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.