• question_answer If the vectors $\vec{a}=\hat{i}+a\hat{j}+{{a}^{2}}\hat{k},\vec{b}=\hat{i}+b\hat{j}+{{b}^{2}}\hat{k},$$\vec{c}=\hat{i}+c\hat{j}+{{c}^{2}}\hat{k}$are three non-coplanar vectors and $\left| \begin{matrix} a & {{a}^{2}} & 1+{{a}^{3}} \\ b & {{b}^{2}} & 1+{{b}^{3}} \\ c & {{c}^{2}} & 1+{{c}^{3}} \\ \end{matrix} \right|=0,$ then the value of aback is A)  0                             B)         1                             C)  2                             D)         - 1

Since, $\vec{a}=\hat{i}+a\hat{j}+{{a}^{2}}\hat{k},\vec{b}=\hat{i}+b\hat{j}+{{b}^{2}}\hat{k}$ and        $\vec{c}=\hat{i}+c\hat{j}+{{c}^{2}}\hat{k}$                 are coplanar vectors                 $\therefore$  $[\vec{a}\,\vec{b}\,\vec{c}]\ne 0$                 $\Rightarrow$               $\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|\ne 0$                 $\Rightarrow$               $\Delta \ne 0$                 where   $\Delta =\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|$                 Now,     $\left| \begin{matrix} a & {{a}^{2}} & 1+{{a}^{3}} \\ b & {{b}^{2}} & 1+{{b}^{3}} \\ c & {{c}^{2}} & 1+{{c}^{3}} \\ \end{matrix} \right|=0$                 $\Rightarrow$               $\left| \begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix} \right|+\left| \begin{matrix} a & {{a}^{2}} & {{a}^{3}} \\ b & {{b}^{2}} & {{b}^{3}} \\ c & {{c}^{2}} & {{c}^{3}} \\ \end{matrix} \right|=0$                 $\Rightarrow$               $\left| \begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix} \right|+abc\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|=0$                 $\Rightarrow$               $\Delta +abc\,\Delta =0$                 $\Rightarrow$               $\Delta (1+abc)=0$                 $\Rightarrow$               $1+abc=0$                        $(\because \,\Delta \ne 0)$ $\Rightarrow$               $abc=-1$