JEE Advanced 2026 Paper 2, Mathematics Section 2 Q3: Lines in Space

JEE Advanced2026Paper 2Mathematics Section 2

Let L be the straight line joining the points $P (1, 2, - 1)$ and $Q (2, 3, 1)$ . Let S be the foot of the perpendicular drawn from the point $R (4, - 1, 5)$ to the line L. Another line passing through R intersects L at a point T such that the point S divides the line segment PT internally in the ratio $∣ P S ∣ : ∣ S T ∣ = 1 : 2$ , where $∣ P S ∣$ and $∣ S T ∣$ are the lengths of the line segments PS and ST, respectively.

Then which of the following statements is (are) TRUE?

(A) The orthocentre of the triangle PRT is $(\frac{23}{5}, - 4, \frac{31}{5})$

(B) The orthocentre of the triangle PRT is $(4, 3, 5)$

(D) The area of the triangle PRT is $185$

A
The orthocentre of the triangle PRT is $(\frac{23}{5}, - 4, \frac{31}{5})$
B
The orthocentre of the triangle PRT is $(4, 3, 5)$
C
The area of the triangle PRT is $65$
D
The area of the triangle PRT is $185$

Solution

The line $L$ through $P (1, 2, - 1)$ and $Q (2, 3, 1)$ has direction $d = \hat{i} + \hat{j} + 2 \hat{k}$ , with parametric form $(1 + λ, 2 + λ, - 1 + 2 λ)$ .

Finding S: require $R S \cdot d = 0$ where $R = (4, - 1, 5)$ . Then $R S = (λ - 3, λ + 3, 2 λ - 6)$ , so $(λ - 3) + (λ + 3) + 2 (2 λ - 6) = 0$ , giving $6 λ = 12$ , $λ = 2$ . Thus $S = (3, 4, 3)$ .

Finding T: S divides PT in ratio $1 : 2$ . With $P = (1, 2, - 1)$ and $T = (x_{1}, y_{1}, z_{1})$ , the section formula gives $3 = \frac{x _{1} + 2}{3}$ , so $x_{1} = 7$ ; similarly $y_{1} = 8$ , $z_{1} = 11$ . Thus $T = (7, 8, 11)$ .

Area of $△ P R T$ : $R P = - 3 \hat{i} + 3 \hat{j} - 6 \hat{k}$ , $R T = 3 \hat{i} + 9 \hat{j} + 6 \hat{k}$ . Their cross product is $\hat{i} (18 + 54) - \hat{j} (- 18 + 18) + \hat{k} (- 27 - 9) = 72 \hat{i} - 36 \hat{k} = 36 (2 \hat{i} - \hat{k})$ . Magnitude $= 365$ , so area $= \frac{1}{2} \cdot 365 = 185$ . Option (D) is correct.

Orthocentre H: since $R S ⊥ P T$ , the altitude from $R$ in $△ P R T$ lies along line $R S$ . Parametrise $H = R + μ (S - R) = (4 - μ, - 1 + 5 μ, 5 - 2 μ)$ . Require $P H \cdot R T = 0$ . With $P H = (3 - μ, - 3 + 5 μ, 6 - 2 μ)$ and $R T = (3, 9, 6)$ , the dot product is $3 (3 - μ) + 9 (- 3 + 5 μ) + 6 (6 - 2 μ) = 18 + 30 μ$ . Setting this to zero gives $μ = - \frac{3}{5}$ . Substituting back: $H = (\frac{23}{5}, - 4, \frac{31}{5})$ . Option (A) is correct.

$∴$ Correct options: (A), (D).

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JEE Advanced 2026 Paper 2, Mathematics Section 2 Q3: Lines in Space

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