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…

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 vectors $\left(1, a, a^2\right),\left(a, b, b^2\right)$ and $\left(a, c, c^2\right)$ are non-coplanar, then the product abc equals
  1. 0
  2. 2
  3. $-1$
  4. 1

Solution

$\left|\begin{array}{lll}\mathrm{a} & \mathrm{a}^2 & 1+\mathrm{a}^3 \\ \mathrm{~b} & \mathrm{~b}^2 & 1+\mathrm{b}^3 \\ \mathrm{c} & \mathrm{c}^2 & 1+\mathrm{c}^3\end{array}\right|=0 \Rightarrow\left|\begin{array}{lll}\mathrm{a} & \mathrm{a}^2 & 1 \\ \mathrm{~b} & \mathrm{~b}^2 & 1 \\ \mathrm{c} & \mathrm{c}^2 & 1\end{array}\right|+\left|\begin{array}{ccc}\mathrm{a} & \mathrm{a}^2 & \mathrm{a}^3 \\ \mathrm{~b} & \mathrm{~b}^2 & \mathrm{~b}^3 \\ \mathrm{c} & \mathrm{c}^2 & \mathrm{c}^3\end{array}\right|=0$ $(a-b)(b-c)(c-a)+a b c(a-b)(b-c)(c-a)=0$ $(a b c+1)[(a-b)(b-c)(c-a)]=0$ As $\left|\begin{array}{lll}1 & \mathrm{a} & \mathrm{a}^2 \\ 1 & \mathrm{~b} & \mathrm{~b}^2 \\ 1 & \mathrm{c} & \mathrm{c}^2\end{array}\right| \neq 0$ (given condition) $\quad \therefore \mathrm{abc}=-1$

Asked in: JEE Main 2003

Practice more Vectors questions on Aicharya