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Maths Question 15 – JEE-MAIN 2025

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The distance of the point (7,10,11) from the line x−41=y−40=z−23 along the line x−92=y−133=z−176 is

The problem asks for the distance between a point P and a line L1 measured along another line L2. This means we need to find a point Q on L1 such that the line segment PQ is parallel to L2.

Step 1: Represent the point on the first line✦ Active

The given point is P(7,10,11). The first line L1 is x−41=y−40=z−23. Let a general point Q on L1 be (4+λ,4,2+3λ). The direction vector of L2 is d2→=⟨2,3,6⟩.

Step 2: Form the vector PQ and use parallelism condition○ Expand

The vector PQ→=Q−P=(4+λ−7,4−10,2+3λ−11)=(λ−3,−6,3λ−9). Since PQ→ is parallel to L2, PQ→ must be proportional to d2→. Thus, we set up the proportionality:

λ−32=−63=3λ−96

From −63=−2, we have λ−32=−2⟹λ−3=−4⟹λ=−1. (This is consistent with the third component: 3(−1)−96=−126=−2).

💡 Teacher's Secret Hint

Ensure all components of the vector PQ are proportional to the direction vector of L2.

Step 3: Find point Q and calculate the distance○ Expand

Substitute λ=−1 into the coordinates of Q: Q(4−1,4,2−3)=Q(3,4,−1). The distance PQ is calculated using the distance formula:

PQ=(7−3)2+(10−4)2+(11−(−1))2
PQ=42+62+122=16+36+144=196=14
💡 Teacher's Secret Hint

Double-check the arithmetic for the distance calculation.

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