Question

Let \(\alpha \in R\) and the three vectors \(\vec{a} = \alpha\hat{i} + \hat{j} + 3\hat{k}\), \(\vec{b} = 2\hat{i} + \hat{j} - \alpha\hat{k}\) and \(\vec{c} = \alpha\hat{i} - 2\hat{j} + 3\hat{k}\). Then the set \(S = \{\alpha : \vec{a}, \vec{b}\) and \(\vec{c}\) are coplanar\(\}\) :

• A. contains exactly two numbers only one of which is positive
• B. is singleton
• C. contains exactly two positive numbers
• D. is empty

Hint:

Coplanar vectors have scalar triple product (determinant) zero; show the resulting equation has no real root.

Answer 1

The Correct Option is D

Coplanarity: \(\begin{vmatrix}\alpha & 1 & 3\\ 2 & 1 & -\alpha\\ \alpha & -2 & 3\end{vmatrix} = 0\).
Expand: \(\alpha(3 - 2\alpha) - 1(6 + \alpha^2) + 3(-4 - \alpha)\).
\(= 3\alpha - 2\alpha^2 - 6 - \alpha^2 - 12 - 3\alpha = -3\alpha^2 - 18\).
\(-3\alpha^2 - 18 = 0 \Rightarrow \alpha^2 = -6\), which has no real solution.
Hence S is empty.
Answer: (4) is empty