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

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q14 – JEE-MAIN 2026

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Home Maths 3D GeometryQuestion #9

(2 α + 1, α^{2} - 3 α, \frac{α - 1}{2})

is the image of

(α, 2 α, 1)

in the line

\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}

, then the possible value(s) of

α

is (are)

Clue 1 of 3: Geometric Properties of Image

The image of a point in a line has two key geometric properties: the midpoint of the segment connecting the point and its image lies on the line, and the segment connecting the point and its image is perpendicular to the line.

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Step 1: Identify Points and Line Properties✦ Active

Let the given point be $P (α, 2 α, 1)$ and its image be $P^{'} (2 α + 1, α^{2} - 3 α, \frac{α - 1}{2})$ . The equation of the line $L$ is $\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}$ . The direction vector of line $L$ is $\vec{d} = (3, 2, 1)$ .

Step 2: Apply Midpoint Condition○ Expand

The midpoint $M$ of $P P^{'}$ must lie on the line $L$ . The coordinates of $M$ are:

M = (\frac{α + (2 α + 1)}{2}, \frac{2 α + (α^{2} - 3 α)}{2}, \frac{1 + \frac{α - 1}{2}}{2}) = (\frac{3 α + 1}{2}, \frac{α^{2} - α}{2}, \frac{α + 1}{4})

Substituting $M$ into the line equation $\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}$ gives:

\frac{\frac{3 α + 1}{2} - 2}{3} = \frac{\frac{α^{2} - α}{2} - 1}{2} = \frac{\frac{α + 1}{4}}{1}

This simplifies to $\frac{3 α - 3}{6} = \frac{α^{2} - α - 2}{4} = \frac{α + 1}{4}$ . Equating the first and third parts: $\frac{α - 1}{2} = \frac{α + 1}{4} ⟹ 2 α - 2 = α + 1 ⟹ α = 3$ . Equating the second and third parts: $\frac{α^{2} - α - 2}{4} = \frac{α + 1}{4} ⟹ α^{2} - 2 α - 3 = 0 ⟹ (α - 3) (α + 1) = 0 ⟹ α = 3$ or $α = - 1$ . For all parts to be equal, $α = 3$ is the only solution from the midpoint condition.

💡 Teacher's Secret Hint

Ensure all components of the midpoint satisfy the line equation simultaneously.

Step 3: Apply Perpendicularity Condition and Find Common Solutions○ Expand

The vector $P P^{'}$ must be perpendicular to the direction vector of line $L$ . The vector $P P^{'}$ is:

P P^{'} = ((2 α + 1) - α, (α^{2} - 3 α) - 2 α, \frac{α - 1}{2} - 1) = (α + 1, α^{2} - 5 α, \frac{α - 3}{2})

The dot product $P P^{'} \cdot \vec{d} = 0$ :

3 (α + 1) + 2 (α^{2} - 5 α) + 1 (\frac{α - 3}{2}) = 0

Multiplying by 2 to clear the fraction: $6 (α + 1) + 4 (α^{2} - 5 α) + (α - 3) = 0$ . This simplifies to $4 α^{2} - 13 α + 3 = 0$ . Factoring the quadratic equation gives $(4 α - 1) (α - 3) = 0$ , so $α = 3$ or $α = \frac{1}{4}$ . For $P^{'}$ to be the image of $P$ , both the midpoint condition and the perpendicularity condition must be satisfied. The common value of $α$ from both conditions is $α = 3$ . Thus, the only possible value of $α$ is 3.

💡 Teacher's Secret Hint

Remember that both conditions (midpoint on line and perpendicularity) must hold true for a point to be the image.

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Maths Question 14 – JEE-MAIN 2026

Clue 1 of 3: Geometric Properties of Image

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