# Vector Triple Product

## Vector Triple Product

The vector triple product of three vectors $$\overrightarrow{a}$$, $$\overrightarrow{b}$$ and $$\overrightarrow{c}$$ is the vector $$\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}.\overrightarrow{b} \right)\overrightarrow{c}$$.

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

In general, $$\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)\ne \left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \overrightarrow{c}$$.

If $$\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \overrightarrow{c}$$, then the vectors $$\overrightarrow{a}$$ and $$\overrightarrow{c}$$ are collinear.

$$\overrightarrow{p}=\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)$$ is a vector perpendicular to $$\overrightarrow{a}$$ and $$\overrightarrow{b}\times \overrightarrow{c}$$ but $$\overrightarrow{b}\times \overrightarrow{c}$$ is a vector perpendicular to the plane of $$\overrightarrow{b}$$ and $$\overrightarrow{c}$$.

Hence, vector $$\overrightarrow{p}$$ must lie in the plane of $$\overrightarrow{b}$$ and $$\overrightarrow{c}$$.

$$\overrightarrow{p}=\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=x.\overrightarrow{b}+y.\overrightarrow{c}$$ … (i)

Multiplying (i) scalarly by $$\overrightarrow{a}$$, we have

$$\overrightarrow{p}.\overrightarrow{a}=x\left( \overrightarrow{a}.\overrightarrow{b} \right)+y\left( \overrightarrow{a}.\overrightarrow{c} \right)$$ … (ii)

But $$\overrightarrow{p}\bot \overrightarrow{a}$$ ⇒ $$\overrightarrow{p}.\overrightarrow{a}=0$$. Therefore,

$$x\left( \overrightarrow{a}.\overrightarrow{b} \right)=-y\left( \overrightarrow{c}.\overrightarrow{a} \right)$$, i.e, $$\frac{x}{\overrightarrow{c}.\overrightarrow{a}}=\frac{-y}{\overrightarrow{a}.\overrightarrow{b}}=\lambda$$.

∴ $$x=\lambda \left( \overrightarrow{c}.\overrightarrow{a} \right)$$, $$y=-\lambda \left( \overrightarrow{a}.\overrightarrow{b} \right)$$ … (iii)

Substituting x and y from (iii) in (i), $$\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\lambda \left[ \left( \overrightarrow{c}.\overrightarrow{a} \right)\overrightarrow{b}-\left( \overrightarrow{a}.\overrightarrow{b} \right)\overrightarrow{c} \right]$$ … (iv)

The simplest way to determine λ is by taking specific vectors $$\overrightarrow{a}=\hat{i}$$, $$\overrightarrow{b}=\hat{i}$$, $$\overrightarrow{c}=\hat{j}$$.

We have from (iv), $$\hat{i}\times \left( \hat{i}\times \hat{j} \right)=\lambda \left[ \left( \hat{i}.\hat{j} \right)\hat{i}-\left( \hat{i}.\hat{i} \right)\hat{j} \right]$$, i.e, $$\hat{i}\times \hat{k}=\lambda \left[ 0\hat{i}-1\hat{j} \right]$$, i.e, $$-\hat{j}=-\lambda \hat{j}$$.

∴ λ = 1

Substituting λ in (iv),

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

Lagrange’s Identity:

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

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

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

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

This is called Lagrange’s Identity.