Question

If the vectors $\overrightarrow{\alpha }=\hat{i}+a\hat{j}+a^2\hat{k},\overrightarrow{\beta }=\hat{i}+b\hat{j}+b^2\hat{k}\text{and}\overrightarrow{\gamma }=\hat{i}+c\hat{j}+c^2\hat{k}$ 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. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is C $-1$

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

$\therefore abc+1=0$