Question
If $$\vec a,\,\vec b,\,\vec c$$ arenon coplanar vectors and $$\lambda $$ is a real number then $$\left[ {\lambda \left( {\vec a + \vec b} \right){\lambda ^2}\vec b\,\lambda \vec c} \right] = \left[ {\vec a\,\vec b + \vec c\,\vec b} \right]$$ for :
A.
exactly one value of $$\lambda $$
B.
no value of $$\lambda $$
C.
exactly three values of $$\lambda $$
D.
exactly two values of $$\lambda $$
Answer :
no value of $$\lambda $$
Solution :
$$\eqalign{
& \left[ {\lambda \left( {\vec a + \vec b} \right){\lambda ^2}\vec b\,\lambda \vec c} \right] = \left[ {\vec a\,\vec b + \vec c\,\vec b} \right] \cr
& \Rightarrow {\lambda ^4}\left[ {\vec a + \vec b\,\,\vec b\vec c} \right] = \left[ {\vec a\,\vec b + \vec c\,\vec b} \right] \cr
& \Rightarrow {\lambda ^4}\left\{ {\left[ {\vec a\,\vec b\,\vec c} \right] + \left[ {\vec b\,\vec b\,\vec c} \right]} \right\} = \left[ {\vec a\,\vec b\,\vec b} \right] + \left[ {\vec a\,\vec c\,\vec b} \right] \cr
& \Rightarrow {\lambda ^4}\left[ {\vec a\,\vec b\,\vec c} \right] = - \left[ {\vec a\,\vec b\,\vec c} \right] \cr
& \Rightarrow {\lambda ^4} = - 1 \cr
& \Rightarrow \lambda \,\,\,{\text{has no real values}} \cr} $$