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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2019 · Mathematics · Three Dimensional Geometry

The equation of the plane passing through the intersection of the...

The equation of the plane passing through the intersection of the planes 2x+y+2z=9,4x−5y−4z=12 x+y+2 z=9,4 x-5 y-4 z=1 and the point (3,2,1)(3,2,1) is

Options

  1. A10x−2y+2z=2810 \mathrm{x}-2 \mathrm{y}+2 \mathrm{z}=28
  2. B10x+2y+2z=2810 \mathrm{x}+2 \mathrm{y}+2 \mathrm{z}=28
  3. C10x+2y−2z=2810 \mathrm{x}+2 \mathrm{y}-2 \mathrm{z}=28
  4. D10x−2y−2z=2410 x-2 y-2 z=24

Correct answer

A. 10x−2y+2z=2810 \mathrm{x}-2 \mathrm{y}+2 \mathrm{z}=28

Step-by-step solution

The equation of plane passing through the intersection of planes 2x+y+2z=9and4x−5y−4z=12 x+y+2 z=9 and 4 x-5 y-4 z=1 is (2x+y+2z−9)+λ(4x−5y−4z−1)=0(2 x+y+2 z-9)+\lambda(4 x-5 y-4 z-1)=0 Given that this plane passes through (3,2,1)(3,2,1)⇒2(3)+2+2(1)−9+λ[4(3)−5(2)−4(1)−1]=0⇒1+λ(−3)=0⇒λ=13\begin{aligned} & \Rightarrow 2(3)+2+2(1)-9+\lambda[4(3)-5(2)-4(1)-1]=0 \\ & \Rightarrow 1+\lambda(-3)=0 \\ & \Rightarrow \lambda=\frac{1}{3}\end{aligned}∴\therefore Equation is (2x+y+2z−9)+13(4x−5y−4z−1)=0(2 x+y+2 z-9)+\frac{1}{3}(4 x-5 y-4 z-1)=0⇒6x+3y+6z−27+4x−5y−4z−1=0⇒10x−2y+2z−28=0⇒10x−2y+2z=28\begin{aligned} & \Rightarrow 6 x+3 y+6 z-27+4 x-5 y-4 z-1=0 \\ & \Rightarrow 10 x-2 y+2 z-28=0 \\ & \Rightarrow 10 x-2 y+2 z=28\end{aligned}$$
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