Vector Triple Product
Here we consider three vectors \( {\bf A},\ {\bf B}\), and \({\bf C}\). How is “\({\bf A} \times ({\bf B} \times {\bf C})\)” calculated? Let me jump to the conclusion. We can obtain the following relational expressions: \begin{eqnarray} {\bf A} \times ( {\bf B} \times {\bf C} ) &=& ({\bf C}\cdot {\bf A}){\bf B} - ({\bf A}\cdot {\bf B}){\bf B} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\ ({\bf A} \times {\bf B}) \times {\bf C} &=& ({\bf A}\cdot {\bf C}){\bf B} - ({\bf B}\cdot {\bf C}){\bf A} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\\ {\bf A} \times ( {\bf B} \times {\bf C} ) + {\bf B} \times ( {\bf C} &\times &{\bf B} ) + {\bf C} \times ( {\bf A} \times {\bf B} ) = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \end{eqnarray} These equations are called the vector triple product. In particular, Equation (3) is called the Jacobi identity. Let me derive the vector triple product.We can derive Equations (2) and (3) by demonstrating Equation (1). Thus, we start with Equation (1). Here, we consider only the x component of Equation (1). The \(x\) component of the \({\bf A} \times {\bf B}\) is expressed as follows: \begin{eqnarray} & & A_y ({\bf B}\times {\bf C})_z - A_z ({\bf B}\times {\bf C})_y \\ &=& A_y (B_x C_y - B_y C_x) - A_z (B_zC_x - B_x C_z) \\ &=& A_y B_x C_y - A_y B_y C_x - A_z B_z C_x + A_z B_x C_z\ \ \ \ \ \ \ \ \ \ \ \ (4) \end{eqnarray} Moreover, we calculate the right-hand side of Equation (1): \begin{eqnarray} & & (A_x C_x + A_y C_y + A_z C_z)B_x - (A_x B_x + A_y B_y A_z B_z)C_x \\ &=& A_y B_x C_y - A_y B_y C_x - A_z B_z C_x + A_z B_x C_z\ \ \ \ \ \ \ \ \ \ \ \ (5) \end{eqnarray} Since Equation (4) is consistent with Equation (5), we demonstrate Equation (1). Equation (2) can also be derived similarly. The Jacobi identity (Equation (3)) can be derived using Equation (1). The vector triple product is frequently used in Plasma Physics, which considers Coulomb’s law and the Lorenz’s force. The Scalar triple vector product is described in another page.
Sponsored link