JEE Main 2025 Jan 22 Shift 2, Mathematics Q9: Lines in Space

JEE Main2025Jan 22, Shift 2Mathematics

The perpendicular distance, of the line $\frac{x - 1}{2} = \frac{y + 2}{- 1} = \frac{z + 3}{2}$ from the point $P (2, - 10, 1)$ , is :

Solution

Parametrise points on the line: $A (λ) = (2 λ + 1, - λ - 2, 2 λ - 3)$ with direction $n = (2, - 1, 2)$ .

$P A = A - P = (2 λ - 1, - λ + 8, 2 λ - 4)$ .

Foot of perpendicular requires $P A \cdot n = 0$ :

$2 (2 λ - 1) - (- λ + 8) + 2 (2 λ - 4) = 0$

$4 λ - 2 + λ - 8 + 4 λ - 8 = 0$

$9 λ - 18 = 0 \Rightarrow λ = 2$ .

Foot $A = (5, - 4, 1)$ . Then $P A = (3, 6, 0)$ and $∣ P A ∣ = 9 + 36 + 0 = 45 = 35$ .

Concept behind this question

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