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:
If point \(P'\) is the image of point \(P\) over a line, the midpoint of the segment \(PP'\) must lie exactly on that given line. By deriving the coordinates of the midpoint and substituting them into the line equation, we can find the viable values of \(\alpha\).
Step 2: Key Formula or Approach:
Midpoint \(M\) between \(P\) and \(P'\):
\(M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2} \right)\)
Line equation: \(\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}\)
Step 3: Detailed Explanation:
Let \(P = (\alpha, 2\alpha, 1)\) and \(P' = \left(2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2}\right)\).
The midpoint \(M\) is:
\[ M = \left(\frac{3\alpha + 1}{2}, \frac{\alpha^2 - \alpha}{2}, \frac{\alpha + 1}{4}\right) \] Substitute \(M\) into the line's Cartesian equation:
\[ \frac{\frac{3\alpha + 1}{2} - 2}{3} = \frac{\frac{\alpha^2 - \alpha}{2} - 1}{2} = \frac{\frac{\alpha + 1}{4}}{1} \] \[ \frac{3\alpha - 3}{6} = \frac{\alpha^2 - \alpha - 2}{4} = \frac{\alpha + 1}{4} \] Equate the first and third expressions:
\[ \frac{\alpha - 1}{2} = \frac{\alpha + 1}{4} \implies 2\alpha - 2 = \alpha + 1 \implies \alpha = 3 \] Check if \(\alpha = 3\) validates the second expression:
\[ \frac{3^2 - 3 - 2}{4} = \frac{4}{4} = 1 \] And \(\frac{3 - 1}{2} = 1\). The equations are perfectly balanced.
It is mathematically sufficient to check only this root since intersection forces a single rigid geometrical relationship. No other parameters align across all three dimensional coordinates simultaneously.
Step 4: Final Answer:
The only possible value of \(\alpha\) is 3.