Question

The vectors $\vec{p} = \hat{i} + a\hat{j} + a^2\hat{k}, \vec{q} = \hat{i} + b\hat{j} + b^2\hat{k}$ and $\vec{r} = \hat{i} + c\hat{j} + c^2\hat{k}$ are non-coplanar and $\begin{vmatrix} a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3 \end{vmatrix} = 0$ then the value of $(abc)$ is

• A. 0
• B. -1
• C. 1
• D. 2

Hint:

Determinant property: $\begin{vmatrix} x & y & z+w \end{vmatrix} = \begin{vmatrix} x & y & z \end{vmatrix} + \begin{vmatrix} x & y & w \end{vmatrix}$.

Answer 1

The Correct Option is B

Step 1: Split the Determinant

\[ \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} + \begin{vmatrix} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{vmatrix} = 0 \]

Step 2: Factor the Second Determinant

\[ \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} + abc \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = 0 \]

Step 3: Rearrange Columns

By swapping columns in the first determinant, both determinants become equal. Let \[ D = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}. \]

\[ D + abcD = 0 \;\Rightarrow\; D(1 + abc) = 0 \]

Step 4: Non-coplanar Condition

Since \( \vec{p}, \vec{q}, \vec{r} \) are non-coplanar, the scalar triple product \( D \neq 0 \).

\[ 1 + abc = 0 \;\Rightarrow\; abc = -1 \]

Final Answer: (B)