Question
If $$\left[ {\vec a \times \vec b\,\,\vec b \times \vec c\,\,\vec c \times \vec a} \right] = \lambda {\left[ {\vec a\,\vec b\,\vec c} \right]^2}$$ then $$\lambda $$ is equal to :
A.
$$0$$
B.
$$1$$
C.
$$2$$
D.
$$3$$
Answer :
$$1$$
Solution :
$$\eqalign{
& {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \left( {\vec a \times \vec b} \right).\left[ {\left( {\vec b \times \vec c} \right) \times \left( {\vec c \times \vec a} \right)} \right] \cr
& = \left( {\vec a \times \vec b} \right).\left[ {\left( {\vec b \times \vec c.\vec a} \right)\vec c - \left( {\vec b \times \vec c.\vec c} \right)\vec a} \right] \cr
& = \left( {\vec a \times \vec b} \right).\left[ {\left[ {\vec b\,\vec c\,\vec a} \right]\vec c} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because \,\vec b \times \vec c.\vec c = 0} \right] \cr
& = \left[ {\vec a\,\vec b\,\vec c} \right].\left( {\vec a \times \vec b.\vec c} \right) \cr
& = {\left[ {\vec a\,\vec b\,\vec c} \right]^2} \cr
& \left[ {\vec a \times \vec b\,\,\vec b \times \vec c\,\,\vec c \times \vec a} \right] = {\left[ {\vec a\,\vec b\,\vec c} \right]^2} \cr
& {\text{So }}\lambda = 1 \cr} $$