question_answer

If and the vectors \[\overrightarrow{A}=(1,\,\,a,\,\,{{a}^{2}});\] \[\overrightarrow{B}=(1,\,\,b,\,\,{{b}^{2}});\] \[\overrightarrow{C}=(1,\,\,c,\,\,{{c}^{2}})\] are non-coplanar then the product\[abc=\]

A) \[0\]                             

B) \[1\]

C) \[-1\]                           

D)  None

Correct Answer: C

Solution :

\[=(1+abc)(a-b)(b-c)(c-a)=0\]           .... (1) also, , since \[A,\,\,B,\,\,C\] are not coplanar\[=(a-b)(b-c)(c-a)\ne 0\]     ... (2) From (1) & (2),\[abc=-1\]