Vectors 3 Question 24
24. If $\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \ b & b^{2} & 1+b^{3} \ c & c^{2} & 1+c^{3}\end{array}\right|=0$ and the vectors $\overrightarrow{\mathbf{A}}=\left(1, a, a^{2}\right), \overrightarrow{\mathbf{B}}=\left(1, b, b^{2}\right), \overrightarrow{\mathbf{C}}\left(1, c, c^{2}\right)$ are non-coplanar, then the product a b c=. .
(1985, 2M)
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Answer:
Correct Answer: 24. $(-1)$
Solution:
- Since, $\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \ b & b^{2} & 1+b^{3} \ c & c^{2} & 1+c^{3}\end{array}\right|=0$ $\Rightarrow \quad\left|\begin{array}{lll}a & a^{2} & 1 \ b & b^{2} & 1 \ c & c^{2} & 1\end{array}\right|+\left|\begin{array}{ccc}a & a^{2} & a^{3} \ b & b^{2} & b^{3} \ c & c^{2} & c^{3}\end{array}\right|=0$
$\Rightarrow \quad(1+a b c)\left|\begin{array}{lll}1 & a & a^{2} \ 1 & b & b^{2} \ 1 & c & c^{2}\end{array}\right|=0$
$\Rightarrow$ Either $(1+a b c)=0$ or $\left|\begin{array}{lll}1 & a & a^{2} \ 1 & b & b^{2} \ 1 & c & c^{2}\end{array}\right|=0$
But $\left(1, a, a^{2}\right),\left(1, b, b^{2}\right),\left(1, c, c^{2}\right)$ are non-coplanar.
$$ \begin{array}{ll} \Rightarrow \quad & \left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| \neq 0 \\ \therefore \quad & a b c=-1 \end{array} $$
