Question

If the vectors $\overrightarrow{a}=\hat{i}+a\hat{j}+a^2\hat{k},\overrightarrow{b}=\hat{i}+b\hat{j}+b^2\hat{k},\overrightarrow{c}=\hat{i}+c\hat{j}+c^2\hat{k}$ are three non-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 $-1$

Since $z\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are non-coplanar vectors, therefore $[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\ne 0$
$\Rightarrow$ $\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|\ne 0$

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

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