The vector triple product

For three vectors ${\bf a}$, ${\bf b}$, and ${\bf c}$, the vector triple product is defined ${\bf a}\times({\bf b} \times {\bf c})$. The brackets are important because ${\bf a}\times({\bf b} \times {\bf c}) \neq ({\bf a}\times{\bf b} )\times {\bf c}$. In fact, it can be demonstrated that

$${\bf a}\times({\bf b} \times {\bf c}) \equiv ({\bf a}\cdot {\bf c}) {\bf b} - ({\bf a}\cdot {\bf b}) {\bf c}$$ (51)

and

$$({\bf a}\times{\bf b}) \times {\bf c}\equiv ({\bf a}\cdot {\bf c}) {\bf b} - ({\bf b}\cdot {\bf c}) {\bf a}.$$ (52)

Let us try to prove the first of the above theorems. The left-hand side and the right-hand side are both proper vectors, so if we can prove this result in one particular coordinate system then it must be true in general. Let us take convenient axes such that the $x$-axis lies along ${\bf b}$, and ${\bf c}$ lies in the $x$-$y$ plane. It follows that ${\bf b} = (b_x, 0, 0)$, ${\bf c} = (c_x, c_y, 0)$, and ${\bf a} = (a_x, a_y, a_z)$. The vector ${\bf b}\times {\bf c}$ is directed along the $z$-axis: ${\bf b}\times{\bf c} = (0, 0, b_x c_y)$. It follows that ${\bf a}\times({\bf b} \times {\bf c})$ lies in the $x$-$y$ plane: ${\bf a}\times({\bf b}\times{\bf c}) = (a_y b_x c_y, -a_x b_x c_y, 0)$. This is the left-hand side of Eq. (51) in our convenient axes. To evaluate the right-hand side, we need ${\bf a}\cdot {\bf c} = a_x c_x + a_y c_y$ and ${\bf a}\cdot {\bf b} = a_x b_x$. It follows that the right-hand side is

$${\rm RHS} = ( [a_x c_x + a_y c_y] b_x, 0, 0) - (a_x b_x c_x, a_x b_x c_y, 0)$$

$$= (a_y c_y b_x, -a_x b_x c_y, 0 ) = {\rm LHS},$$ (53)

which proves the theorem.