JEE Advanced 2024 — Paper 2 — MATH-II solution

JEE Advanced2024Paper 2MATH-II

A straight line drawn from the point $P (1, 3, 2)$ , parallel to the line $\frac{x - 2}{1} = \frac{y - 4}{2} = \frac{z - 6}{1}$ , intersects the plane $L_{1} : x - y + 3 z = 6$ at the point $Q$ . Another straight line which passes through $Q$ and is perpendicular to the plane $L_{1}$ intersects the plane $L_{2} : 2 x - y + z = - 4$ at the point $R$ . Then which of the following statements is(are) TRUE?

A
The length of the line segment $P Q$ is $6$
B
The coordinates of $R$ are $(1, 6, 3)$
C
The centroid of the triangle $P QR$ is $(\frac{4}{3}, \frac{14}{3}, \frac{5}{3})$
D
The perimeter of the triangle $P QR$ is $2 + 6 + 11$

Solution

Finding $Q$ : The line through $P (1, 3, 2)$ parallel to direction $⟨ 1, 2, 1 ⟩$ is parametrised as $(1 + λ, 3 + 2 λ, 2 + λ)$ . Substituting into $L_{1} : x - y + 3 z = 6$ :

$(1 + λ) - (3 + 2 λ) + 3 (2 + λ) = 6 \Rightarrow 4 + 2 λ = 6 \Rightarrow λ = 1.$

Hence $Q = (2, 5, 3)$ .

Finding $R$ : Normal to $L_{1}$ is $⟨ 1, - 1, 3 ⟩$ . Line through $Q$ along this normal: $(2 + μ, 5 - μ, 3 + 3 μ)$ . Substituting into $L_{2} : 2 x - y + z = - 4$ :

$2 (2 + μ) - (5 - μ) + (3 + 3 μ) = - 4 \Rightarrow 2 + 6 μ = - 4 \Rightarrow μ = - 1.$

Hence $R = (1, 6, 0)$ , so (B) is false.

Checking each:

(A) $P Q = (2 - 1)^{2} + (5 - 3)^{2} + (3 - 2)^{2} = 1 + 4 + 1 = 6 .$ True.

(C) Centroid $= (\frac{1 + 2 + 1}{3}, \frac{3 + 5 + 6}{3}, \frac{2 + 3 + 0}{3}) = (\frac{4}{3}, \frac{14}{3}, \frac{5}{3}) .$ True.

(D) $QR = 1 + 1 + 9 = 11$ , $P R = 0 + 9 + 4 = 13$ . Perimeter $= 6 + 11 + 13 \neq = 2 + 6 + 11 .$ False.

$∴$ Correct options are (A) and (C).

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