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AP EAMCET · Maths · Three Dimensional Geometry

The equation of the plane \(\pi\) through the line of intersection of the planes \(\pi_1 \equiv x+3 y-6=0\), and \(\pi_2 \equiv 3 x-y+4 z=0\) is \(\pi_1+\lambda \pi_2=0\). If the plane \(\pi\) is at unit distance from the origin, then an equation of the plane \(\pi\) is

  1. A \(2 x+y+2 z-3=0\)
  2. B \(2 x-y-2 z+3=0\)
  3. C \(2 x+y+2 z+3=0\)
  4. D \(x+2 y+2 z+3=0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2 x+y+2 z-3=0\)

Step-by-step Solution

Detailed explanation

Given equation of plane are \(\begin{aligned} & \pi_1=x+3 y-6=0 \\ & \pi_2=3 x-y+4 z=0 \end{aligned}\) Given, \(\pi_1+\lambda \pi_2=0\) \(\begin{aligned} (x+3 y-6)+\lambda(3 x-y+4 z) & =0 \\ (1+3 \lambda) x+(3-\lambda) y+4 \lambda z-6 & =0 \quad \ldots (i) \end{aligned}\)…