Vector Triple Product Of Vectors

As has already been mentioned in the previous section, the vector product $\vec v$ of three vectors is  $\vec a,\;\;\vec b,\;\;\vec c$ defined as

$$\vec v = \vec a \times (\vec b \times \vec c)$$

The most important property that the vector triple product satisfies is this:

$$\boxed{\begin{align}&\vec v = \vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec c)\;\vec b - (\vec a \cdot \vec b)\vec c \hfill \\\\& \,\,\,\,\,\,\,\;\,{\text{I}}\,\,\,\,\,\,\,{\text{II}}\,\,\,\,{\text{III}}\,\,\,\,\,\,\,{\text{(I}} \cdot {\text{III)}}\,\,\,{\text{II -- }}\,{\text{(I}} \cdot {\text{II)III}}\hfill\\\end{align}}\qquad\qquad\qquad....\left( 1 \right)$$

Let us see how this property comes about. Since $\vec v$ is perpendicular to $(\vec b \times \vec c)$ (and $\vec a$ too), $\vec v$ must lie in the plane containing $\vec b$and $\vec c$ (convince yourself about this). We may $\vec b\,\,{\text{and}}\,\,\vec c$ assume and to be non-collinear (since if they are collinear, (1) automatically holds because both sides will be zero). Thus, $\vec b\,\,{\text{and}}\,\,\vec c$ form a basis of the plane in which they lie.

$$\vec v = \lambda \vec b + \mu \vec c\,\,for{\text{ }}some{\text{ }}\lambda ,\;\mu  \in \mathbb{R}\qquad\qquad\qquad...\left( 2 \right)$$

Now, since  $\vec v$ is perpendicular to $\vec a$ too, we have

$$\begin{align}&\quad\qquad\vec v \cdot \vec a = 0 \hfill \\\\& \Rightarrow  \quad (\lambda \vec b + \mu \vec c) \cdot \vec a = 0 \hfill \\\\& \Rightarrow \quad  \lambda (\vec a \cdot \vec b) + \mu (\vec a \cdot \vec c) = 0 \hfill \\\\&  \Rightarrow\quad   \frac{\lambda }{{(\vec a \cdot \vec c)}} = \frac{\mu }{{ - (\vec a \cdot \vec b)}} = l({\text{say}}) \hfill \\\\&\Rightarrow \quad   \lambda  = l(\vec a \cdot \vec c),\;\;\mu  =  - l(\vec a \cdot \vec b) \hfill \\ \end{align}$$

Using the values of $\lambda \,and\,\mu$ in (2), we have

$$\vec v = l\{ (\vec a \cdot \vec c)\vec b - (\vec a \cdot \vec b)\vec c\} \qquad\qquad\qquad ...\left( 3 \right)$$

The only task that now remains is to find out the value of l. This can be done by taking particular values of $\vec a,\,\,\vec b\,\,{\text{and}}\,\vec c$ ((3) must hold for all arbitrary vectors $\vec a,\vec b,\vec c$ ).

Let $\vec a = \hat i,\;\;\vec b = \hat i + \hat j\;\;{\text{and}}\;\;\vec c = \hat k$

Thus,

$$\begin{align}&\qquad\vec v = \vec a \times (\vec b \times \vec c) \hfill \\\\&\;\qquad= \hat i \times \left\{ {(\hat i + \hat j) \times \hat k} \right\} \hfill \\\\&\;\qquad = \hat i \times \left\{ { - \hat j + \hat i} \right\} \hfill \\\\&\;\qquad=  - \hat k \hfill \\\\&and\quad l\left\{ {(\vec a \cdot \vec c)\vec b - (\vec a \cdot \vec b)\vec c} \right\} = l\left\{ {0 - \hat k} \right\} \hfill \\\\&\;\qquad  =  - l\hat k \hfill \\ \end{align}$$

This gives l = 1. Thus, we see that the relation in (1) holds.

It should be obvious that the vector triple product is a non-associative operation. For example, $\vec a \times (\vec b \times \vec c)$ lies in the plane of $\vec b$ and $\vec c$ while  $(\vec a \times \vec b) \times \vec c = ( - \vec c) \times (\vec a \times \vec b)$ lies in the plane of $\vec a$ and $\vec b$ .

In general, therefore,

$$\vec a \times (\vec b \times \vec c) \ne (\vec a \times \vec b) \times \vec c$$