JEE Main 2025 Jan 24 Shift 1, Mathematics Q18: Lines in Space

JEE Main2025Jan 24, Shift 1Mathematics

Let the line passing through the points $(- 1, 2, 1)$ and parallel to the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ intersect the line $\frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1}$ at the point $P$ . Then the distance of $P$ from the point $Q (4, - 5, 1)$ is :

A
$5$
B
$10$
C
$56$
D
$55$

Solution

Line through $(- 1, 2, 1)$ with direction $(2, 3, 4)$ : parametrise as $(2 λ - 1, 3 λ + 2, 4 λ + 1)$ .

Second line: $(3 μ - 2, 2 μ + 3, μ + 4)$ .

Set equal coordinate-wise:

$2 λ - 1 = 3 μ - 2 (i)$

$3 λ + 2 = 2 μ + 3 (ii)$

$4 λ + 1 = μ + 4 (iii)$

From (iii): $μ = 4 λ - 3$ . Substitute in (i): $2 λ - 1 = 3 (4 λ - 3) - 2 = 12 λ - 11 \Rightarrow 10 λ = 10 \Rightarrow λ = 1$ . Check (ii): $5 = 2 (1) + 3 = 5$ . $✓$

Hence $P = (1, 5, 5)$ and

$∣ P Q ∣ = (4 - 1)^{2} + (- 5 - 5)^{2} + (1 - 5)^{2} = 9 + 100 + 16 = 125 = 55$ .

$∴$ Option (4).

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JEE Main 2025 Jan 24 Shift 1, Mathematics Q18: Lines in Space

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