Question

If the vectors $a=i+aj+a^{2}k,b=i+bj+b^{2}k,$ and $c=i+cj+c^{2}k$ 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

$-1$

Explanation for the correct option :

Find the value of $abc$,

As the vectors $a=i+aj+a^{2}k,b=i+bj+b^{2}k,$$c=i+cj+c^{2}k$ are non-coplanar vectors so $\left[abc\right]\ne 0$. Thus,

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

Now it is given that $\left|\begin{array}{ccc}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|=0$, split the left hand side into two determinants and simplify.

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

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

$\begin{array}{rcl}1+abc& =& 0\\ abc& =& -1\end{array}$

Hence, the correct option is D.