JEE Main 2025 Jan 23 Shift 2, Mathematics Q6: Lines in Space

JEE Main2025Jan 23, Shift 2Mathematics

The distance of the line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ from the point $(1, 4, 0)$ along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$ is :

A
$17$
B
$14$
C
$15$
D
$13$

Solution

Travel from $(1, 4, 0)$ along the second line and find where it meets the first line; the distance traveled is the answer.

Parametrise the path: $(1 + s, 4 + 2 s, 3 s)$ for $s \in R$ . Parametrise the target line: $(2 + 2 t, 6 + 3 t, 3 + 4 t)$ .

Equate components: $1 + s = 2 + 2 t, 4 + 2 s = 6 + 3 t, 3 s = 3 + 4 t$ . The first two give $s = 2 t + 1$ and $2 s = 3 t + 2$ , hence $2 (2 t + 1) = 3 t + 2 \Rightarrow t = 0$ , $s = 1$ .

(Check the third: $3 (1) = 3 + 4 (0)$ $✓$ .) The point of intersection is $(2, 6, 3)$ .

Distance $= (2 - 1)^{2} + (6 - 4)^{2} + (3 - 0)^{2} = 1 + 4 + 9 = 14 .$

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JEE Main 2025 Jan 23 Shift 2, Mathematics Q6: Lines in Space

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