JEE Advanced 2025 — Paper 1 — MATH-II solution

JEE Advanced2025Paper 1MATH-II

Let $L_{1}$ be the line of intersection of the planes given by the equations

$2 x + 3 y + z = 4$ and $x + 2 y + z = 5.$

Let $L_{2}$ be the line passing through the point $P (2, - 1, 3)$ and parallel to $L_{1}$ . Let $M$ denote the plane given by the equation

$2 x + y - 2 z = 6$

Suppose that the line $L_{2}$ meets the plane $M$ at the point $Q$ . Let $R$ be the foot of the perpendicular drawn from $P$ to the plane $M$ .

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

A
The length of the line segment $P Q$ is $93$
B
The length of the line segment $QR$ is $15$
C
The area of $△ P QR$ is $\frac{3}{2} 234$
D
The acute angle between the line segments $P Q$ and $P R$ is $cos^{- 1} (\frac{1}{2 3})$

Solution

Direction of $L_{1}$ . The cross product of the normals $(2, 3, 1)$ and $(1, 2, 1)$ gives a direction vector $(1, - 1, 1)$ for $L_{1}$ , hence also for $L_{2}$ .

Finding $Q$ . Parametrise $L_{2}$ as $(2 + t, - 1 - t, 3 + t)$ . Substituting into $2 x + y - 2 z = 6$ : $2 (2 + t) + (- 1 - t) - 2 (3 + t) = 6 \Rightarrow - t - 3 = 6 \Rightarrow t = - 9$ . So $Q = (- 7, 8, - 6)$ .

$P Q = 9^{2} + 9^{2} + 9^{2} = 93$ . (A) is correct.

Finding $R$ . The perpendicular from $P$ to $M$ has direction $(2, 1, - 2)$ . Parametrise as $(2 + 2 s, - 1 + s, 3 - 2 s)$ and substitute into $M$ : $2 (2 + 2 s) + (- 1 + s) - 2 (3 - 2 s) = 6 \Rightarrow 9 s - 3 = 6 \Rightarrow s = 1$ . So $R = (4, 0, 1)$ .

$QR = 1 1^{2} + 8^{2} + 7^{2} = 121 + 64 + 49 = 234 .$ (B) is false.

Area of triangle. Using $P Q = - 9 (\hat{i} - \hat{j} + \hat{k})$ and $P R = (2, 1, - 2)$ , $∣ P Q \times P R ∣ = 9 \cdot ∣ (\hat{i} - \hat{j} + \hat{k}) \times (2 \hat{i} + \hat{j} - 2 \hat{k}) ∣ = 9 1 + 16 + 9 = 926 .$ Hmm, recompute directly. With $QP = (9, - 9, 9)$ and $QR = (11, - 8, 7)$ , the cross product equals $(- 9 \cdot 7 - 9 \cdot (- 8), 9 \cdot 11 - 9 \cdot 7, 9 \cdot (- 8) - (- 9) \cdot 11) = (9, 36, 27)$ , with magnitude $81 + 1296 + 729 = 2106 = 3234$ . Therefore Area $= \frac{1}{2} \cdot 3234 = \frac{3}{2} 234 .$ (C) is correct.

Angle between $P Q$ and $P R$ . $cos θ = \frac{∣ P Q \cdot P R ∣}{∣ P Q ∣ \cdot ∣ P R ∣} = \frac{∣ - 9 ( 2 ) + 9 ( 1 ) + ( - 9 ) ( - 2 ) ∣}{9 3 \cdot 3} = \frac{9}{27 3} = \frac{1}{3 3} .$ This is not $\frac{1}{2 3}$ . (D) is false.

$∴$ The correct options are (A) and (C).

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