JEE Main 2026 · 4 Apr · Shift 2 · Mathematics Q14

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MathematicsSingle correct+4 / -1Question 14 of 75

If $\left(2\alpha+1,\ \alpha^2-3\alpha,\ \dfrac{\alpha-1}2\right)$ is the image of $(\alpha,2\alpha,1)$ in the line $\dfrac{x-2}3=\dfrac{y-1}2=\dfrac z1$ , then the possible value(s) of $\alpha$ is (are)

Only 3

Only 3 and −1

Only 3,

\dfrac14

and −1

Only 3 and

\dfrac14

Answer and worked solution

Official answer: A

Perpendicularity. $\overrightarrow{PP'}=\left(\alpha+1,\ \alpha^2-5\alpha,\ \frac{\alpha-3}2\right)$ must be ⟂ $(3,2,1)$ : $4\alpha^2-13\alpha+3=0$ , so $\alpha=3$ or $\frac14$ .

Midpoint on the line.

$\alpha=3$ : the midpoint $(5,3,1)$ satisfies $\frac{5-2}3=\frac{3-1}2=1$ ✓.
$\alpha=\frac14$ : the midpoint $\left(\frac78,-\frac3{32},\frac5{16}\right)$ does not.

Only 3.

(A)

Topics:Equation of a line in space

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JEE Main 2026 · 4 Apr · Shift 2, Mathematics question 14