Question

If \(\left(2\alpha+1,\;\alpha^2-3\alpha,\;\frac{\alpha-1}{2}\right)\) is the image of \((\alpha,2\alpha,1)\) in the line \[ \frac{x-2}{3}=\frac{y-1}{2}=\frac{z}{1}, \] then the value of \(\alpha\) is:

Answer 1

Correct Answer: 3

Concept: If point \(B\) is the image of point \(A\) in a line, then: • The midpoint of \(A\) and \(B\) lies on the line. • The line joining \(A\) and \(B\) is perpendicular to the given line.
Step 1: Let the points \[ A(\alpha,2\alpha,1) \] \[ B\left(2\alpha+1,\;\alpha^2-3\alpha,\;\frac{\alpha-1}{2}\right) \]
Step 2: Find midpoint \[ P=\left(\frac{\alpha+(2\alpha+1)}{2},\; \frac{2\alpha+(\alpha^2-3\alpha)}{2},\; \frac{1+\frac{\alpha-1}{2}}{2}\right) \] \[ P=\left(\frac{3\alpha+1}{2},\;\frac{\alpha^2-\alpha}{2},\;\frac{\alpha+1}{4}\right) \]
Step 3: Line equation \[ \frac{x-2}{3}=\frac{y-1}{2}=\frac{z}{1} \] Direction ratios: \[ (3,2,1) \] Since midpoint lies on the line, \[ \frac{\frac{3\alpha+1}{2}-2}{3} = \frac{\frac{\alpha^2-\alpha}{2}-1}{2} = \frac{\frac{\alpha+1}{4}}{1} \]
Step 4: Solve From first and third terms: \[ \frac{\alpha-1}{2}=\frac{\alpha+1}{4} \] \[ 2(\alpha-1)=\alpha+1 \] \[ \alpha=3 \]
Step 5: Verification For \(\alpha=3\), vector \(AB\) becomes perpendicular to direction vector \((3,2,1)\). Thus \(B\) is the image of \(A\) in the given line.