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

Foot of Perpendicular 3D: a^2 + b^2 + alpha^2 = 1 | JEE Main 2026

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

Question
Let the point AA be the foot of perpendicular drawn from the point P(a,b,0)P(a,b,0) on the line x−12=y−21=z−α3\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-\alpha}{3}. If the mid point of the line segment PAPA is (0,34,−14)\left(0,\dfrac34,-\dfrac14\right), then the value of a2+b2+α2a^2+b^2+\alpha^2 is equal to
A11correct
B22
C66
D99
Solution
Step 1: A general point on the line is A=(2r+1, r+2, 3r+α)A=(2r+1,\ r+2,\ 3r+\alpha) for parameter rr. Step 2: Midpoint of PAPA equals (0,34,−14)\left(0,\tfrac34,-\tfrac14\right) coordinate-wise:
(2r+1)+a2=0 ⇒ 2r+a=−1,\frac{(2r+1)+a}{2}=0\ \Rightarrow\ 2r+a=-1,
(r+2)+b2=34 ⇒ r+b=−12,\frac{(r+2)+b}{2}=\frac34\ \Rightarrow\ r+b=-\frac12,
3r+α2=−14 ⇒ 3r+α=−12.\frac{3r+\alpha}{2}=-\frac14\ \Rightarrow\ 3r+\alpha=-\frac12.
 Step 3: PAPA is perpendicular to the line, so PA→⋅(2,1,3)=0\overrightarrow{PA}\cdot(2,1,3)=0. Writing PA→=A−P=(2r+1−a, r+2−b, 3r+α)\overrightarrow{PA}=A-P=(2r+1-a,\ r+2-b,\ 3r+\alpha) and using the midpoint relations, this reduces to
2(−1−2r)+(−r−12)=0 ⇒ −5r−52=0 ⇒ r=−12.2(-1-2r)+\left(-r-\tfrac12\right)=0\ \Rightarrow\ -5r-\tfrac52=0\ \Rightarrow\ r=-\frac12.
 Step 4: Back-substitute r=−12r=-\tfrac12:
a=−1−2r=0,b=−12−r=0,α=−12−3r=1.a=-1-2r=0,\qquad b=-\tfrac12-r=0,\qquad \alpha=-\tfrac12-3r=1.
 Step 5:
a2+b2+α2=0+0+1=1.a^2+b^2+\alpha^2=0+0+1=1.
 Correct answer: (1)
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