JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Three Dimensional Geometry

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 :

Answer: (B)

Solution

The given line is

L_{1} : \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}

, passing through

(2, 6, 3)

.

The line through

(1, 4, 0)

with direction ratios

(1, 2, 3)

(matching the second given line) is

(x, y, z) = (1 + t, 4 + 2 t, 3 t)

Find its intersection with

L_{1}

, written as

(2 + 2 s, 6 + 3 s, 3 + 4 s)

1 + t = 2 + 2 s \Rightarrow t = 1 + 2 s

4 + 2 t = 6 + 3 s \Rightarrow 2 t = 2 + 3 s

Substituting:

2 (1 + 2 s) = 2 + 3 s \Rightarrow 4 s + 2 = 3 s + 2 \Rightarrow s = 0

, giving

t = 1

.

The point of intersection is

(2, 6, 3)

.

Distance from

(1, 4, 0)

(2, 6, 3)

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

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option