JEE Main 2025 Apr 3 Shift 2, Mathematics Q19: Lines in Space

JEE Main2025Apr 3, Shift 2Mathematics

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

A
$18$
B
$14$
C
$12$
D
$16$

Solution

Any point on the first line: $Q = (λ + 4, 4, 3 λ + 2)$ .

The vector $P Q$ from $P = (7, 10, 11)$ is $(λ - 3, - 6, 3 λ - 9)$ .

This must be parallel to the direction $(2, 3, 6)$ of the second line:

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

From the middle ratio: $- 2$ , so $λ - 3 = - 4 \Rightarrow λ = - 1$ . Check: $3 λ - 9 = - 12 = - 2 \cdot 6$ . Consistent.

So $Q = (3, 4, - 1)$ and $∣ P Q ∣ = (7 - 3)^{2} + (10 - 4)^{2} + (11 + 1)^{2} = 16 + 36 + 144 = 196 = 14$ .

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JEE Main 2025 Apr 3 Shift 2, Mathematics Q19: Lines in Space

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