & A_2\\A_3 & 0 & -A_1\\-A_2 & A_1 & 0\end{bmatrix}\begin{bmatrix}B_1\\B_2\\B_3\end{bmatrix}=\begin{bmatrix}A_2B_3-A_3B_2\\A_3B_1-A_1B_3\\A_1B_2-A_2B_1\end{bmatrix}=(A_2B_3- A_3B_2)\hat{i} + (A_3B_1- A_1B_3)\hat{j} + (A_1B_2-A_2B_1)\hat{k}\)
   
   Similarly, the Matrix Product \([\vec{B}_{\mathbf{\times}}]\vec{A}\) for Cross Product \(\vec{B} \times \vec{A}\) is calculated as
   
   \(\vec{B} \times \vec{A}=[\vec{B}_{\mathbf{\times}}]\vec{A}=\begin{bmatrix}0 & -B_3 & B_2\\B_3 & 0 & -B_1\\-B_2 & B_1 & 0\end{bmatrix}\begin{bmatrix}A_1\\A_2\\A_3\end{bmatrix}=\begin{bmatrix}A_3B_2-A_2B_3\\A_1B_3-A_3B_1\\A_2B_1-A_1B_2\end{bmatrix}=(A_3B_2-A_2B_3)\hat{i} + (A_1B_3-A_3B_1)\hat{j} + (A_2B_1-A_1B_2)\hat{k}\)
   
   Also since \(\vec{A}_{\mathbf{\times}}\) and \(\vec{B}_{\mathbf{\times}}\) are Skew-Symmetric Matrices (i.e. their Negative is their Transpose), the following equations hold true
   
   \(\vec{A} \times \vec{B}= - \vec{B} \times \vec{A} = - [\vec{B}_{\mathbf{\times}}]\vec{A}= {[\vec{B}_{\mathbf{\times}}]}^T\vec{A}\)
   
   \(\vec{B} \times \vec{A}= - \vec{A} \times \vec{B} = - [\vec{A}_{\mathbf{\times}}]\vec{B}= {[\vec{A}_{\mathbf{\times}}]}^T\vec{B}\)
9. Following examples demonstrates the calculation of the Cross Product of Vectors \(\vec{A}\) and \(\vec{B}\) defined below using both Determinant Method and Matrix Product Method
   
   \(\vec{A}=-2\hat{i} + 5\hat{j} + 7\hat{k}\hspace{.5cm}\vec{B}=3\hat{i} - 6\hat{j} - 4\hat{k}\)
   
   Using Determinant Method the Cross Products are calculated as
   
   \(\vec{A} \times \vec{B} =\begin{vmatrix}\mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}}\\-2 & 5 & 7\\3 & -6 & -4\end{vmatrix}=(5\cdot(-4) - 7 \cdot (-6))\mathbf{\hat{i}} + (7 \cdot 3 - (-2) \cdot (-4))\mathbf{\hat{j}} + ((-2)\cdot(-6) - 5 \cdot 3)\mathbf{\hat{k}}=22\mathbf{\hat{i}} + 13\mathbf{\hat{j}} - 3\mathbf{\hat{k}}\)
   
   \(\vec{B} \times \vec{A} =\begin{vmatrix}\mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}}\\3 & -6 & -4\\-2 & 5 & 7\end{vmatrix}=(7 \cdot (-6) - 5\cdot(-4))\mathbf{\hat{i}} + ((-2) \cdot (-4) - 7 \cdot 3)\mathbf{\hat{j}} + ( 5 \cdot 3 - (-2)\cdot(-6))\mathbf{\hat{k}}=-22\mathbf{\hat{i}} - 13\mathbf{\hat{j}} + 3\mathbf{\hat{k}}\)
   
   Using Matrix Product Method the Cross Products are calculated as
   
   \(\vec{A} \times \vec{B} =[\vec{A}_{\mathbf{\times}}]\vec{B}=\begin{bmatrix}0 & -7 & 5\\7 & 0 & 2\\-5 & -2 & 0\end{bmatrix}\begin{bmatrix}3\\-6\\-4\end{bmatrix}=\begin{bmatrix}(-7) \cdot (-6) + 5 \cdot (-4)\\7 \cdot 3 + 2 \cdot (-4)\\(-5) \cdot 3 + (-2) \cdot (-6)\end{bmatrix}=\begin{bmatrix}22\\13\\-3\end{bmatrix}=22\mathbf{\hat{i}} + 13\mathbf{\hat{j}} - 3\mathbf{\hat{k}}\)
   
   \(\vec{B} \times \vec{A} =[\vec{B}_{\mathbf{\times}}]\vec{A} =\begin{bmatrix}0 & 4 & -6\\-4 & 0 & -3\\6 & 3 & 0\end{bmatrix}\begin{bmatrix}-2\\5\\7\end{bmatrix}=\begin{bmatrix}4 \cdot 5 + (-6)\cdot 7\\(-4) \cdot (-2) + (-3) \cdot 7 \\6 \cdot (-2) + 3 \cdot 5\end{bmatrix}=\begin{bmatrix}-22\\-13\\3\end{bmatrix}=-22\mathbf{\hat{i}} - 13\mathbf{\hat{j}} + 3\mathbf{\hat{k}}\)