Vector Triple Product

Vector Triple Product

Let \(\overrightarrow{a}\), \(\overrightarrow{b}\), \(\overrightarrow{c}\) be any three vectors, then the vectors \(\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)\,and\,\left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \overrightarrow{c}\) are called vector triple products of \(\overrightarrow{a}\), \(\overrightarrow{b}\), \(\overrightarrow{c}\). For any three vectors \(\overrightarrow{a}\), \(\overrightarrow{b}\), \(\overrightarrow{c}\)  we have \(\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)\,=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}.\overrightarrow{b} \right).\overrightarrow{c}\).

• The vector triple product \(\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)\) is a linear combination of those two vectors which are within brackets.
• The vectors \(\overrightarrow{r}=\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)\) is perpendicular to \(\overrightarrow{a}\) and lies in the plane of \(\overrightarrow{b}\) and \(\overrightarrow{c}.\)
• The formula \(\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}.\overrightarrow{b} \right)\overrightarrow{c}\) is true only when the vector outside the bracket is on the left most side. If it is not we first shift on left by using the properties of cross product and then apply the same formula.

For example,

\(\left( \overrightarrow{b}\times \overrightarrow{c} \right)\times \overrightarrow{a}=-\left\{ \overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right) \right\}\).

\(\,=-\left\{ \left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}.\overrightarrow{b} \right)\overrightarrow{c} \right\}\).

\(\left( \overrightarrow{a}\times \overrightarrow{b} \right).\left(\overrightarrow{c}\times\overrightarrow{d}\right)=\left|\begin{matrix}\overrightarrow{a}.\overrightarrow{c}&\overrightarrow{a}.\overrightarrow{d}\\ \overrightarrow{b}.\overrightarrow{c} &\overrightarrow{b}.\overrightarrow{d}  \\\end{matrix}\right|\).

Example: Show that \(\overrightarrow{a}\), \(\overrightarrow{b}\), \(\overrightarrow{c}\) are coplanar if \(\overrightarrow{a}\times \overrightarrow{b},\,\,\overrightarrow{b}\times \overrightarrow{c},\,\,\overrightarrow{c}\times \overrightarrow{a}\,\) are coplanar

Solution: \(\overrightarrow{a}\), \(\overrightarrow{b}\), \(\overrightarrow{c}\) are coplanar

\(\left[ \overrightarrow{a},\overrightarrow{b},\overrightarrow{c} \right]=0\).

\({{\left[ \overrightarrow{a},\overrightarrow{b},\overrightarrow{c} \right]}^{2}}=0\).

\(\left[ \overrightarrow{a}\times \overrightarrow{b}\overrightarrow{,b}\times \overrightarrow{c},\overrightarrow{c}\times \overrightarrow{a} \right]=0\).

\(\overrightarrow{a}\times \overrightarrow{b},\overrightarrow{b}\times \overrightarrow{c},\overrightarrow{c}\times \overrightarrow{a}\,\,\text{are}\,\,\text{coplanar}\).

Example: Prove that \(\left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \left( \overrightarrow{c}\times \overrightarrow{d}\, \right)=\left[ \overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{d} \right]\overrightarrow{c}-\left[ \overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c} \right]\overrightarrow{d}\).

Solution: we have

\(\left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \left( \overrightarrow{c}\times \overrightarrow{d}\, \right)\).

\(=\overrightarrow{r}\times \left( \overrightarrow{c}\times \overrightarrow{d} \right),\,where\,\overrightarrow{e}=\overrightarrow{a}\times \overrightarrow{b}\).

\(=\left( \overrightarrow{r.}\overrightarrow{d} \right).\overrightarrow{c}-\left( \overrightarrow{r}.\overrightarrow{c} \right)\overrightarrow{d}\).

\(=\left[ \left( \overrightarrow{a}\times \overrightarrow{b} \right).\overrightarrow{d} \right]\overrightarrow{c}-\left[ \left( \overrightarrow{a}\times \overrightarrow{b} \right).\overrightarrow{c} \right]\overrightarrow{d}\).

