Question

If $\left|\begin{array}{ccc}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|=0$ 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

Answer 1

The correct option is C $-1$

$\overrightarrow{A}=(1,a,a^2)\equiv \hat{i}+a\hat{ȷ}+a^2\hat{k}$

$\overrightarrow{B}=(1,b,b^2)\equiv \hat{i}+b\hat{ȷ}+b^2\hat{k}$

$\overrightarrow{c}=(1,c,c^2)\equiv \hat{ı}+c\hat{ȷ}+c^2\hat{k}$

$\cdot \overrightarrow{A},\overrightarrow{B}$ and $\overrightarrow{C}$ are non coplanar

$\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|\ne 0$

$Now\cdot \left|\begin{array}{ccc}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|=0($ Given $)$

$\Rightarrow \left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|+\left|\begin{array}{ccc}a& a^2& a^3\\ b& b^2& b^3\\ c& c^2& c^3\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|+abc\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|-abc\left|\begin{array}{ccc}a& 1& a^2\\ b& 1& b^2\\ c& 1& c^2\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|+abc\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|(1+abc)=0$

Now we know that $\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|\ne 0$

$1+abc=0\Rightarrow abc=-1$

Answer : option $\left(c\right)$