13.0 Vector Triple Product

It is defined for the three vectors $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ and written as the vector $\overrightarrow a \times (\overrightarrow b \times \overrightarrow c )$. It is the cross product of one vector with the resultant of cross product of other two vectors.

This vector is perpendicular to $\overrightarrow b \times \overrightarrow c $ which means it is coplanar to $\overrightarrow b $ and $\overrightarrow c $ i.e., $$\overrightarrow a \times (\overrightarrow b \times \overrightarrow c ) = l\overrightarrow b + m\overrightarrow c ...(1)$$

Taking scalar product with $\overrightarrow a $ on both sides, we get $$0 = l\overrightarrow a .\overrightarrow b + m\overrightarrow a .\overrightarrow c $$ $$\begin{equation} \begin{aligned} 0 = l\vec a.\vec b + m\vec a.\vec c \\ - l\vec a.\vec b = m\vec a.\vec c \\ l = - m\frac{{\vec a.\vec c}}{{\vec a.\vec b}} \\\end{aligned} \end{equation} $$ Put the value of $l$ in equation $(1)$, we get $$\begin{equation} \begin{aligned} \vec a \times (\vec b \times \vec c) = \left( { - m\frac{{\vec a.\vec c}}{{\vec a.\vec b}}} \right)\vec b + m\vec c \\ \vec a \times (\vec b \times \vec c) = m\left[ { - \left( {\frac{{\vec a.\vec c}}{{\vec a.\vec b}}} \right)\vec b + \vec c} \right] \\ \vec a \times (\vec b \times \vec c) = m\left[ {\frac{{ - \vec b\left( {\vec a.\vec c} \right) + \vec c\left( {\vec a.\vec b} \right)}}{{\vec a.\vec b}}} \right] \\ \vec a \times (\vec b \times \vec c) = \frac{m}{{\vec a.\vec b}}\left[ { - \vec b\left( {\vec a.\vec c} \right) + \vec c\left( {\vec a.\vec b} \right)} \right] \\\end{aligned} \end{equation} $$

As we know that ${\vec a.\vec b}$ will give us a scalar value, we take it out of the bracket and we replace $\lambda = \frac{m}{{\vec a.\vec b}}$, where $\lambda$ is some constant value. We get $$\vec a \times (\vec b \times \vec c) = \lambda \left[ { - \vec b\left( {\vec a.\vec c} \right) + \vec c\left( {\vec a.\vec b} \right)} \right]...(2)$$