\(=\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{d} \right]\overrightarrow{c}-\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]\overrightarrow{d}\).

\(\left( \overrightarrow{a}\times \overrightarrow{b}
\right).\left( \overrightarrow{c}\times \overrightarrow{d}\, \right)=\left|
\begin{matrix}   \overrightarrow{a} & \overrightarrow{b} & \overrightarrow{c} & \overrightarrow{d}  \\   \overrightarrow{a} &
\overrightarrow{b} & \overrightarrow{c} & \overrightarrow{d}  \\\end{matrix} \right|\).

Example: If \(\overrightarrow{a}\), \(\overrightarrow{b}\), \(\overrightarrow{c}\)  are vectors such that \(\left| \overrightarrow{b} \right|=\left| \overrightarrow{c} \right|\) prove that \(\left[ \left( \overrightarrow{a}+\overrightarrow{b} \right)\times \left( \overrightarrow{a}+\overrightarrow{c} \right) \right]\times \left( \overrightarrow{b}\times \overrightarrow{c} \right).\left( \overrightarrow{b}+\overrightarrow{c} \right)=0\).

Solution: we have,

\(\left[ \left( \overrightarrow{a}+\overrightarrow{b} \right)\times \left( \overrightarrow{a}+\overrightarrow{c} \right) \right]\times \left( \overrightarrow{b}\times \overrightarrow{c} \right).\left( \overrightarrow{b}+\overrightarrow{c} \right)\).

\(=\left( \overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{a}+\overrightarrow{b}\times \overrightarrow{c} \right)\times \left( \overrightarrow{b}\times \overrightarrow{c} \right).\left( \overrightarrow{b}+\overrightarrow{c} \right)\).

\(=\left[ \left( \overrightarrow{a}+\overrightarrow{c} \right)\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)+\left( \overrightarrow{b}\times \overrightarrow{a} \right)\times \left( \overrightarrow{b}+\overrightarrow{c} \right)+\left( \overrightarrow{b}+\overrightarrow{c} \right)\times \left( \overrightarrow{b}\times \overrightarrow{c} \right) \right].\left( \overrightarrow{b}+\overrightarrow{c} \right)\).

\(=\left[ \left( \overrightarrow{a}\overrightarrow{c}\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}\overrightarrow{c}\overrightarrow{b} \right)\overrightarrow{c}+\left( \overrightarrow{b}\overrightarrow{a}\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{b}\overrightarrow{a}\overrightarrow{b} \right)\overrightarrow{c} \right].\left( \overrightarrow{b}+\overrightarrow{c} \right)\).

\(=\left[ \left( \overrightarrow{a}\overrightarrow{c}\overrightarrow{b} \right)\overrightarrow{c}+\left( \overrightarrow{b}\overrightarrow{a}\overrightarrow{c} \right)\overrightarrow{b} \right].\left( \overrightarrow{b}+\overrightarrow{c} \right)\).

\(=-\left[ \overrightarrow{a}\overrightarrow{c}\overrightarrow{b} \right]\left[ \overrightarrow{c}.\left( \overrightarrow{b}+\overrightarrow{c} \right) \right]+\left[ \overrightarrow{b}\overrightarrow{a}\overrightarrow{c} \right]\left[ \overrightarrow{b}.\left( \overrightarrow{b}+\overrightarrow{c} \right) \right]\).

\(=\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]\left( \overrightarrow{c}.\overrightarrow{b}+{{\left| \overrightarrow{c} \right|}^{2}} \right)-\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]\left( {{\left| \overrightarrow{b} \right|}^{2}}+\overrightarrow{b}\overrightarrow{c} \right)\).

\(=\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]\left( {{\left| \overrightarrow{c} \right|}^{2}}-{{\left| \overrightarrow{b} \right|}^{2}} \right)\).

\(=\left[ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right]\times 0\,\left[ \because \left| \overrightarrow{b} \right|=\left| \overrightarrow{c} \right| \right]\).

= 0.