JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Three Dimensional Geometry

The distance of the point

(7, 10, 11)

from the line

\frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}

along the line

\frac{x - 9}{2} = \frac{y - 13}{3} = \frac{z - 17}{6}

(A)18
(B)14✓ Correct
(C)12
(D)16

Answer: (B)

Solution

A point on the first line is

Q (4 + λ, 4, 2 + 3 λ)

, so with

P (7, 10, 11)

P Q = (λ - 3, - 6, 3 λ - 9)

. Measuring the distance along the second line means

P Q

is parallel to its direction

(2, 3, 6)

, so

\frac{λ - 3}{2} = \frac{- 6}{3} = \frac{3 λ - 9}{6}

. The middle ratio is

- 2

, giving

λ = - 1

and

Q (3, 4, - 1)

. The required distance is

P Q = (7 - 3)^{2} + (10 - 4)^{2} + (11 + 1)^{2} = 16 + 36 + 144 = 196

, which equals

14

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Single correct option

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