JEE Advanced 2024 Paper 2 Q06 Mathematics Vectors & 3D Geometry 3D Geometry Medium

JEE Advanced 2024 Paper 2 · Q06 · 3D Geometry

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

  1. A. The length of the line segment $PQ$ is $\sqrt{6}$
  2. B. The coordinates of $R$ are $(1, 6, 3)$
  3. C. The centroid of the triangle $PQR$ is $\left(\dfrac{4}{3}, \dfrac{14}{3}, \dfrac{5}{3}\right)$
  4. D. The perimeter of the triangle $PQR$ is $\sqrt{2} + \sqrt{6} + \sqrt{11}$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, C

Solution

Line through $P(1,3,2)$ with direction $(1,2,1)$: $(1+t, 3+2t, 2+t)$. Plug into $L_{1}$: $(1+t) - (3+2t) + 3(2+t) = 6 \Rightarrow 4 + 2t = 6 \Rightarrow t = 1$. So $Q = (2, 5, 3)$ and $|PQ| = \sqrt{1+4+1} = \sqrt{6}$ — (A) TRUE. From $Q$, perpendicular to $L_{1}$ has direction $(1,-1,3)$: $(2+s, 5-s, 3+3s)$. Plug into $L_{2}$: $2(2+s) - (5-s) + (3+3s) = -4 \Rightarrow 6s + 2 = -4 \Rightarrow s = -1$. So $R = (1, 6, 0)$ — (B) FALSE. Centroid $= ((1+2+1)/3, (3+5+6)/3, (2+3+0)/3) = (4/3, 14/3, 5/3)$ — (C) TRUE. $|QR| = \sqrt{1+1+9} = \sqrt{11}$, $|PR| = \sqrt{0+9+4} = \sqrt{13}$. Perimeter $= \sqrt{6}+\sqrt{11}+\sqrt{13} \ne \sqrt{2}+\sqrt{6}+\sqrt{11}$ — (D) FALSE.

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