Vectors & 3D GeometrymediumPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Image of a Point in a Line: Only α = 3 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
If (2α+1, α2−3α, α−12)\left(2\alpha+1,\ \alpha^2-3\alpha,\ \dfrac{\alpha-1}{2}\right) is the image of (α, 2α, 1)(\alpha,\ 2\alpha,\ 1) in the line x−23=y−12=z1\dfrac{x-2}{3}=\dfrac{y-1}{2}=\dfrac{z}{1}, then the possible value(s) of α\alpha is(are)
AOnly 33correct
BOnly 33 and −1-1
COnly 3, 143,\ \dfrac14 and −1-1
DOnly 33 and 14\dfrac14
Solution
Step 1: The midpoint PP of A(α,2α,1)A(\alpha,2\alpha,1) and its image B(2α+1,α2−3α,α−12)B\left(2\alpha+1,\alpha^2-3\alpha,\tfrac{\alpha-1}{2}\right) is
P=(3α+12, α2−α2, α+14).P=\left(\frac{3\alpha+1}{2},\ \frac{\alpha^2-\alpha}{2},\ \frac{\alpha+1}{4}\right).
 PP must lie on the line. Step 2: Substitute into x−23=y−12=z1\dfrac{x-2}{3}=\dfrac{y-1}{2}=\dfrac{z}{1}:
3α+12−23=α2−α2−12=α+141 ⇒ α−12=α+14=α2−α−24.\frac{\frac{3\alpha+1}{2}-2}{3}=\frac{\frac{\alpha^2-\alpha}{2}-1}{2}=\frac{\frac{\alpha+1}{4}}{1}\ \Rightarrow\ \frac{\alpha-1}{2}=\frac{\alpha+1}{4}=\frac{\alpha^2-\alpha-2}{4}.
 Step 3: From α−12=α+14\dfrac{\alpha-1}{2}=\dfrac{\alpha+1}{4}: 2(α−1)=α+1⇒α=32(\alpha-1)=\alpha+1\Rightarrow\alpha=3. Step 4: For α=3\alpha=3, AB→\overrightarrow{AB} is indeed perpendicular to the direction 3i^+2j^+k^3\hat i+2\hat j+\hat k, confirming it is the image. (The other algebraic value −1-1 fails the perpendicularity condition.) Hence only α=3\alpha=3. Correct answer: (1)
Solution working
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