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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2017 · Mathematics · Three Dimensional Geometry

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

The equation of the plane passing through the line of intersection of the planes x+y+z=1,2x+3y+4z=7x+y+z=1,2 x+3 y+4 z=7, and perpendicular to the plane x−5y+3z=5x-5 y+3 z=5 is given by

Options

  1. Ax+2y+3z−6=0x+2 y+3 z-6=0
  2. Bx+2y+3z+6=0x+2 y+3 z+6=0
  3. C3x+4y+5z−8=03 x+4 y+5 z-8=0
  4. D3x+4y+5z+8=03 x+4 y+5 z+8=0

Correct answer

A. x+2y+3z−6=0x+2 y+3 z-6=0

Step-by-step solution

Given planes, p1:x+y+z=1p2:2x+3y+4z=7p_1: x+y+z=1\mathrm{p}_2: 2 \mathrm{x}+3 \mathrm{y}+4 \mathrm{z}=7 So, equation of plane passing through intersection of planes p1andp2p_1 and p_2 is x+y+z−1+k(2x+3y+4z−7)=0⇒x+y+z−1+2kx+3ky+4kz−7k=0⇒x(1+2k)+y(1+3k)+z(1+4k)−1−7k=0\begin{aligned} & x+y+z-1+k(2 x+3 y+4 z-7)=0 \\ & \Rightarrow x+y+z-1+2 k x+3 k y+4 k z-7 k=0 \\ & \Rightarrow x(1+2 k)+y(1+3 k)+z(1+4 k)-1-7 k=0\end{aligned} This is perpendicular to x−5y+3z=5x-5 y+3 z=5. ⇒x−5y+3z−5=0⇒1(1+2k)−5(1+3k)+3(1+4k)=0⇒1+2k−5−15k+3+12k=0⇒−k−1=0⇒k=−1\begin{aligned} & \Rightarrow \mathrm{x}-5 \mathrm{y}+3 \mathrm{z}-5=0 \\ & \Rightarrow 1(1+2 \mathrm{k})-5(1+3 \mathrm{k})+3(1+4 \mathrm{k})=0 \\ & \Rightarrow 1+2 \mathrm{k}-5-15 \mathrm{k}+3+12 \mathrm{k}=0 \\ & \Rightarrow-\mathrm{k}-1=0 \Rightarrow \mathrm{k}=-1\end{aligned}\thereforeEquationofplaneisEquation of plane isx+y+z−1−1(2x+3y+4z−7)=0⇒x+y+z−1−2x−3y−4z+7=0⇒−x−2y−3z+6=0⇒x+2y+3z−6=0\begin{aligned} & x+y+z-1-1(2 x+3 y+4 z-7)=0 \\ & \Rightarrow x+y+z-1-2 x-3 y-4 z+7=0 \\ & \Rightarrow-x-2 y-3 z+6=0 \\ & \Rightarrow x+2 y+3 z-6=0\end{aligned}$
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