Cross Product

Cross Product

Definition 1. Let $A, B$ be vectors. The cross product of $A$ and $B$ is defined as $A \times B = (|A||B| \sin \theta) \hat{\mathbf{e}}_c$ where $\hat{\mathbf{e}}_c$ is the unit vector to be perpendicular to the plane of $A$ and $B$, such that $A, B$, and $C$ form a right-handed system.

Remark. Let $A, B \in \mathbb{R}^3$. Then $$C_i = \sum_{j, k} \varepsilon_{ijk} A_j B_k$$ equivalent to $$C = \begin{vmatrix} \hat{\mathbf{e}}_x & \hat{\mathbf{e}}_y & \hat{\mathbf{e}}_z \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}$$

Scalar Triple Product

Theorem 1. Let $A, B$, and $C$ be vectors in $\mathbb{R}^3$. Then $$A \cdot (B \times C) = \begin{vmatrix} A_x & A_y & A_z \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{vmatrix}.$$

Remark. $$A \cdot B \times C = A \times B \cdot C.$$

Vector Triple Product

Theorem 1. Let $A, B$, and $C$ be vectors in $\mathbb{R}^3$. Then $$A \times (B \times C) = B(A \cdot C) - C(A \cdot B).$$

Proof. $$[A \times (B \times C)]_i = \sum_{j, k} \varepsilon_{ijk} A_j (B \times C)_k \\ = \sum_{j, k} \varepsilon_{ijk} A_j (\sum_{l, m} \varepsilon_{klm} B_l C_m) \\ = \sum_{j, k, l, m} \varepsilon_{ijk} \varepsilon_{lmk} A_j B_l C_m \\ = \sum_{j, l, m} (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}) A_j B_l C_m \\ = \sum_{j, m} [(\delta_{ii} \delta_{jm} - \delta_{im} \delta_{ji}) A_j B_i C_m + (\delta_{ij} \delta_{jm} - \delta_{im} \delta_{jj}) A_j B_j C_m].$$ If $i = j$, then $$\sum_{j, m} [(\delta_{ii} \delta_{jm} - \delta_{im} \delta_{ji}) A_j B_i C_m + (\delta_{ij} \delta_{jm} - \delta_{im} \delta_{jj}) A_j B_j C_m] \\ = \sum_{j, m} [\delta_{im} - \delta_{im}) A_i B_i C_m + (\delta_{im} - \delta_{im}) A_i B_i C_m] = 0$$ If $i \neq j$, then $$\sum_{j, m} [(\delta_{ii} \delta_{jm} - \delta_{im} \delta_{ji}) A_j B_i C_m + (\delta_{ij} \delta_{jm} - \delta_{im} \delta_{jj}) A_j B_j C_m] \\ = \sum_{j, m} (\delta_{jm} A_j B_i C_m - \delta_{im} A_j B_j C_m) = \sum_{j} (A_j B_i C_j - A_j B_j C_i) \\ = B_i \sum_{j} A_j C_j - C_i \sum_j A_j B_j = B_i (A \cdot C) - C_i (A \cdot B). \blacksquare$$