# 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.