JEE Main 2026 Apr 6 Shift 2, Mathematics Q24: Lines in Space

JEE Main2026Apr 6, Shift 2Mathematics

Let the image of the point $P (0, - 5, 0)$ in the line $\frac{x - 1}{2} = \frac{y}{1} = \frac{z + 1}{- 2}$ be the point $R$ , and the image of the point $Q (0, - \frac{1}{2}, 0)$ in the line $\frac{x - 1}{- 1} = \frac{y + 9}{4} = \frac{z + 1}{1}$ be the point $S$ . Then the square of the area of the parallelogram $P QR S$ is _______.

Your answer

Solution

Image of $P$ in line 1. Parametrise foot as $M_{1} = (2 k + 1, k, - 2 k - 1)$ . Perpendicularity to direction $⟨ 2, 1, - 2 ⟩$ :

$2 (2 k + 1) + 1 (k + 5) - 2 (- 2 k - 1) = 0$ $\Rightarrow$ $9 k + 9 = 0$ $\Rightarrow$ $k = - 1$ .

$M_{1} = (- 1, - 1, 1)$ , so $R = 2 M_{1} - P = (- 2, 3, 2)$ .

Image of $Q$ in line 2. Foot $M_{2} = (1 - λ, - 9 + 4 λ, - 1 + λ)$ . Perpendicularity to $⟨ - 1, 4, 1 ⟩$ :

$(- λ) (- 1) + (4 λ - 9 + \frac{1}{2}) (4) + (λ - 1) (1) = 0$ $\Rightarrow$ $18 λ - 36 = 0$ $\Rightarrow$ $λ = 2$ .

$M_{2} = (- 1, - 1, 1)$ , so $S = 2 M_{2} - Q = (- 2, - \frac{3}{2}, 2)$ .

Parallelogram area. The two midpoints $M_{1}, M_{2}$ coincide, so the diagonals $P R$ and $QS$ bisect each other (parallelogram confirmed) and are perpendicular (each lies along its respective line's perpendicular from a point).

$∣ P R ∣ = 4 + 64 + 4 = 72 = 62$ . $∣ QS ∣ = 4 + 1 + 4 = 3$ .

Area $= \frac{1}{2} \cdot ∣ P R ∣ \cdot ∣ QS ∣ = \frac{1}{2} \cdot 62 \cdot 3 = 92$ .

(Area) $^{2} = 162$ .

Concept behind this question

Lines in SpaceNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2026 Apr 6 Shift 2, Mathematics Q24: Lines in Space

Concept behind this question

More previous year questions on Lines in Space

Practice more Mathematics