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Maths Question 15 – JEE-MAIN 2026

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The square of the distance of the point (−2,−8,6) from the line x−11=y−12=z−1 along the line x+51=y+5−1=z2 is equal to:

The phrase 'along the line' implies that the line segment connecting the given point to the first line is parallel to the second line.

Step 1: Define Points and Lines✦ Active

Let the given point be P(−2,−8,6). The first line L1 is given by x−11=y−12=z−1. A general point Q on L1 can be written in parametric form as Q(1+λ,1+2λ,−λ). The second line L2 is given by x+51=y+5−1=z2, and its direction vector is d2→=⟨1,−1,2⟩.

Step 2: Find the Point Q on L1○ Expand

The vector PQ→ from P to Q is (1+λ−(−2),1+2λ−(−8),−λ−6)=(3+λ,9+2λ,−6−λ). Since the distance is measured 'along the line' L2, the vector PQ→ must be parallel to d2→. Thus, PQ→=kd2→ for some scalar k:

(3+λ,9+2λ,−6−λ)=k(1,−1,2)

Equating components gives the system of equations:

3+λ=k(1) 9+2λ=−k(2) −6−λ=2k(3)

Substitute (1) into (2): 9+2λ=−(3+λ)⟹9+2λ=−3−λ⟹3λ=−12⟹λ=−4. Substitute λ=−4 into (1) to get k=3+(−4)=−1. (Check with (3): −6−(−4)=−2 and 2k=2(−1)=−2, which is consistent). Now find point Q using λ=−4: Q(1−4,1+2(−4),−(−4))=Q(−3,−7,4).

💡 Teacher's Secret Hint

Ensure consistency by checking the values of λ and k with all three equations.

Step 3: Calculate the Square of the Distance○ Expand

The vector PQ→ is Q−P=(−3−(−2),−7−(−8),4−6)=(−1,1,−2). The square of the distance is the square of the magnitude of PQ→:

|PQ→|2=(−1)2+(1)2+(−2)2=1+1+4=6
💡 Teacher's Secret Hint

Remember to calculate the square of the distance, not just the distance itself.

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