Question
If $$\vec a,\,\vec b,\,\vec c$$ are vectors such that $$\left[ {\vec a,\,\vec b,\,\vec c} \right] = 4$$ then $$\left[ {\vec a \times \,\vec b\,\,\vec b \times \,\vec c\,\,\vec c \times \,\vec a} \right] = ?$$
A.
$$16$$
B.
$$64$$
C.
$$4$$
D.
$$8$$
Answer :
$$16$$
Solution :
$$\eqalign{
& {\text{We have, }}\left[ {\vec a \times \,\vec b\,\,\vec b \times \,\vec c\,\,\vec c \times \,\vec a} \right] \cr
& = \left( {\vec a \times \,\vec b} \right).\left\{ {\left( {\vec b \times \,\vec c} \right) \times \left( {\vec c \times \,\vec a} \right)} \right\} \cr
& = \left( {\vec a \times \,\vec b} \right).\left\{ {\left( {\vec m.\vec a} \right)\vec c - \left( {\vec m.\vec c} \right)\vec a} \right\}\,\,\,\,\,\,\left( {{\text{where }}\vec m = \vec b \times \,\vec c} \right) \cr
& = \left\{ {\left( {\vec a \times \,\vec b} \right).\vec c} \right\}.\left\{ {\vec a.\left( {\vec b \times \,\vec c} \right)} \right\} \cr
& = {\left[ {\vec a,\,\vec b,\,\vec c} \right]^2} = {4^2} = 16 \cr} $$