JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Three Dimensional Geometry

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

Answer: (A), (C)

Solution

Direction ratios of

L_{1}

(cross product of normals):

b = \hat{i} - \hat{j} + \hat{k}

, so

L_{1}

has direction

⟨ 1, - 1, 1 ⟩

L_{2}

passes through

P (2, - 1, 3)

with the same direction:

\frac{x - 2}{1} = \frac{y + 1}{- 1} = \frac{z - 3}{1} = λ

Point on

L_{2}

(λ + 2, - λ - 1, λ + 3)

. Substituting into plane

M : 2 x + y - 2 z = 6

2 (λ + 2) - λ - 1 - 2 (λ + 3) - 6 = 0 ⟹ λ = - 9

Q (- 7, 8, - 6)

P Q = (- 9)^{2} + 9^{2} + (- 9)^{2} = 93

. (A) is true.

For

R

, perpendicular from

P

M

has direction

⟨ 2, 1, - 2 ⟩

\frac{x - 2}{2} = \frac{y + 1}{1} = \frac{z - 3}{- 2} = μ

Substituting into

M

μ = 1

, so

R (4, 0, 1)

QR = 1 1^{2} + (- 8)^{2} + 7^{2} = 234

. (B) is false.

Area of

△ P QR = \frac{1}{2} ∣ QP \times QR ∣ = \frac{3}{2} 234

. (C) is true.

Angle:

cos θ = \frac{∣ P Q \cdot P R ∣}{∣ P Q ∣∣ P R ∣} = \frac{9}{9 3 \cdot 3} = \frac{1}{3 3}

. (D) is false.

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · One or more correct options