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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Advanced 2024 · Paper 2 · Mathematics · Three Dimensional Geometry

A straight line drawn from the point P(1,3,2) , parallel to the...

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?

Options

  1. AThe length of the line segment PQP Q is 6\sqrt{6}
  2. BThe coordinates of RR are (1,6,3)(1,6,3)
  3. CThe centroid of the triangle PQRP Q R is (43,143,53)\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right)
  4. DThe perimeter of the triangle PQRP Q R is 2+6+11\sqrt{2}+\sqrt{6}+\sqrt{11}

Correct answer

A. The length of the line segment PQP Q is 6\sqrt{6}

Step-by-step solution

 line :x−11=y−32=z−21(x,y,z)=(λ+1,2λ+3,λ+2)\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)\end{aligned}PutinPut inL_1: x-y+3 z=6 (λ+1)−(2λ+3)+3(λ+2)=62λ=2⇒λ=1Q=(2,5,3)\begin{aligned} & (\lambda+1)-(2 \lambda+3)+3(\lambda+2)=6 \\ & 2 \lambda=2 \Rightarrow \lambda=1 \\ & Q=(2,5,3)\end{aligned}line:line :\frac{x-2}{1}=\frac{y-5}{-1}=\frac{z-3}{3} (x, y, z)=(t+2,5-t, 3 t+3)PutinPut inL_2: 2 x-y+z=-4 2(t+4)−(5−t)+(3t+3)=−46t=−6⇒t=−1R=(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}  Perimeter =6+13+11 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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