If we take vectors as $$\begin{equation} \begin{aligned} \overrightarrow a = {a_1}\widehat i \\ \overrightarrow b = {b_1}\widehat i + {b_2}\widehat j \\ \overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k \\\end{aligned} \end{equation} $$ We get $${a_1}(\widehat i) \times \left[ {\left( {{b_1}\hat i + {b_2}\hat j} \right) \times \left( {{c_1}\hat i + {c_2}\hat j + {c_3}\hat k} \right)} \right] = \lambda \left[ { - \left( {{b_1}\hat i + {b_2}\hat j} \right)\left\{ {\left( {{a_1}(\widehat i)} \right).\left( {{c_1}\hat i + {c_2}\hat j + {c_3}\hat k} \right)} \right\} + \left( {{c_1}\hat i + {c_2}\hat j + {c_3}\hat k} \right)\left\{ {\left( {{a_1}(\widehat i)} \right).\left( {{b_1}\hat i + {b_2}\hat j} \right)} \right\}} \right]$$ We solve left hand side i.e., $$\begin{equation} \begin{aligned} {a_1}(\widehat i) \times \left[ {\left( {{b_1}\hat i + {b_2}\hat j} \right) \times \left( {{c_1}\hat i + {c_2}\hat j + {c_3}\hat k} \right)} \right] = {a_1}(\widehat i) \times \left[ {{b_1}{c_2}\left( {\hat k} \right) - {b_1}{c_3}\left( {\hat j} \right) - {b_2}{c_1}\left( {\hat k} \right) + {b_2}{c_3}\left( {\hat i} \right)} \right] \\ {\text{ }} = \left( {\hat j} \right)\left[ {{a_1}{b_2}{c_1} - {a_1}{b_1}{c_2}} \right] - \left( {\hat k} \right)\left[ {{a_1}{b_1}{c_3}} \right] \\\end{aligned} \end{equation} $$ Now, solving the right hand side, i.e., $$\begin{equation} \begin{aligned} = \lambda \left[ { - \left( {{b_1}\hat i + {b_2}\hat j} \right)\left\{ {\left( {{a_1}(\widehat i)} \right).\left( {{c_1}\hat i + {c_2}\hat j + {c_3}\hat k} \right)} \right\} + \left( {{c_1}\hat i + {c_2}\hat j + {c_3}\hat k} \right)\left\{ {\left( {{a_1}(\widehat i)} \right).\left( {{b_1}\hat i + {b_2}\hat j} \right)} \right\}} \right] \\ = \lambda \left[ { - \left( {{b_1}\hat i + {b_2}\hat j} \right)\left\{ {{a_1}{c_1}} \right\} + \left( {{c_1}\hat i + {c_2}\hat j + {c_3}\hat k} \right)\left\{ {{a_1}{b_1}} \right\}} \right] \\ = \lambda \left[ {\left( {\hat i} \right)\left\{ {{c_1}{a_1}{b_1} - {a_1}{b_1}{c_1}} \right\} + \left( {\hat j} \right)\left\{ {{a_1}{b_1}{c_2} - {a_1}{b_2}{c_2}} \right\} + \left( {\hat k} \right)\left\{ {{a_1}{b_1}{c_3}} \right\}} \right] \\ = \lambda \left[ {\left( {\hat j} \right)\left\{ {{a_1}{b_1}{c_2} - {a_1}{b_2}{c_2}} \right\} + \left( {\hat k} \right)\left\{ {{a_1}{b_1}{c_3}} \right\}} \right] \\\end{aligned} \end{equation} $$ Comparing left hand and right hand side, we get $$\lambda = - 1$$

Therefore, by putting the value of $\lambda$ in equation $(2)$, we get $$\begin{equation} \begin{aligned} \vec a \times (\vec b \times \vec c) = - 1\left[ { - \vec b\left( {\vec a.\vec c} \right) + \vec c\left( {\vec a.\vec b} \right)} \right] \\ \vec a \times (\vec b \times \vec c) = \left( {\vec a.\vec c} \right)\vec b - \left( {\vec a.\vec b} \right)\vec c \\\end{aligned} \end{equation} $$

Note: In general, $$\overrightarrow a \times (\overrightarrow b \times \overrightarrow c ) \ne (\overrightarrow a \times \overrightarrow b ) \times \overrightarrow c $$ It is equal if some or all the three vectors $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ are zero vectors or $\overrightarrow a $ and $\overrightarrow c $ are collinear.

Property 1: If $\overrightarrow r = \overrightarrow a \times (\overrightarrow b \times \overrightarrow c )$, then $\overrightarrow r $ is perpendicular to $\overrightarrow a $ and lie in the plane of $\overrightarrow b $ and $\overrightarrow c $.

Property 2: Vector triple product is $$\begin{equation} \begin{aligned} \overrightarrow a \times (\overrightarrow b \times \overrightarrow c ) = (\overrightarrow a .\overrightarrow c )\overrightarrow b - (\overrightarrow a .\overrightarrow b )\overrightarrow c \\ (\overrightarrow a \times \overrightarrow b ) \times \overrightarrow c = (\overrightarrow c .\overrightarrow a )\overrightarrow b - (\overrightarrow c .\overrightarrow b )\overrightarrow a \\\end{aligned} \end{equation} $$For memory: $$I \times (II \times III) = (I.III)II - (I.II)III$$