Question

Three lines through origin having direction ratios $(1,a,a^2),$ $(1,b,b^2)$ and $(1,c,c^2)$ are non- coplanar. But the lines with direction ratios $(a,a^2,1+a^3),$ $(b,b^2,1+b^3)$ and $(c,c^2,1+c^3)$ are coplanar. Then the value of $abc$ is

• A. $2$
• B. $-1$
• C. $1$
• D. $-2$

Answer 1

The correct option is B $-1$

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

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

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

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

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

$\Rightarrow abc=-1$ $\left[\text{From}\right(1\left)\right]$