Question

If the vectors $\overrightarrow{a}=\hat{i}+a\hat{j}+a^2\hat{k},\overrightarrow{b}=\hat{i}+b\hat{j}+b^2\hat{k},\overrightarrow{c}=\hat{i}+c\hat{j}+c^{2}k$ are three none-coplanar vectors and $\left|\begin{array}{ccc}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|=0$, then the value of abc is

• A. 0
• B. 1
• C. 2
• D. -1

Answer 1

The correct option is D -1

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

As the given vectors are non coplanar

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

So option $C$ is correct.