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QJEEAdv24Ma06a8: JEE Advanced 2024 Mathematics - Class 12 JEEAdv24Ma06a8 Solution: A straight line drawn from the point , parallel to the line ,…

A straight line drawn from the point P(1,3,2)P(1,3,2) , parallel to the line x−21=y−42=z−61\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1} , intersects the plane L1:x−y+3z=6L_1: x-y+3 z=6 at the point QQ . Another straight line which passes through QQ and is perpendicular to the plane L1L_1 intersects the plane L2:2x−y+z=−4L_2: 2 x-y+z=-4 at the point RR . Then which of the following statements is (are) TRUE?

Multiple Choice Options

  • A The length of the line segment PQP Q is 6\sqrt{6}
  • B The coordinates of RR are (1,6,3)(1,6,3)
  • C The centroid of the triangle PQRP Q R is (43,143,53)\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right)
  • D The perimeter of the triangle PQRP Q R is 2+6+11\sqrt{2}+\sqrt{6}+\sqrt{11}

Step-by-Step Verified Solution

amp; line :x−11=y−32=z−21amp;(x,y,z)=(λ+1,2λ+3,λ+2)amp; Put in L1:x−y+3z=6amp;(λ+1)−(2λ+3)+3(λ+2)=6amp;2λ=2⇒λ=1amp;Q=(2,5,3)\begin{aligned}& \text { line }: \frac{x-1}{1}=\frac{y-3}{2}=\frac{z-2}{1} \\& (x, y, z)=(\lambda+1,2 \lambda+3, \lambda+2) \\& \text { Put in } L_1: x-y+3 z=6 \\& (\lambda+1)-(2 \lambda+3)+3(\lambda+2)=6 \\& 2 \lambda=2 \Rightarrow \lambda=1 \\& Q=(2,5,3)\end{aligned}

line: x−21=y−5−1=z−33\frac{x-2}{1}=\frac{y-5}{-1}=\frac{z-3}{3}

(x,y,z)=(t+2,5−t,3t+3)(\mathrm{x}, \mathrm{y}, \mathrm{z})=(\mathrm{t}+2,5-\mathrm{t}, 3 \mathrm{t}+3)

Put in L2:2x−y+z=−4L_2: 2 x-y+z=-4

amp;2(t+4)−(5−t)+(3t+3)=−4amp;6t=−6⇒t=−1amp;R=(1,6,0)\begin{aligned}& 2(t+4)-(5-t)+(3 t+3)=-4 \\& 6 t=-6 \Rightarrow t=-1 \\& R=(1,6,0)\end{aligned}

JEE Advanced 2024 Paper 2 Online Mathematics - 3D Geometry Question 1 English Explanation

amp; Perimeter =6+13+11amp; Centroid =(43,143,53)\begin{aligned}& \text { Perimeter }=\sqrt{6}+\sqrt{13}+\sqrt{11} \\& \text { Centroid }=\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right)\end{aligned}

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