Question

If \((2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2})\) is the image of \((\alpha, 2\alpha, 1)\) in the line \(\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}\), then the possible value(s) of \(\alpha\) is (are):

• A. Only 3
• B. Only 3 and \(-1\)
• C. Only \(3, \frac{1}{4}\) and \(-1\)
• D. Only \(3\) and \(\frac{1}{4}\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The midpoint of the point and its image must lie on the line, and the vector connecting the point to its image must be perpendicular to the line's direction.

Step 2: Key Formula or Approach:
1. Midpoint \(M = \frac{P + P'}{2}\) satisfies the line equation.
2. \(\vec{PP'} \cdot \vec{d}_{line} = 0\).

Step 3: Detailed Explanation:
Let \(P = (\alpha, 2\alpha, 1)\) and \(P' = (2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2})\).
Midpoint \(M = (\frac{3\alpha + 1}{2}, \frac{\alpha^2 - \alpha}{2}, \frac{\alpha + 1}{4})\).
Check \(\vec{PP'} = (\alpha + 1, \alpha^2 - 5\alpha, \frac{\alpha - 3}{2})\).
Perpendicularity: \(3(\alpha + 1) + 2(\alpha^2 - 5\alpha) + 1(\frac{\alpha - 3}{2}) = 0\).
\(6\alpha + 6 + 4\alpha^2 - 20\alpha + \alpha - 3 = 0 \implies 4\alpha^2 - 13\alpha + 3 = 0\).
\((4\alpha - 1)(\alpha - 3) = 0 \implies \alpha = 3, 1/4\).
Testing \(\alpha = 3\) in line equation for midpoint \(M\):
For \(\alpha = 3, M = (5, 3, 1)\). Line: \(\frac{5-2}{3} = \frac{3-1}{2} = \frac{1}{1} \implies 1 = 1 = 1\). Works.
For \(\alpha = 1/4, M\) does not satisfy the line equation.

Step 4: Final Answer:
The only possible value is 3.