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MathHardMain 2026 • 4th Apr - Shift 2

Possible Values of Parameter for Image of Point in Line

If (2α+1,α2−3α,α−12)\left(2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2}\right) is the image of (α,2α,1)(\alpha, 2\alpha, 1) in the line x−23=y−12=z1\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}, then the possible value(s) of α\alpha is (are)

Options

A

Only 33

Correct
B

Only 33 and −1-1

C

Only 3,143, \frac{1}{4} and −1-1

D

Only 33 and 14\frac{1}{4}

Topics & Concepts

Three Dimensional GeometryLine in 3D

Step-by-Step Solution

To find the possible value(s) of α\alpha, we utilize the geometric properties of an image of a point with respect to a line in 3D space:

  1. The midpoint MM of the segment joining the point P(α,2α,1)P(\alpha, 2\alpha, 1) and its image P′(2α+1,α2−3α,α−12)P'\left(2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2}\right) must lie on the given line.
  2. The vector PP′⃗\vec{PP'} must be perpendicular to the direction vector of the line.

Step 1: Midpoint Condition

The coordinates of the midpoint MM of PP and P′P' are: M=(α+(2α+1)2,2α+(α2−3α)2,1+α−122)=(3α+12,α2−α2,α+14)M = \left( \frac{\alpha + (2\alpha + 1)}{2}, \frac{2\alpha + (\alpha^2 - 3\alpha)}{2}, \frac{1 + \frac{\alpha - 1}{2}}{2} \right) = \left( \frac{3\alpha + 1}{2}, \frac{\alpha^2 - \alpha}{2}, \frac{\alpha + 1}{4} \right)

The given line is: L:x−23=y−12=z1L: \frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}

Since MM lies on line LL, its coordinates must satisfy the equation of the line: 3α+12−23=α2−α2−12=α+141\frac{\frac{3\alpha + 1}{2} - 2}{3} = \frac{\frac{\alpha^2 - \alpha}{2} - 1}{2} = \frac{\frac{\alpha + 1}{4}}{1}

Simplifying each term: 3α−36=α2−α−24=α+14\frac{3\alpha - 3}{6} = \frac{\alpha^2 - \alpha - 2}{4} = \frac{\alpha + 1}{4}   ⟹  α−12=α2−α−24=α+14\implies \frac{\alpha - 1}{2} = \frac{\alpha^2 - \alpha - 2}{4} = \frac{\alpha + 1}{4}

Equating the second and third expressions: α2−α−2=α+1\alpha^2 - \alpha - 2 = \alpha + 1 α2−2α−3=0\alpha^2 - 2\alpha - 3 = 0 (α−3)(α+1)=0  ⟹  α=3 or α=−1(\alpha - 3)(\alpha + 1) = 0 \implies \alpha = 3 \text{ or } \alpha = -1

Now, we verify these values against the first expression α−12\frac{\alpha - 1}{2}:

  • For α=3\alpha = 3: 3−12=1and3+14=1(Satisfies the line equation)\frac{3 - 1}{2} = 1 \quad \text{and} \quad \frac{3 + 1}{4} = 1 \quad \text{(Satisfies the line equation)}
  • For α=−1\alpha = -1: −1−12=−1and−1+14=0(−1≠0,does not satisfy)\frac{-1 - 1}{2} = -1 \quad \text{and} \quad \frac{-1 + 1}{4} = 0 \quad (-1 \neq 0, \text{does not satisfy})

Thus, from the midpoint condition, we get α=3\alpha = 3.

Step 2: Perpendicularity Condition

The vector PP′⃗\vec{PP'} is: PP′⃗=P′−P=((2α+1)−α,(α2−3α)−2α,α−12−1)=(α+1,α2−5α,α−32)\vec{PP'} = P' - P = \left( (2\alpha + 1) - \alpha, (\alpha^2 - 3\alpha) - 2\alpha, \frac{\alpha - 1}{2} - 1 \right) = \left( \alpha + 1, \alpha^2 - 5\alpha, \frac{\alpha - 3}{2} \right)

The direction vector of line LL is d⃗=(3,2,1)\vec{d} = (3, 2, 1).

Since PP′⃗⊥d⃗\vec{PP'} \perp \vec{d}, their dot product must be zero: PP′⃗⋅d⃗=0\vec{PP'} \cdot \vec{d} = 0 3(α+1)+2(α2−5α)+1⋅(α−32)=03(\alpha + 1) + 2(\alpha^2 - 5\alpha) + 1 \cdot \left( \frac{\alpha - 3}{2} \right) = 0

Multiplying the entire equation by 22 to clear the fraction: 6(α+1)+4(α2−5α)+(α−3)=06(\alpha + 1) + 4(\alpha^2 - 5\alpha) + (\alpha - 3) = 0 6α+6+4α2−20α+α−3=06\alpha + 6 + 4\alpha^2 - 20\alpha + \alpha - 3 = 0 4α2−13α+3=04\alpha^2 - 13\alpha + 3 = 0

Factoring the quadratic equation: (4α−1)(α−3)=0  ⟹  α=3 or α=14(4\alpha - 1)(\alpha - 3) = 0 \implies \alpha = 3 \text{ or } \alpha = \frac{1}{4}

Conclusion

For P′P' to be the image of PP, both conditions must hold simultaneously. Taking the common value from both steps: α=3\alpha = 3

Thus, the correct option is A (Only 33).

Possible Values of Parameter for Image of Point in Line | Mathematics PYQ Solution - JEE Challenger