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Maths Question 15 – JEE-MAIN 2026

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Let the point A be the foot of perpendicular drawn from the point P(a, b, 0) on the line x−12=y−21=z−α3. If the midpoint of the line segment PA is (0,34,−14), then the value of a2+b2+α2 is equal to :

Understand that the foot of the perpendicular A, the point P, and the midpoint M of PA are collinear, and M lies on the segment PA.

Step 1: Express Point A and relate P, A, M using Midpoint Formula✦ Active

Let the given line be L:x−12=y−21=z−α3=λ. Any point A on the line can be written as A(2λ+1,λ+2,3λ+α). The point P is (a,b,0) and the midpoint M of PA is (0,34,−14). Using the midpoint formula M=P+A2, we can write:

0=a+(2λ+1)2⟹a=−2λ−1(1) 34=b+(λ+2)2⟹b=32−λ−2=−λ−12(2) −14=0+(3λ+α)2⟹α=−3λ−12(3)
Step 2: Apply Perpendicularity Condition○ Expand

Since A is the foot of the perpendicular from P to the line L, the vector PA→ must be perpendicular to the direction vector of the line, d→=(2,1,3). The coordinates of A can also be expressed as A(2Mx−Px,2My−Py,2Mz−Pz)=(−a,32−b,−12). Thus, PA→=A−P=(−2a,32−2b,−12). The dot product PA→⋅d→=0 gives:

(−2a)(2)+(32−2b)(1)+(−12)(3)=0 −4a+32−2b−32=0 −4a−2b=0⟹2a+b=0(4)

Substitute equations (1) and (2) into equation (4):

2(−2λ−1)+(−λ−12)=0 −4λ−2−λ−12=0 −5λ−52=0⟹−5λ=52⟹λ=−12
💡 Teacher's Secret Hint

Ensure correct calculation of vector PA and its dot product with the direction vector.

Step 3: Calculate a,b,α and the final expression○ Expand

Substitute λ=−12 back into equations (1), (2), and (3):

a=−2(−12)−1=1−1=0 b=−(−12)−12=12−12=0 α=−3(−12)−12=32−12=1

Finally, calculate a2+b2+α2:

a2+b2+α2=(0)2+(0)2+(1)2=0+0+1=1
💡 Teacher's Secret Hint

Double-check the arithmetic for a,b,α and the final sum of squares.

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