Question

If \[ (2\alpha+1,\; \alpha^2-3\alpha,\; \frac{\alpha-1}{2}) \] is the image of the point \[ (\alpha,\; 2\alpha,\; 1) \] in the line \[ \frac{x-2}{3}=\frac{y-1}{2}=\frac{z}{1}, \] then find the possible values of \(\alpha\).

Answer 1

Concept:
If a point \(P'\) is the image of point \(P\) in a line, then: \begin{itemize} \item The midpoint of \(PP'\) lies on the given line. \item The line joining \(P\) and \(P'\) is perpendicular to the given line. \end{itemize} Step 1: Write the given points. \[ P = (\alpha, 2\alpha, 1), \quad P' = (2\alpha+1, \alpha^2-3\alpha, \frac{\alpha-1}{2}) \] Step 2: Find midpoint of \(PP'\). \[ M = \left(\frac{\alpha + (2\alpha+1)}{2},\; \frac{2\alpha + (\alpha^2-3\alpha)}{2},\; \frac{1 + \frac{\alpha-1}{2}}{2}\right) \] \[ M = \left(\frac{3\alpha+1}{2},\; \frac{\alpha^2 - \alpha}{2},\; \frac{\alpha+1}{4}\right) \] Step 3: Parametric form of given line. \[ \frac{x-2}{3}=\frac{y-1}{2}=\frac{z}{1}=t \] \[ x=2+3t,\quad y=1+2t,\quad z=t \] Step 4: Since midpoint lies on the line, equate coordinates. \[ \frac{3\alpha+1}{2}=2+3t \quad ...(1) \] \[ \frac{\alpha^2-\alpha}{2}=1+2t \quad ...(2) \] \[ \frac{\alpha+1}{4}=t \quad ...(3) \] Step 5: Substitute \(t\) from (3) into (1) and (2). From (3): \[ t=\frac{\alpha+1}{4} \] Substitute in (1): \[ \frac{3\alpha+1}{2}=2+3\cdot\frac{\alpha+1}{4} \] \[ \frac{3\alpha+1}{2}=\frac{8+3\alpha+3}{4} =\frac{3\alpha+11}{4} \] \[ 2(3\alpha+1)=3\alpha+11 \] \[ 6\alpha+2=3\alpha+11 \] \[ 3\alpha=9 \Rightarrow \alpha=3 \] Substitute in (2): \[ \frac{\alpha^2-\alpha}{2}=1+2\cdot\frac{\alpha+1}{4} \] \[ \frac{\alpha^2-\alpha}{2}=1+\frac{\alpha+1}{2} \] \[ \alpha^2-\alpha=2+\alpha+1 \] \[ \alpha^2-2\alpha-3=0 \] \[ (\alpha-3)(\alpha+1)=0 \] \[ \alpha=3,\; -1 \] Step 6: Check consistency. Both satisfy midpoint condition, hence valid. \[ \boxed{\alpha=3,\; -1} \]