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JEE Main 2026 · 2 Apr · Shift 2, Mathematics question 15

Let the point A 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 midpoint of the line segment PA is (0,34,−14)\left(0,\dfrac34,\dfrac{-1}4\right), then the value of a2+b2+α2a^2+b^2+\alpha^2 is equal to :

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Answer and worked solution

Official answer: A

Point A. A=2M−P=(−a,32−b,−12)A=2M-P=\left(-a,\frac32-b,-\frac12\right).

Perpendicularity. PA→=(−2a,32−2b,−12)\overrightarrow{PA}=(-2a,\frac32-2b,-\frac12) is perpendicular to (2,1,3)(2,1,3): −4a−2b=0-4a-2b=0, so b=−2ab=-2a.

A on the line. −a−12=32−b−2=t\frac{-a-1}2=\frac32-b-2=t gives −a−1=−1+4a-a-1=-1+4a, so a=0=ba=0=b and t=−12t=-\frac12. Then −12−α=3t-\frac12-\alpha=3t gives α=1\alpha=1.

Value. a2+b2+α2=1a^2+b^2+\alpha^2=1.

(A)

Topics:Equation of a line in spaceAngles and distances
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